If you’re willing to wait, you don’t even need to invest any time or energy at all: you’ll eventually just hear who won by word of mouth. Pretty hard not to know that Obama won in 2008, if you’re walking around in the world of 2010 for any length of time. Probably the best way to follow politics is to wait ’til you’re on your death bed and then skim a history textbook that covers most your own lifetime. Only the important stuff will make it in, and if it’s not important, then you don’t really need to know about it.

So why do people spend so much time and energy following politics? To answer that question, we need only compare it with sports. Sports has the same game theory as politics: watching a game live is theoretically a waste of time, since you could just wait ’til it’s finished and look up the score. But of course, there’s more to sports than merely learning who wins. It’s a social ritual, an opportunity to appease those parts of our brain evolved in previous times, yearning to compete with the other tribes. It’s fun.

Politics is the same as sports. We root for our party the same way we root for our football team. It’s fun to watch the latest (mis)adventures of Sarah Palin. Without politics to poke fun at, some of the greatest comedians of the twentieth and twenty-first centuries would have had nothing.

From a game theory perspective, the most efficient way to follow politics is to not follow it all, and just glean the really important stuff from everyday conversations. But that completely ignores the whole fun of it. So, provided you’re really interested in politics, and you do it for fun, then how does the answer change? Again, we look to sports. With sports, the most efficient way to follow is to follow like a maniac. There’s no point being a “moderate” sports fan. Either you love the Yankees or you hope they’ll burn in Hell. If you were to take the middle ground, then that would be no fun and you might as well tune out of the game completely. So you wear the big rubber hat, you paint your chest, you buy banners and caps, you go all out. And you don’t bother with rationality: a Sox fan doesn’t argue rationally why his team is better; it just is, and anybody who disagrees is obviously lame!

So if politics is your thing, treat it like you’d treat a sports team. Drop the front of reasoning and facts, as though anything could realistically ever get you to switch parties. You quote Ayn Rand because it’s fun to piss everybody off, not because there’s any actual merit to her selfish screed. And more power to you– I approve, if we didn’t have libertarians to kick around, half the humor on the internet would vanish! When you protest the obviously Satanist president (I heard he has some things in common with Adolf Hitler), it’s a competition who can design the most over-the-top protest signs. It’s a festival! Never give quarter to the fans of the enemy party. A Yankees fan would never give quarter to a Bostonite. If your candidate is in favor of kitten sacrifices, then by god, kitten sacrifices are a fundamental and irreplaceable part of our sacred heritage, and don’t you forget it or let anybody else! If the enemy candidate loves apple pie, then by god, apple pie is the entire problem with this country, an abominable mockery of all that is good and true!

But at the end of the day, remember that everybody else is playing the same game. The Buckeye father and the Michigan son (may God have mercy on his soul) still love and respect each other outside the realm of sports. They know that when they taunt and mock each others’ teams, it’s all for fun. Same with politics. While you may be a member of the vile, lockstep, evil enemy hivemind, against which my noble and selfless party is waging a valiant underdog resistance, I still respect you as a person, and accept and love you exactly the way you are.

But your team still sux ballz.

FURTHER READING

How to Escape an Echo Chamber
Seven Signs we Live in a Dystopia
Why the U.S. Should Implement Basic Income
Worsen the World for a Billion Dollars?

Metaskills are abstract skills which have to do with other skills. For example, an autodidact is a person who has the metaskill of being able to self-teach him or herself new skills without an outside teacher. A teacher is a person who has the metatalent to teach skills to others; here I speak of someone who is a teacher of a wide variety of things, not necessarily a teacher who focuses on one single topic. Teaching one single topic, like calculus, is a skill, but the ability to learn an arbitrary skill and then teach it to others, that is a meta-skill. Generalization is a metaskill where you look at a wide variety of skills and figure out the common underlying patterns. Specialization is one where you can take a skill and focus it more precisely, to get a new skill which is a special case of the broader original skill.

Skills Training and Metaskills Training

Skills training involves performing a skill over and over. As you perform a skill, your subconscious mind constantly tries to figure out how it can help you. At first, it doesn’t know how to help you at all, and you have to consciously think about every littlest detail. In time, the subconscious takes over more and more of the workload, allowing you to perform the skill with less and less conscious attention. This is sometimes referred to as muscle memory; you can do the skill without even thinking about it. To get to this point, you have to perform the skill quite a bit. Each time, it comes a little easier. Once your subconscious has completely taken over the performance of the skill, it shifts toward finding ways to optimize and improve the skill, and that’s how you evolve from a mere expert into a master.

Training a metaskill is the same. Just because a skill is meta, doesn’t make it any different from any other skill. The difference is that we don’t usually consciously train our meta-skills because most people don’t even recognize them as skills. Besides that, training a meta-talent is more difficult than training a skill, because you can’t as easily fall into a pattern of repetition. Whereas you can do basketball training by throwing a basketball through a hoop a whole lot of times, you can’t, for example, teach yourself calculus a whole lot of times. In order to train the meta ability of being an autodidact, you must consciously seek out new things to teach yourself. If mastering chess requires playing ten thousand games, then mastering autodidacticism requires teaching yourself ten thousand different skills.

The benefit of mastering a skill is that you get to use that one skill. It makes a contribution to your overall intelligence by giving you that much more referential material from which to draw patterns and analogies. By learning Japanese, I’ve gained the ability to talk to Japanese people in their native tongue. The benefit of mastering a metaskill is that you can get new regular skills more easily, or make better use of the regular skills you already have. When you train a skill, you are making a long term investment; when you train a meta-skill, you are making a “long long term” investment. You’re making an investment into your ability to make or profit from other long term investments.

