Zorn’s Mystery

I am teaching topology this semester, and as I get older, I more and more dislike the consequences of Zorn’s Lemma, and thus of the general Axiom of choice.

Assume you have a sequence of sequences of number in [0,1]. The n-th sequence looks like


If you cannot imagine this, you are in good companion. Drawing the graphs of the sequences in the plane does not really help, since the possible choice for the sequence of sequences is richer than we can imagine.

However, Zorn’s Lemma and its consequence, the Tihonov compactness theorem tells us that there is a point wise convergent subsequence. I.e., we can choose indices

\(n_1 < n_2 < n_3 < \ldots\)

such that all sequences

\(x_{i,n_1},x_{i,n_2},x_{i,n_3},\ldots \)

converge, where i is any natural number. For this, the sequence must not even be completely bounded, but only point wise bounded, which is of course equivalent, but makes matters even more frightening.

Trying to conduct a constructive proof of this fails. Yes, if we had only m-tupels

\((x_{1,n},x_{2,n},x_{3,n},\ldots,x_{m,n}) \)

instead of sequences, we can prove the result in the following way: Since the sequence of the first components

\(x_{1,1},x_{1,2},x_{1,3},\ldots \)

is bounded, we can select a convergent subsequence of this. From this subsequence, we can select a subsequence of the subsequence, such that the second component converges, and so on until the m-th component. Since subsequences of convergent subsequences converge, we are done.

For sequences we can start the same, but our resulting subsequences may become ever more sparse, so that we are left with nothing.

Moreover, choosing a convergent subsequence of a bounded sequence of real numbers is far from being a really constructive procedure. We can subdivide the interval over and over again, and in each step choose the part, which contains infinitely many members of the sequence. This looks like a constructive procedure, but it requires an overview over the complete sequence. By the way, it is an interesting fact, that each sequence contains either a monotone increasing or a monotone decreasing subsequence. The proof of this is also pseudo constructive.

So let us simplify the setup a bit. We restrict the members of our sequence to 0 or 1. So our sequences look like

\(1,1,0,0,1,0,0,1,\ldots \)

Convergence in the discrete space {0,1} is very easy. The sequence must be eventually identical to 0 or 1. So we are looking for a subsequence of our sequences, such that the i-th place is remains at 0 or 1 starting from one index K depending on i. E.g.

\(x_{i,n_k} = 1, \quad k \ge K_i \)

This looks familiar, if we represent the 0-1 sequences by the dual representation of real numbers. E.g.,

\(x_n = 0.1101001\ldots \)

Effectively, we want to select a convergent subsequence of this sequence of real numbers. The existence of such a subsequence follows from the completeness of the real numbers. We can avoid Zorn’s Lemma in this case.

The same we can do for all sequences with members in a point wise finite range. Since large finite sets can fill out interval [0,1] quite well, we are very close to the general result, which uses Zorn’s Lemma.

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