PHYS771 Lecture 2: Sets

Scott Aaronson

Thursday's class started out with a brief presentation by Rahul Jain about atomist ideas in Jainism (circa 500BC). It seems the Jain (the ancient ones, not Rahul) were barking up more or less the same tree as Democritus, but their ideas (like many of the pre-Socratics') were mixed with generous helpings of mysticism.

I mentioned another example of East-West convergence: apparently, several ancient cultures independently came up with the same proof that A = π r2. It's obvious that the area of a circle should go like the radius2; the question is why the constant of proportionality (π) should be the same one that relates circumference to diameter. Proof by pizza: Cut a circle of radius r into thin pizza slices, and then "Sicilianize" (i.e. stack the slices into a rectangle of height r and length π r). QED.

One thing I forgot to share with you on Tuesday was a quote from Democritus:

"I would rather discover a single cause than become king of the Persians."

Something to keep in mind when you consider those job offers from Microsoft or Google...

Today we're gonna talk about sets. What will these sets contain? Other sets! Like a bunch of cardboard boxes that you open only to find more cardboard boxes, and so on all the way down.

You might ask, "how is this relevant to a class on quantum computing?" I can give three answers:

When I gave that puzzle on Tuesday (which by the way, we're going to "answer" today), some of you asked what "countable" means. OK, dude. Math is the foundation of all human thought, and set theory -- countable, uncountable, etc. -- that's the foundation of math. So even if this class was about Sanskrit literature, it should still probably start with set theory. I have a hidden agenda: I'm told we have some physicists here, and I intend to browbeat you into thinking like mathematicians. I mean, what you do in the lab is your own business, but now you're in theorem country. There actually is a tenuous connection between quantum computing and set theory, which I'll touch on in the next lecture. To give a sneak preview, the connection is that quantum mechanics applied to finite-dimensional systems (like qubits) seems like an interesting "intermediate" case between a continuous and a discrete theory. That is, it involves quantities (the amplitudes) that vary continuously, but that are not directly observable. In this way, it seems to avoid the "paradoxes" associated with the continuum in a way that other continuous physical theories do not. But what are those paradoxes? Well, welcome to my haunted house horror tour of the continuous and the transfinite...

So let's start with the empty set and see how far we get.

THE EMPTY SET.

Any questions so far?

Actually, before we talk about sets, we need a language for talking about sets. The language that Frege, Russell, and others developed is called first-order logic. It includes Boolean connectives (and, or, not), the equals sign, parentheses, variables, predicates, quantifiers ("there exists" and "for all") -- and that's about it. So for example, here are the Peano axioms for the nonnegative integers (where S(x) is the successor function, intuitively S(x)=x+1, and I'm assuming functions have already been defined):

Zero Exists: There exists a z such that for all x, S(x) is not equal to z.

There exists a z such that for all x, S(x) is not equal to z. Every Integer Has At Most One Predecessor: If S(x)=S(y) then x=y.

The nonnegative integers themselves are called a model for the axioms (though interestingly, they're not the only model).

Writing down these axioms seems like pointless hairsplitting -- and indeed, as someone pointed out in class, there's an obvious chicken-and-egg problem. How can we state axioms that will "put the integers on a more secure foundation," when the very symbols and so on that we're using to write down the axioms presuppose that we already know what the integers are? Well, precisely because of this point, I don't think that axioms and formal logic can be used to "place arithmetic on a more secure foundation" (whatever that would mean). But this stuff is still extremely interesting for at least three reasons:

The situation will change once we start talking not about integers, but about different sizes of infinity. There, writing down axioms and working out their consequences is pretty much all we have to go on! Once we've formalized everything, we can then program a computer to reason for us: Premise 1: For all x, if A(x) is true then B(x) is true.

For all x, if A(x) is true then B(x) is true. Premise 2: There exists an x such that A(x) is true.

There exists an x such that A(x) is true. Conclusion: There exists an x such that B(x) is true. Well, you get the idea. The point is that deriving the conclusion from the premises is purely a syntactic operation -- one that doesn't require any understanding of what the statements mean. Besides having a computer find proofs for us, we can also treat proofs themselves as mathematical objects, which opens the way to metamathematics.

Anyway, enough pussyfooting around. Let's see some axioms for set theory. (I'll state the axioms in English; converting them to first-order logic is left as an "exercise for the reader.")

Empty Set: There exists an empty set.

There exists an empty set. Extensionality: If two sets have the same members then they're equal.

If two sets have the same members then they're equal. Pairing: For all sets x,y there exists a set {x,y}.