The Reflexive Nature of Metaskills

The great thing about a meta-skill is that it’s reflexive. It’s something you apply to skills; but it is a skill, therefore, you can apply it to itself. For example, consider a master teacher who can skilfully teach every skill she possesses: in particular, she can teach how to teach. A master autodidact can, in principle, teach himself any skill: in particular, he can teach himself any metaskill. (In a very real sense, master autodidacts are like gods. They can do basically anything. I consider myself something of an intermediate level autodidact.)

Here’s another example of a metaskill. Skills analysis is the ability to take any skills you know, and break them down, analyzing them and figuring out exactly how they work. For different skills, it requires a different mastery of skills analysis to break them down. For example, just about anyone can analyze the “skill” of flipping coins. But it would take a very good skills analyst to analyze the skill of playing the harp. Skills analysis is itself just another skill, so in theory, a good enough skills analyst could break it down and analyze it.

The novel “Cheaper By The Dozen” tells the tale of the family of Frank Gilbreth, a self-described “Efficiency Expert”. He devoted his life to finding ways to make various tasks more efficient. He even invented a general system of “therbligs”, small undecomposable units of work, for analyzing general tasks. In fact, he was pioneering the “time and motion study” metaskill, which takes skills and finds ways to make them more efficient. What if someone was so good at time and motion study that they could apply it to itself, and find ways to make time and motion study itself more efficient? Then they could apply it to itself more efficiently, and make it even more efficient, and so on. How efficient could it get?

The Continuum Between Skills and Metaskills

I’ve actually been speaking of just “skills” and “meta skills” to simplify the discussion. There’s actually an entire continuous spectrum between the two. Take computer programming, for example. Programming computer games in Java is a specific skill. Programming arbitrary java applets is a slightly more meta skill, which includes the ability to program games, if you’re so inclined. Being able to program websites in arbitrary languages, learning the languages as you need ‘em, is a more meta skill. Going even more meta, you have the skill of programming any program, not just websites, in any language, learning the languages as you go.

One way to train a metaskill is to figure out the spectrum below it, and start low on the spectrum and work your way up. For example, if you want to learn to be a master teacher, you might start by simply learning how to teach your favorite subject, say, singing. Once you’re good at teaching people how to sing, you might generalize it to teaching people performance art in general. And from there, it’s not as big a jump to teaching people any arbitrary skill that you yourself possess. The master of a meta-ability probably got that way by applying the technique “by accident”, without actually being consciously aware of what was going on.

FURTHER READING

How Children Understand Language
Autodidact: Be A Self-Teacher
Spaced Repetition Systems
How to Stand on the Shoulders of Giants

(Skills and Metaskills was originally published January 3, 2009)

I can see why people like to speculate this sort of thing: it makes a compelling narrative. It’s just not accurate. If somebody proves P=NP, then maybe cryptography will break and computation power will jump by a few millenia. But maybe not. First of all, a proof of P=NP does not have to be constructive (granted, it’d be a little surprising if it were non-constructive, but P=NP would already be surprising by itself). Maybe the proof would use contradiction, or even the axiom of choice. In that case, it would give no progress toward actually obtaining polynomial time algorithms for problems in NP.

Even if someone came up with a constructive proof of P=NP, the algorithms it produces are not, a priori, going to be any good in practice. An algorithm which runs in time complexity O(n^9999999999) isn’t all that more practical than one which runs in O(2^n). In theory, sure, but in practice, neither algorithm is any good.

Robertson’s graph minor theorem provides polynomial time algorithms for checking whether arbitrary graphs contain a given graph as a minor. But these algorithms are generally not useful in reality. For one thing, they require knowledge of a finite obstruction set, which the graph minor theorem guarantees exists, but doesn’t give any construction for. This obstruction set can be enormous anyway, so even if it were known, the polynomial time algorithms are only of academic interest.

“P=NP is a statement about real-life computers”

This misconception is related to the previous one. When excitedly talking about real-life implications of computer science, there’s a tendency to forget that real-world computers are NOT Turing machines. I admit this is pedantic, but every algorithm used in a real-world computer runs in time complexity O(1). This is because a real-world computer program can only possibly take finitely many inputs (an absolute upper bound would be 2^n where n is the number of bits in the computer’s total memory). Any problem with only finitely many inputs, has an O(1) solution.

Of course, this is silly and pedantic, and tells us nothing about the real performance of algorithms. But then, neither would P=NP, necessarily.

“If it turned out P were =NP, that would be the most amazing possible event in computer science”

No wonder complexity theorists say this– they’re understandably biased. Fact is, if we could figure out a way to solve the Halting Problem in real life, that would be much more interesting and would have much more profound implications.

Even if someone figured out that P=NP and figured out a constructive way to come up with very nice O(n) algorithms with small constants for problems in NP, that would still merely save time. Anything we could deduce with the wonder-algorithms, we could have eventually deduced with more naive algorithms, just it would’ve taken longer than the lifetime of the universe. But if a method were discovered for solving the Halting Problem, that would allow us to deduce things which we could never deduce otherwise, no matter how many universe-lifetimes we had to spend.

P=NP would (in the best case scenario) help us keep the trains running on time. A real-life Halting Problem oracle would let us see into the very mind of God.

Betting on a Proof of P=NP or P!=NP

A recent flawed attempt at proving P!=NP generated some discussion about the economics of betting on the validity of such a proof. Inevitably, people began asking, how would a perfectly rational actor bet on such a thing?

*Record Needle Scratch* Hold on a second. There’s one thing the rational-actor-economists forget when talking about betting on proofs. A “perfectly rational actor” could simply read the proof and assess its validity right away. Because, you know, he’s “perfectly rational”. In fairness, this is more of a fatal flaw in conventional economics than it is a misconception about P=NP.

