Plausibility argument for Zorn's Lemma

512 Views Asked by At

In "Mathematical Physics" by Robert Geroch, the following 'plausibility argument' is given for Zorn's Lemma [If every totally ordered subset of a partially ordered set $S$ is bounded above, $S$ has a maximal element]:

It is not difficult to convince onselef that this statement [Zorn's Lemma] seems "true". Pick $s_1$ in $S$. If this $s_1$ is not a maximal element, there is an $s_2$, different from $s_1$, with $s_1 \leq s_2$. If $s_2$ is not maximal, there is an $s_3$, different from $s_2$, with $s_2 \leq s_3$, etc. Either we eventually obtain a maximal element (in which case we are done), or we obtain $s_1 \leq s_3 \leq s_3 \leq ... $ . In this latter case, $\{s_1,s_2,...\}$ is a totally ordered subset of $S$, whence it is bounded above, say by $s_1'$. If $s_1'$ is not a maximal element, there is $s_2'$ with $s_1' \leq s_2'$, etc. (We "go right through infinity.") If we find no maximal element in $s_1', s_2', ...$, we obtain, again, a totally ordered set, which must be bounded above, say by $s_1''$, etc. Suppose we are unlucky and go through an infinite sequence of such sequences without finding a maximal element. Then we obtain totally ordered $s_{m}^{n} \{m,n = 1,2,...\}$, which must be bounded above, say by $t1$, etc. We "continue in this way. At every stage we have a totally ordered subset, which must be bounded above, so we add in this upper bound to obtain a new totally ordered subset. We can go through infinity, an infinity of infinities, etc. We never get stuck; we can always go on. We should obtain eventually a maximal element."

I don't see how this argument makes Zorn's Lemma plausible, however. If anything, it seems to say that it is possible that we can go through infinities of elements forever, always finding upper bounds and forming totally ordered sets with them, and doing this on and on, without ever finding a maximal element. In particularly, how does the last sentence above "We should obtain eventually a maximal element" at all follow from the argument?

2

There are 2 best solutions below

8
On BEST ANSWER

This is not quite a mathematical argument. And since it is not really a proof, Zorn's lemma doesn't quite follow from it. What it is is an intuitive explanation trying to convince the reader to accept Zorn's lemma as an axiom.

The point that the paragraph is trying to make is that if we have a partial order satisfying the conditions in Zorn's lemma, then we can define this recursive process which will construct a chain, and if there is no maximal element then this process cannot terminate.

But when we say "terminate" we don't mean that in the "applied" kind of way. We mean that in the set theoretical way. Saying that we in fact defined an injective function from a proper class into a set, which is a contradiction. That what it means "go through infinity, an infinite of infinites etc.", that we go through the first $\omega$ steps, but we can continue, and we go through more and more and more steps and so on.

One reason to give this sort of... half-assed explanation lies in the third word of your question "Physics". To fully understand Zorn's lemma one has to understand some set theory first, to understand the axiom of choice, to understand the well-ordering principle, and so on. One can just write it up as a black box, or convince themselves that they understand how Zorn's lemma works. But I won't expect a physicist to sit down and understand the mechanics of transfinite induction, and the axiom of choice.

So in order to convince a physicist that Zorn's lemma is plausible, one has to resort to furious handwaving like in the quoted paragraph.

The paragraph is, in fact, a broad stroke of how to prove Zorn's lemma from the axiom of choice, using transfinite recursion. The "through infinity" part is what is known as a limit stage in transfinite recursion.


Finally, let me point that of course one has to be convinced about the plausibility of Zorn's lemma. It is unprovable without the axiom of choice (and indeed assuming Zorn's lemma we can prove the axiom of choice) and that makes the lemma very non-constructive. And one can see that from the formulation of the lemma. There exists a maximal element. We are not told how that element looks like, or what sort of properties it may have. It just exists.

2
On

The main idea, which is not intuitive at all without some set theory background, is that if this process of going through infinity, infinities of infinities, etc., did not eventually result in a maximal element, then the original $S$ would be too large to be a set (it would be a proper class).