For all sets x,y there exists a set {x,y}. Union: For all sets x, there exists a set equal to the union of all sets in x.

For all sets x, there exists a set equal to the union of all sets in x. Existence of Infinite Sets: There exists a set x that contains the empty set and that contains y∪{y} for every y∈x.

There exists a set x that contains the empty set and that contains y∪{y} for every y∈x. Power Set: For all sets x there exists a set consisting of the subsets of x.

For all sets x there exists a set consisting of the subsets of x. Replacement (for every function A): For all sets x, there exists a set {A(y) | y∈x}.

For all sets x, there exists a set {A(y) | y∈x}. Foundation: All nonempty sets x have a member y such that for all z, either z∉x or z∉y. (This is a technical axiom, whose point is to rule out sets like {{{{...}}}}.)

These axioms -- called the Zermelo-Fraenkel axioms -- are the foundation for basically all of math. So I thought you should see them at least once in your life.

Alright, one of the most basic questions we can ask about a set is, how big is it? What's its size, its cardinality? You might say, just count how many elements it has. But what if there are infinitely many? Are there more integers than even integers? This brings us to Georg Cantor (1845-1918), and the first of his several enormous contributions to human knowledge. He says, two sets have the same cardinality if and only if their elements can be put in one-to-one correspondence. Period. And if, whenever you try to pair off the elements, one set always has elements left over, the set with the elements left over is the bigger set.

What possible cardinalities are there? Of course there are finite ones, and then there's the first infinite cardinality, the cardinality of the integers, which Cantor called ℵ 0 ("Aleph-Zero"). The rational numbers have the same cardinality ℵ 0 , a fact that's also expressed by saying that the rational numbers are "countable" (i.e., can be placed in one-to-one correspondence with the integers).

What's the proof that the rational numbers are countable? You haven't seen it before? Oh, alright. First list all the rational numbers where the sum of the numerator and the denominator is 2. Then list all the rational numbers where the sum of the numerator and the denominator is 3. And so on. It's clear that every rational number will eventually appear in this list. Hence there's only a countable infinity of them. QED.

But Cantor's biggest contribution was to show that not every infinity is countable -- so for example, the infinity of real numbers is greater than the infinity of integers. More generally, just as there are infinitely many numbers, there are also infinitely many infinities.

You haven't seen the proof of that either? Alright, alright. Let's say you have an infinite set A. We'll show how to produce another infinite set, B, which is even bigger than A. This B will simply be the set of all subsets of A (which is guaranteed to exist by the Zermelo-Fraenkel axioms). How do we know B is bigger than A? Well, suppose we could pair off every element a∈A with an element f(a)∈B, in such a way that no elements of B were left over. Then we could define a new subset S⊆A, consisting of every a that's not contained in f(a). Notice that this S can't have been paired off with any a∈A -- since otherwise, a would be contained in f(a) if and only if it wasn't contained in f(a), contradiction. Therefore B is larger than A, and we've ended up with a bigger infinity than the one we started with.

This is certainly one of the four or five greatest proofs in all of math -- again, good to see at least once in your life.

Besides cardinal numbers, it's also useful to discuss ordinal numbers. Rather than defining these, it's easier to just illustrate them. We start with the natural numbers:

1, 2, 3, ...

Then we say, let's define something that's greater than every natural number:

ω

What comes after omega?

ω+1, ω+2, ...

Now, what comes after all of these?

2ω

Alright, we get the idea:

3ω, 4ω, ...

Alright, we get the idea:

ω2, ω3, ...

Alright, we get the idea:

ωω, , ...

We could go on for quite a while! Basically, for any set of ordinal numbers (finite or infinite), we stipulate that there's a first ordinal number that comes after everything in that set.

The set of ordinal numbers has the important property of being well-ordered, which means that every subset has a minimum element.

Now, here's something interesting. All of the ordinal numbers I've listed have a special property, which is that they have at most countably many predecessors (i.e., at most ℵ 0 of them). What if we consider the set of all ordinals with at most countably many predecessors? Well, that set also has a successor, call it α. But does α itself have ℵ 0 predecessors? Certainly not, since otherwise α wouldn't be the successor to the set; it would be in the set! The set of predecessors of α has the next possible cardinality, which is called ℵ 1 .

What this sort of argument proves is that the set of cardinalities is itself well-ordered. After the infinity of the integers, there's a "next bigger infinity," and a "next bigger infinity after that," and so on. You never see an infinite decreasing sequence of infinities, as you do with the real numbers.