(In a related matter, the economist’s hypothetical “perfectly rational actor” already knows whether P=NP or P!=NP, unless it turns out to be unprovable either way. Because, if either result has a proof, then the perfectly rational actor can systematically enumerate all possible proofs until he finds one or the other. Sure, such a strategy is absolutely impractical in the real world, but that shouldn’t matter to an economist’s rational actor. It never has before!)

Comparison to The Riemann Hypothesis

As an undergrad, I once got in a playful argument with a physics student about whose major was more useful. I pointed out that if mathematicians managed to prove the Generalized Riemann Hypothesis, it would have profound applications to cryptography since it could help factor primes faster (e.g. by a result of H.K. Lenstra). I’ve since realized how dumb a statement that was. Any algorithm for factoring, whose proof relies on some deep conjecture like RH, can be run whether or not we have a proof of that conjecture. Whether the algorithm works or not doesn’t hinge on us knowing that the conjecture holds. In the unlikely event the algorithm fails, then we’ve disproved the conjecture, which is arguably a better outcome than whatever we were attempting to compute in the first place.

Similarly, any application which follows from P!=NP, does not rely on us having a proof of P!=NP. If computer scientists figure out a way to colonize the sun, but it relies on P!=NP, then we can go ahead and start colonizing.

FURTHER READING

The Halting Problem
Goodstein Sequences
Busy Beaver Numbers
Unlimited Register Machines

And here’s what makes this unique to children, what sets it aside from, say, an adult learning a 2nd language: I never *knew* that I didn’t know what was going on. If you’d asked me then, I’d've told you I was a Star Wars expert (this was before the internet caught on, so being a Star Wars expert wasn’t stigmatized yet). I never *knew* that Casa Blanca had the potential to be more than a two-hour snorefest. One time when my mother read a novel to us kids, I mistakenly had the impression the main character was a dog (he was actually a human), and didn’t revise that idea until his own pet dog died; nevertheless, I still enjoyed the story perfectly well!

As adults, we always seek to analyze and understand. Even when we’re playing a game or watching a movie strictly for fun, we have to understand and analyze. This explains why grown-ups write Wikipedia articles about “The Lord of the Rings” as if they’re actual historical events of great real world importance! The other night when my girlfriend and I were watching Iron Man, she paused it to ask me what the mini Arc Reactor was; a kid wouldn’t care, all that matters is it’s glowing and cool and wedged in Nick Stark’s freakin’ chest.

It is this ability to suspend analysis which allows children to learn languages so effortlessly. Take a child with a five year old’s grasp of English and make him watch Star Wars, and it’s fun. Do the same to an adult with the same language skills, and suddenly it’s an arduous chore. Is it any wonder adults suck so hard at picking up new languages?? When I was playing through Final Fantasy VI in Japanese, I was constantly pausing the emulator to look up difficult kanji and words. A nine year old Japanese kid playing the same ROM would have just skipped right past them without a care in the world. Question is, who’s the real winner? Maybe I managed to analyze more characters and create more SRS cards, but the kid had a lot more fun blasting Kefka with Ultima. A Buddhist zen-master might say the kid is playing the game with presence.

If any kids are reading this article right now, you’ve probably skimmed through most of it ‘cuz you were bored, so here’s the Kid Summary: Adults suck, kids rock, Kefka, Ultima, light sabers, kanji, Iron Man, Zelda, and the Arc Reactor. Now, back to the adults: if you aren’t already familiar with 100% of the movies and games I’ve referenced here, then reading this article is technically a language-learning experience. By reading it, you’ve exposed yourself to certain new words like “Ultima” or “Arc Reactor” and thus made yourself a tiny bit better at English. Here’s the crowning moment of irony: you could’ve gotten that same exposure by skimming the article and just reading the Kid Summary.

FURTHER READING

Ten Reasons why English is a Hard Language
Three Ways to Be More Present
How the Mind Learns
Right Action

How does one swim for fitness? It’s dead easy. Put on a swimsuit. Jump in the pool. Swim back and forth for N laps. That’s it! I suppose form still matters, but not nearly as much as with weight-training. Basically, as long as you don’t drown, your body is going to speed-evolve like a primitive monkey who just discovered the black monolith from 2001: Space Odyssey.

I’ve been applying the principle of Progressive Training in the pool, and it’s working really well. I started with 20 laps (which is less impressive than it sounds– we’re just talking a dinky apartment pool here, maybe 30 feet across at the most). That was a few weeks ago. Now I’m up to 60 (!). That’s a 3x increase, and it’s never been particularly difficult. I guess I am blessed with an above-average pain tolerance, and that’s been very helpful with other physical training I’ve done, but it just hasn’t been an issue with fitness swimming. The familiar “good soreness” is there, but somehow the way it’s balanced across the whole body seems to dull it significantly. It’s like I took the old saying “no pain no gain” and stole all its lunch money then kicked it down into the dirt.

I’ve never been the jock type, and have always gotten stuck at a plateau when trying to train for appearance. I’m naturally the slender body type and have never managed to pull off the Schwarzenegger look, even when I was working out every day for half a year. I guess weight training doesn’t play to my strengths; swimming does. Lately I’ve found myself literally standing in front of the mirror, shirt off, admiring myself O_O Needless to say, my girlfriend is thrilled

Another benefit is I can eat whatever I want, with no guilt. Do you know about the Michael Phelps diet? The guy who won eight gold medals for swimming in the 2008 Olympics? Yeah… he consumes 12,000 calories a day. And we’re not talking salads here. The guy’s a pig, and based on his diet alone, you’d expect he’d weigh a million pounds. But hitting the water has some kind of magical effect on the body, shaving everything off.

Once the quarter starts back up, I’ll resume regular weight training, and see how that gels with swimming for fitness. Maybe all the water training will enable me to push past the massive plateau effect I’ve been stuck on on the bench press for the last eternity. My goal is still to press 200lbs…

One thing’s for sure, you aren’t gonna find me running on the treadmills very often from now on O_O It would be like an air force switching from Stealth Fighters back to Sopwith Camels. I wish someone had told me about this when I was, like, 16 Better late than never, I guess… in fact, that’s just another bonus the pool holds over the gym. You could be an 80 year old woman and still swim for fitness and get tons of benefits. The pool does not discriminate…

If you’ve got a pool at your apartment or at your university gym or somewhere, don’t let it just sit there! That water is begging you to jump in! I heard that pool bad-mouthing your mom, you’d better get in there and kick the crap out of it! Seriously, though… I cannot praise the pool too highly.

FURTHER READING

The Plateau Effect
Progressive Training
30-Day Workout-A-Day Challenge Completed!

Zorn’s Lemma says: Given a nonempty partial order, if it has the property that every totally ordered subset is bounded, then there is a maximal element.

When I first started grad school in math, I was a little skeptical that such a strange axiom can have such far-ranging and amazing power. A quarter or two of advanced algebra and analysis very quickly changed my mind! Here are some examples to change everyone else’s mind as well:

Every Vector Space Has A Basis

This is probably the prototypical application of Zorn’s Lemma. Anyway it’s the first one I ever encountered.

• Theorem: Every vector space has a basis (except the trivial vector space with only one element).
• Proof:
  • Let V be a vector space with at least one nonzero element. Consider the set X of linearly independent subsets of V. This is nonempty since we’re assuming there’s at least one nonzero element.
  • If S and T are two elements of X (i.e., two linearly independent subsets of V), declare S≤T if S is a subset of T. Note that this defines a partial order on X.
  • I claim that if C is any totally ordered subset of X, then UC, the union of sets in C, is an upper bound on everything in C. Certainly, everything in C is a subset of this union– but we need to know that UC is actually an element of X (i.e., a linearly independent subset of V).
  • Suppose not, so UC is linearly dependent. Then there are elements c₁,…,c_k in UC and scalars a₁,…,a_k such that a₁c₁+…+a_kc_k=0. Each c_i is an element of an element C_i of C. Since C is totally ordered under ≤, and there are only finitely many of the C₁,…,C_k, we can find some j such that every C_i≤C_j. So C₁, …, C_k are all subsets of C_j. Thus, C_j contains all the bad elements c₁,…,c_k. This is a contradiction: C_j is supposed to be linearly independent!
  • So by contradiction, UC is indeed an element of X, and so it is an upper bound of C.
  • This shows that every totally ordered subset of X, has an upper bound. We are all set up to invoke Zorn.
  • By Zorn’s Lemma, there is a maximal element B of X. I claim that B spans all of V, and is thus a basis.
  • If B does not span all of V, then there is some vector v which is not spanned by B. Then BU{v}, the result of adding v to B, must be linearly independent: if there were a relationship of the form a₁b₁+…+a_kb_k+a_k+1v, where a₁,…,a_k were scalars and b₁,…,b_k were elements of B, then– since the scalars are a field– we could divide by a_k+1 and solve the equation for v, contradicting that v was not in the span of B.
  • But wait! The whole point is that B is maximal! Assuming B doesn’t span all of V, I’ve gone and constructed an even bigger linearly independent set, BU{v}. This contradicts the maximality of B. So indeed, B is the basis we needed to find.

Here’s the intuition in the above proof. To get a basis, first pick a nonzero vector. If it spans the whole space, we are done. If not, then take a vector which isn’t spanned, and add that to our potential basis. If it spans the space now, we are done. If not, add a vector which isn’t spanned yet… continue this process as long as needed. Unfortunately, with infinite-dimensional vector spaces, our job will take infinitely many steps to complete, and so the construction never terminates. Zorn’s Lemma is used to short-circuit the infinite process.

The Axiom of Choice

Here’s how to prove the Axiom of Choice from Zorn’s Lemma.

• Theorem (The Axiom of Choice): For any set of nonempty sets, there exists a choice function.
• Proof:
  • Let S be any set of nonempty sets. We’ll construct a choice function on S. If S is empty, the empty function is its choice function; assume S is nonempty.
  • Let X be the set of pairs (T,f) where T is a subset of S and f is a choice function on T. X is not empty: since S is nonempty, S has some element s, and {s} is a subset of S. {s} clearly has a choice function, namely, the function f whose domain is {{s}}, defined by f({s})=s. Thus ({s},f) is an element of X, showing X is non-empty, which means we at least have some prayer we can apply Zorn’s Lemma eventually…
  • Define (T,f)≤(R,g) if T is a subset of R and f is a restriction of g. Note that this is a partial order on X.
  • Suppose C is any totally ordered subset of X. I’ll construct a bound, in X, for all the things in C. Let R={s:sεT for some (T,f)εC}. (In words: R is the set of sets which are elements of the left component of anything in C) I’ll define a choice function f on R. Given an element r of R, by definition of R, that means r is in T for some pair (T,g) in C. So I’ll define f(r)=g(r).
  • But hold on a second! Is f well-defined? Given an r in R, there might be multiple pairs, say (T,g), (Q,h) in C, with r in T and r in Q. In that case, isn’t my definition ambiguous: should f(r) be g(r) or should it be h(r)? Fortunately, it turns out it doesn’t matter. Since C is totally ordered, we have (T,g)≤(Q,h) or (Q,h)≤(T,g). So g is a restriction of h, or h is a restriction of g. Either way, h(r)=g(r). So this f I’ve defined is well-defined (whew!) It inherits the property of being a choice function from the choice functions it’s built up from.
  • The (R,f) I just built is an upper bound, in X, of all the things in C.
  • I’ve shown every totally ordered subset of X has an upper bound in X. And X is nonempty. By Zorn’s Lemma, X has a maximal element, call it (P,g), where P is a subset of S and g is a choice function on P. I claim that, in fact, P=S, so that g is the choice function I needed to construct, proving the Axiom of Choice.
  • Suppose not: P≠S. Then we can find a set s which is in S but not P. Since S is a set of nonempty sets, s is nonempty, so it contains an element σ. Extend g to a new function g’ by defining g’(s)=σ. Then g’ is a choice function on PU{s}. Thus, (PU{s},g’) is bigger than (P,g), but this contradicts the maximality of (P,g). The theorem is proved.