So, starting from ℵ 0 (the cardinality of the integers), we've seen two different ways to produce "bigger infinities than infinity." One of those ways yields the cardinality of sets of integers (or equivalently, the cardinality of real numbers), which we denote 2 . The other way yields ℵ 1 . Is 2 equal to ℵ 1 ? Or to put it another way: is there any infinity of intermediate size between the infinity of the integers and the infinity of the reals?

(Note: No sooner had I revealed that there were more reals than integers, than a student actually asked this. He claimed never to have heard of the question before; he thought he was just asking for a technical clarification.)

Well, the question of whether there are any "intermediate" infinities between the integers and the reals was David Hilbert's first problem in his famous 1900 address. It stood as one of the great math problems for over half a century, until it was finally "solved" (in a rather disappointing way, as you'll see).

Cantor himself believed there were no intermediate infinities, and called this conjecture the Continuum Hypothesis. Cantor was extremely frustrated with himself for not being able to prove it.

Besides the Continuum Hypothesis, there's another statement about these infinite sets that no one could prove or disprove from the Zermelo-Fraenkel axioms. This statement is the infamous Axiom of Choice. It says that, if you have a (possibly infinite) set of sets, then it's possible to form a new set by choosing one item from each set. Sound reasonable? Well, if you accept it, you also have to accept that there's a way to cut a solid sphere into a finite number of pieces, and then rearrange those pieces into another solid sphere a thousand times its size. (That's the "Banach-Tarski paradox." Admittedly, the "pieces" are a bit hard to cut out with a knife...)

Why does the Axiom of Choice have such dramatic consequences? Basically, because it asserts that certain sets exist, but without giving any rule for forming those sets. As Bertrand Russell put it: "To choose one sock from each of infinitely many pairs of socks requires the Axiom of Choice, but for shoes the Axiom is not needed." (What's the difference?)

The Axiom of Choice turns out to be equivalent to the statement that every set can be well-ordered: in other words, the elements of any set can be paired off with the ordinals 1, 2, ..., ω, ω+1, ... 2ω, 3ω, ... If you think (for example) about the set of real numbers, this seems far from obvious.

It's easy to see that well-ordering implies the Axiom of Choice: just well-order the whole infinity of socks, then choose the sock from each pair that comes first in the ordering.

Do you want to see the other direction: why the Axiom of Choice implies that every set can be well-ordered? Yes?

OK! We have a set A that we want to well-order. For every proper subset B⊂A, we'll use the Axiom of Choice to pick an element f(B)∈A\B. Now we can start well-ordering A, as follows: first let s 0 = f({}), then let s 1 = f({s 0 }), s 2 = f({s 0 ,s 1 }), and so on.

Can this process go on forever? No, it can't. For if it did, then by a process of "transfinite induction," we could stuff arbitrarily large infinite cardinalities into A. And while admittedly A is infinite, it has at most a fixed infinite size! So the process has to stop somewhere. But where? At a proper subset B of A? No, it can't do that either -- since if it did, then we'd just continue the process by adding f(B). So the only place it can stop is A itself. Therefore A can be well-ordered.

OK, should we come back to the puzzle from Tuesday? We have a box, [0,1]2. To each real number x∈[0,1], we associate a countable subset S(x)⊂[0,1]. Now, can we choose S in such a way that for every (x,y) pair, either y∈S(x) or x∈S(y)? What do you think?

I'll give you two answers: that it isn't possible, and that it is possible. Which answer do you want to see first?

Alright, we'll start with why it isn't possible. For this I'll assume that the Continuum Hypothesis is false. Then there's some proper subset A⊂[0,1] that has cardinality ℵ 1 . Let B be the union of S(x) over all x∈A. Then B also has cardinality ℵ 1 . So, since we assumed that ℵ 1 is less than 2 , there must be some y∈[0,1] not in B. Now observe that there are ℵ 1 real numbers x∈A, but none of them satisfy y∈S(x), and only ℵ 0 < ℵ 1 of them can satisfy x∈S(y).

Now let's see why it is possible. For this I want to assume both the Axiom of Choice and the Continuum Hypothesis. By the Continuum Hypothesis, there are only ℵ 1 real numbers in [0,1]. So by the Axiom of Choice, we can well-order those real numbers, and do it in such a way that every number has at most ℵ 0 predecessors. Now put y in S(x) if and only if y≤x, where ≤ means with respect to the well-ordering (not the usual ordering on real numbers). Then for every (x,y), clearly either y∈S(x) or x∈S(y).

Today's puzzle is about the power of self-esteem and positive thinking. Is there any theorem that you can only prove by assuming as an axiom that the theorem can be proved?

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