The intuition here is similar to the vector space application. To get a choice function on S, start with a choice function on a one-element subset of S. If that’s all of S, we’re done. If not, extend to a choice function on a two-element subset of S. If that’s all of S, we’re done. Continue this process as long as needed. Unfortunately, S may have infinitely many sets in it, so Zorn’s Lemma is needed to short-circuit the process.

The Well-Ordering Theorem

Here’s how to prove the Well-Ordering Principle (every set can be well-ordered) from Zorn’s Lemma.

• Theorem: Every set can be well-ordered.
• Proof:
  • Let S be a set. I’ll construct a well-order on S.
  • Let X be the set of pairs (T,<) where T is a subset of S and < is a well order of T. X is nonempty because, if nothing else, it contains (emptyset,emptyorder). Partially order X by saying (R,<₁)≤(T,<₂) if R is a subset of T and <₁ is a restriction of <₂. (Does this sound familiar at all?) Since X is nonempty, there is some hope we can apply Zorn’s Lemma…
  • Let C be any totally ordered subset of X. I claim X contains something which bounds everything above in C. Let R={r:rεT for some (T,<)εC}, or in words, R is the set of elements of left components of things in C. I’ll define a well-order <_R of R. Given some elements x,y in R, their presence in R means there are (T,<_T) and (Q,<_Q) in C such that xεT and yεQ. Since C is totally ordered, either (T,<_T)≤(Q,<_Q) or (Q,<_Q)≤(T,<_T). Assume the former; the latter case is similar. So T is a subset of Q, and thus xεQ. Finally, define x<_Ry iff x<_Qy.
  • But wait! This definition of <_R seems ambiguous. What if there are (Q,<_Q) and (P,<_P) in C, both containing x and y? Then is x<_Ry iff x<_Qy, or is x<_Ry iff x<_Py? Turns out that it doesn’t matter. Since C is totally ordered, one of <_R or <_P is a restriction of the other. Since they’re well-orders, this implies x<_Ry iff x<_Py.
  • Evidently the (R,<_R) I’ve constructed is a bound, in X, of everything in C. By the arbitrariness of C, we can invoke Zorn’s Lemma. Zorn says X has a maximal element, call it (T,<). As you probably can guess, I claim T=S, so that < is the well-order of S which we seek.
  • If not, then pick some element s in S but not in T. Extend < to an order <’ on TU{s} by defining t<’s for every t in T. (In other words, put s “after” everything in T) This is a well-order. Thus X contains (TU{s},<’), which violates the maximality of (T,<). The theorem is proved.

The intuition here is: pick an element and call it smallest. If that’s everything, we’re done. Otherwise, pick another element and call it 2nd smallest. If that’s everything, we’re done. Otherwise… continue the process as long as necessary. Of course, if the original set if infinite, this process will never end. That’s why we use Zorn’s Lemma to short circuit it.

FURTHER READING

Applications to Conway’s Game of Life
Konig’s Lemma
Three Applications of Higher Math
Uncountably Many Bloodlines
Nonstandard Worlds

Recognize the Echo Chamber for what it is

The first step is to break free of denial, and acknowledge that the community in question is one. Here are the telltale signs of a community which is stubbornly fixed in its assumptions:

Every Development Supports Us

Consider the political parties in the U.S., two prime examples of echo chambers. If Republicans control the legislature, and ram through some horrid law banning puppies, then the Democrats cry: “See how horrible our opponents are! This proves we’re right!” But if the Repubs do the exact opposite, if they let the old puppy ban expire, then the cry on the left is: “See, even the Republicans are gradually coming around to agreeing with us!” Either way, the development is twisted into supporting Dems. Similarly if the roles are reversed: if Dems outlaw kittens, then such a vile act proves how right Repubs were all along; if Dems let the kitten ban expire, it shows how even they are coming around and seeing the light the Pubs have been preaching.

This phenomenon is common in advocacy groups of all stripes. Take feminists for example. If a new study comes out showing that men make more money than women, feminists will say: “See, this proves everything we’ve been saying!” and pop open the champagne. But if that same study shows the opposite, that women make more money than men, then they say: “See, our efforts are starting to pay off, this is a move in the right direction!” and pop open the champagne. Either way, the news is twisted in their favor, and champagne is (figuratively) opened. It doesn’t even matter if two studies come out on the same day, with contradictory findings: each will be twisted into victory.

All of this is, of course, good news for John McCain.

Under what circumstances would the group change its mind?

As a mental exercise, once you know the group and its various members a little bit, ask yourself what it would take to make them question their assumptions. If an academic paper were published which called the assumptions into question, would that be enough? If so, then this might not be an echo chamber after all. But in many cases, the paper would be ignored, or its authors maligned. Then you’ve got an echo chamber.

How extreme a situation must you imagine, before the community would question its assumptions? Would it require a massive public scandal in which the leader was revealed to be a fraud? A mathematical proof that the assumptions were contradictory? A message directly from God? The more extreme refutations you imagine the group weathering, the more likely it’s a ‘chamber. Readers can draw comparisons with the notion of falsifiability in science. What hypothetical situation would it take in order for astrologers to question astrology? What hypothetical situation would it take in order for capitalists to question capitalism?

Some groups go so far as to explicitly forbid the questioning of the assumptions. Certain religions come to mind.

Observe with an Objective Mind

Having realized a hivemind you’re in is an echo chamber, the next step is to turn your critical thinking skills back on. When a relevant news story hits the New York Times, and the group starts discussing it, inevitably finding it supports pre-existing beliefs, try to evaluate it as a neutral judge. The ability to twist the article and read what the group wants out of it, is actually kind of useful, since it lets you analyze the article from that perspective, you just don’t want to make that the only perspective you can analyze it from.

Predict Them

After you’ve been observing with an objective mind for awhile, start trying to make predictions. When your RSS feed shows the group is discussing the new puppy-banning editorial in the New York Times, try to predict what the group is going to say about it before opening the discussion. If it’s an echo chamber, and you’ve been stuck there for awhile, you’ll find it’s shockingly easy to get very accurate predictions about what’ll be said. (HINT: it probably won’t involve questioning the assumptions behind the group!)

Once you can accurately predict what “Citizens for Our Way Of Thinking” is going to say in varying scenarios, then congratulations, you’ve escaped them: you can look down on them from a higher plane, chuckling as you predict their every comment. You’ll no longer have any need to read what they say, since you can predict it all in your head faster than your eyes could take in the words anyway. Now you are ready to move on to other communities, ones you cannot yet predict, and learn the new knowledge they have to offer.

If you still sympathize with the echo chamber for legitimate reasons, then by all means, continue to contribute. You can even use your new-found wisdom to try and nudge the group toward greater self-awareness. Just don’t be surprised, if you ever try to play the devil’s advocate, the echo chamber won’t appreciate a veteran trying to undermine established conventions.

FURTHER READING

Smarter than Teacher
Very Serious People
The English Double Negative
Seven Cheesy Real-Life Cliches

His “Baire space” and “Baire Category Theorem” help us understand topological spaces whose complements are very bare.

George Cantor

Taught us how to count infinite sets.

Carl Friedrich Gauss

Made a pretty good guess at how many primes there are below x. Also, some stuff about statistics.

Kurt Gödel

Played God by applying mathematics to deep philosophical questions.

Oliver Heaviside

His “Heaviside function” has a graph which is very heavy on one side.

Robert Hooke

Told us how to calculate the force between two objects hooked together by a spring.

Stephen Kleene

His “Kleene normal form theorem” provides a very clean way to represent mathematical computer programs.

Pierre Pigeonhole

His “Pigeonhole Principle” tells us about what happens if we put pigeons inside holes. (1)

Jules Poincaré

One of the founding fathers of topology, he certainly took good care of his points.

John Poynting

The fact that they named the “Poynting vector” after this guy seems like they were deliberately trying to make a pun out of it.

Alan Turing

Invented Turing Machines, which compute recursive functions by turning spools of tape.

(1) No. Sorry, but there is no mathematician named Pierre Pigeonhole. Unfortunately, I made that one up.

FURTHER READING

How to Stand on the Shoulders of Giants
The Hope Function

Example 1: The Real Numbers

Every real number has a size, namely, its absolute value. Now, it’s true that the set of real numbers contains arbitrarily large numbers. Given any possible finite size, there is a real number whose size is bigger. The set of reals does not, however, have any infinite element. There is no individual number whose size is infinity.

Example 2: The Real Intervals

A real interval is a set X of real numbers with “no gaps”, that is, it must have the property that if it contains two points x and y, then it also contains everything between x and y. The set of all real intervals does contain something with infinite size: the whole real line itself, for example, is a real interval (it has no gaps), so is an element of the set of all real intervals, and it certainly has infinite size.

These two examples are kind of silly and trivial, but the next one is more subtle and seems to cause a lot of confusion.

Example 3: Branches in a Tree

Given a mathematical tree, a branch is a sequence of nodes, starting with the root, such that each node in the sequence (except for the first one) is a child of the previous node. A priori, branches are allowed to be infinitely long, but for a specific tree, there may or may not exist an infinite branch.

If the tree itself is finite, then immediately it cannot have an infinite branch, nor can it have arbitrarily long branches. No branch can contain more nodes than the tree itself contains.

People often fall prey to the trap of thinking that if a particular tree does have arbitrarily long branches, then voila, it must have an infinite branch. This is not true. The following tree is a counterexample:

The root in this tree has infinitely many children. The root’s first child has no children. The root’s 2nd child has a child and no further descendants. The root’s 3rd child has a child and a grandchild and no deeper descendants. And so on. If you cut the root, the graph breaks into infinitely many finite graphs. The tree has no infinite branch. Such a branch would have to begin with the root, and then go to some child of the root, and then pass through infinitely many descendants of that child. But each child of the root has only finitely many descendants, so no can do. On the other hand, the tree does have arbitrarily long branches.

Konig’s Lemma provides a sufficient condition for a tree to have an infinite branch. It says that if a tree is infinite, and every node has only finitely many children, then there’s an infinite branch. The theorem confuses some people, who assume it’s absolutely trivial because they equate “tree has arbitrarily long branches” with “tree has an infinite branch”. BTW, the tree above fails the hypotheses of Konig’s Lemma because the root has infinitely many children.

The Topping Game

Here’s a way to intuitively test whether a set has an infinite element. Fix a set S of things which have sizes ranging among the nonnegative integers and maybe “infinity”. Let’s play a little game. In this game, you pick an element from S, and then I try to pick a bigger element. If I manage to pick a bigger element, I win. If not, you win.

If S contains something with infinite size, then you have a winning strategy. Namely, pick something with infinite size. I can’t pick something bigger, because the biggest thing I can possibly pick would be another size “infinity” thing. I’m beaten.

If S contains arbitrarily large things, but nothing with size infinity, then I have a winning strategy. Namely, watch what you pick, and then pick something bigger. I can do this because there are arbitrarily big things, and the thing you picked was finite size, hence cannot be a bound on the sizes of everything else, or else they wouldn’t be arbitrarily big.

If S contains neither arbitrarily large things nor infinite things, then you have a trivial winning strategy: pick something with maximum size.

Advanced Example: Sets Within Sets

One of the axioms of ZFC, the Axiom of Foundation, implies that you can’t find an infinite sequence s₁, s₂, …, such that each s_i+1 is an element of s_i. In other words, you can’t find an infinite chain of sets-within-sets. So a class of sets (satisfying ZFC) does not have such infinite chains. But any such class does have arbitrarily long chains of sets-within-sets. For example: 0 is inside {0} is inside {{0}} is inside {{{0}}} is inside… (repeat the process any finite number of times).

What happens if we try to repeat this process infinitely often? Start with 0, then go to {0}, and keep going, adding on infinitely many new braces on each side. If this resulted in a set, say X={{{…{0}…}}}, where the … indicate infinitely many omitted brackets, then X would have the property that X={X}, and hence, X would be an element of itself, which violates the axiom of foundation.

In this example, things get really bizarre when we start playing with the Compactness Theorem.

Take the language L of ZFC and add a new constant symbol, s. Now take the ZFC axioms and add a bunch of new axioms in the extended language: “s is an infinite sequence”, “s(0) contains s(1)”, “s(1) contains s(2)”, … Call this extended axiom system ZFC+. The Compactness Theorem says that if a set of sentences has the property that every finite subset of those sentences is satisfied by a model, then the entire set of sentences has a model. If we take a finite subset of the axioms in ZFC+, then since it’s finite, it only contains finitely many of the “s(i) contains s(i+1)” axioms. Hence, it’s satisfiable. By the compactness theorem, ZFC+ itself is satisfiable and has a model.

This means there is a model of ZFC– satisfying the Axiom of Foundation and all– containing an infinite sequence s such that s(0) contains s(1) contains s(2) contains … forever. Isn’t this a paradox?

The thing is, we can see that this perverse model of ZFC contains an infinite chain of sets-within-sets, but the model itself cannot see it. We specified that s(i) contains s(i+1) for i=0,1,2,…, that is, for i ranging over the standard natural numbers. The strange model yielded by Compactness Theorem must contain nonstandard natural numbers– or else the sequence s would break the Axiom of Foundation. This model contains natural numbers which are bigger than all the “standard” natural numbers, but in the model’s eyes, these nonstandard naturals are still finite. (In order for the model to see that one of the nonstandard naturals is infinite, the model would have to contain a function which injects the model’s copy of the natural numbers into that number, and this is impossible)

This provides an example where, externally and metamathematically, we can see an infinite chain within a model, but if we work within that model itself, then we only see arbitrarily long chains and no infinite one.

FURTHER READING

Konig’s Lemma
Nonstandard Worlds
The Higher Infinite
Goodstein Sequences

The Game of Life

The game is played on an infinite grid. At any point in time, there are “live” cells and “dead” cells. The player’s only input is the choice of initial live cells. Once the initial live cells have been chosen, the game plays out with no additional input needed. The cells “evolve” from generation to generation by the following rules:

1. If a live cell has <2 live neighbors, then this cell will become dead in the next generation.
2. If a live cell has >3 live neighbors, then this cell will become dead in the next generation.
3. If a live cell has exactly 2 or 3 live neighbors, it will live on to the next generation.
4. If a dead cell has exactly 3 live neighbors, it will come to life in the next generation.

If you’re not familiar with the game, then read more at the Wikipedia article.

Worlds

A world is a directed graph, where an edge pointing from node A to node B is thought of as indicating that B is a child of A. Much intuitive vocabulary follows (“parent”, “ancestor”, “descendant”, and so on) with the obvious definitions. There are four axioms which may or may not hold for any given world:

• Genesis: There is a finite set of nodes, called “xatriarchs”, such that no xatriarch has a parent, and such that any ascending chain of parents eventually reaches a xatriarch.
• N-Biology: Every non-xatriarch node has at least N parents, and no node is its own descendant.
• Feebleness: No node has infinitely many children.
• Humanism: There are infinitely many nodes.

If a world satisfies the above four axioms, I’ll call it an (N-parent) axiom-conforming world. The requirement of at least instead of exactly N parents is a departure from earlier articles, required here to smooth the application to Conway. The proof of Uncountably Many Bloodlines requires some minor alteration to work with the modified definition (the “subworlds” defined there must be taken to respect “number of parents” to make the Zorn’s Lemma argument go through).

The World Corresponding to a Game of Life

Given an initial configuration for the Game of Life, there is a corresponding world, which is 3-Parent axiom-conforming if the initial configuration is finite and doesn’t die out. I’ll describe a world W corresponding to the given initial configuration.

A person in W will be a pair (c,g) where c is (the location of) a cell on the Life board, and g is a generation. To be specific, W will be the set of all (c,g) pairs such that cell c is live in generation g. It remains to define the parent-child relations. If (c,g) is a node in W (and g>1), that means cell c is alive in generation g. The way Life is defined, this implies that there are at least 3 (and at most 4) nodes of the form (d,g-1) in W where d=c or d is a neighbor of c. These three or four nodes are declared to be the parents of (c,g).

• Lemma: If the given initial configuration has finitely many live cells, and the cells in the resulting Game do not all die out, then the world W is 3-Parent axiom-conforming.
• Proof:
  • Genesis: The xatriarchs of W are the nodes (c,1) where c is live initially. These are finitely many by assumption. The way parents are defined, if (d,h) is a parent of (c,g), then h=g-1. This implies any ascending chain of parents eventually reaches a xatriarch.
  • 3-Biology: By construction, every non-xatriarch has at least 3 parents. No xatriarch can be its own descendant since children are always a later generation than their parents.
  • Feebleness: If (e,h) is the parent of (c,g), then by the way W is constructed, c=e or c neighbors e. So a parent cannot have more than nine children.
  • Humanism: If W had only finitely many nodes, that would force the Game of Life to die out, which we’re assuming is not the case with the given initial conditions.

Application 1: Uncountably Many Lifelines

A bloodline is an infinite sequence, starting with a xatriarch, such that each person in the sequence is a parent of the next. In this article, I showed that any (N-parent where N>1) axiom-conforming world has uncountably many bloodlines.

If an initial Life configuration has finitely many live cells, and the resulting species doesn’t ever die out completely, then by the above lemma, the corresponding world W is 3-Parent axiom-conforming. Thus it has uncountably many bloodlines.

• Definition: Given an initial Life configuration, a lifeline is an infinite sequence S of cells such that for every n>1:
  1. The nth cell in S is live in generation n, and
  2. The nth cell in S is equal to, or is a neighbor of, the (n+1)th cell in S.
• Theorem: Given an initial Life configuration with finitely many cells live initially, either there are uncountably many lifelines, or all the cells eventually die.
• Proof: A bloodline in W corresponds to a lifeline in an obvious way.

Application 2: Two Forbidden Directions

A lifeline (defined above) can be thought of as a “walk” through the Game of Life, from live cell to live cell, where each step is a step into the next generation. There are nine directions a step can take: north, south, east, west, northeast, northwest, southeast, southwest, or stand still. E.g., if you’re on a cell in generation g and you want to go north, the north cell must be alive in generation g+1. Here’s an awesome fact: Suppose the initial configuration is finite and the game doesn’t die out; suppose any two of the nine directions are forbidden; then it’s possible to find an infinite lifeline which never takes either of the two forbidden directions. For example it’s possible to find an infinite lifeline which never goes northeast nor northwest.

To prove this fact, we use a modified version of the Royal Male Konig’s Lemma from here. Here’s the result we’ll use, modified for Conway, the proof is similar to the proof of the original version:

• Theorem (Royal Male Konig’s Lemma, modified): Suppose W is an N-parent axiom-conforming world, N>1. To every non-xatriarch node p in W, assign a nonempty set of fathers from among the parents of p. There is an infinite bloodline, starting with a xatriarch, such that each person in the bloodline is a father of the next.

Since the theorem doesn’t specify how the fathers are chosen, that gives us great liberty of choice, and we can prove the claimed result:

• Theorem: Consider a Game of Life species with a finite initial configuration and which never dies out. Suppose two directions are forbidden. There exists an infinite lifeline which never goes in either of those two directions.
• Proof:
  • Let x and y be the forbidden directions (without loss of generality, assume they’re distinct). Let x’ be the “opposite” direction of x (e.g., if x is north, then x’ is south, if x is “stand still” then so is x’), and y’ the opposite direction of y. Let W be the world corresponding to this game of life species.
  • Given a non-xatriarch node (c,g) in W, I’ll define the fathers of (c,g). They shall be those parents (d,g-1) such that d is not the cell in the x’ direction or y’ direction from c.
  • [For example, if x=north and y=east, then the fathers of (c,g) are those parents (d,g-1) where d is neither south nor west of c. If x=north and y=stand still, then the fathers of (c,g) are those parents (d,g-1) where d is neither south of c nor equal to c.]
  • Since every non-xatriarch has at least three parents, each corresponding to a distinct cell, by the pigeonhole principle every non-xatriarch has at least one father, so Male Konig’s applies. There is an infinite bloodline starting with a xatriarch, such that each node is a father of the next.
  • This bloodline corresponds to an infinite lifeline in the Game of Life. The lifeline never goes in direction x or y: suppose it went, say, in direction x at, say, generation n. Then the new cell, in generation n+1, is an x-step from the parent cell. Thus the parent cell is an x’-step from the new cell. This violates the choice of fathers.

Example: The Glider

The following animation shows a “glider” species which “glides” toward the southeast. Since it generally moves in a southeast direction, it’s surprising there’s a lifeline which never goes south nor southeast. One such lifeline is indicated in blue. (There’s also, e.g., a lifeline which never goes southeast nor east, and a lifeline which never goes south nor east, and so on.)

(Thanks to Wikipedia for the original uncolored animation, and to Mike for helping with editing)

This glider moves one unit southeast per four generations, for an aggregate speed of 1/4 cells/generation. One might ask, is it possible to find a faster glider, which still moves southeast (and leaves no trail)? By the two-forbidden-directions result, we cannot find a glider which moves southeast (or any other diagonal direction) faster than 1/3 cells/generation. This is because such a glider will always have a lifeline which never goes S or SE. In order for the lifeline to move a unit southeast, requires at least 3 steps: SW, E, E, for example. The lifeline can only tend southeast at 1/3 cells/generation at most, so if a pattern moves southeast (without leaving detritus behind it), the pattern itself has 1/3 cells/generation as an upper speed limit. Similarly, 1/2 cells/generation is a speed limit for such patterns moving in a cardinal direction.

FURTHER READING

Nonstandard Worlds
Periodic Konig’s Lemma
Uncountably Many Bloodlines
Mathematical Royalty
Unlimited Register Machines