By Tychonoff Theorem $\prod_{\mathbb R} [0,1]$ is compact and since $\mathbb R=2^{\omega}$, if for $\alpha \in 2^{\omega}$, $x_n(\alpha)=\alpha(n)$ then if we consider a subsequence $x_{n_0}, x_{n_1}, x_{n_2},…$ where $ n_0 < n_1< n_2<… $ and where for $\alpha \in 2^{\omega}, \alpha(n_0)=1, \alpha(n_1)=0, \alpha(n_2)=1,…$, then $(\alpha(n_0), \alpha(n_1), \alpha(n_2),…)=(0,1,0,…)$ does not converge. This is of course an example of a compact space which not sequentially compact.
But is it possible to do it without the axiom of choice? I would like to show an explicit example of a sequence with an explicit limit point that does not have a convergent subsequence in $\prod_{\mathbb R} [0,1]$. Thx.
Best Answer
I think this question is fairly clear now, in its edited form. Assuming AC, $[0,1]^{\mathbb R}$ is not sequentially compact, witnessed by the sequence specified in the question. What happens without AC? Is there a sequence with a limit point that has no convergent subsequence? I think this is an interesting question.
I cannot give a complete answer to your question right now, but I hope the following clarifies things.
For simplicity, I take the exponent to be $2^\omega$ instead of $\mathbb R$ (the two sets have the same size even without choice). Take the sequence $(x_n)$ in $[0,1]^{2^\omega}$ defined by $x_n(y)=y(n)$ for $n\in\omega$ and $y\in 2^\omega$.
Now, if $z\in[0,1]^{2^\omega}$ is a limit point of the sequence, then the collection of all sets of the form $\{n\in\omega:x_n\in U\}$, $U$ an open neighborhood of $z$, generates a free ultrafilter on $\omega$.
The non-existence of a free ultrafilter on $\omega$ is consistent with ZF.
Hence, the sequence $(x_n)$ has no convergent subsequence (we don't need AC for this) but it might also not have limit point.
I don't know whether you can construct without AC a sequence in $[0,1]^{2^\omega}$ that has no convergent subsequence but a limit point. My guess would be no, and an argument might go like this: Let $(x_n)$ be a sequence without convergent subsequence.
No projection of the sequence to a countable set of coordinates can have this property. Figure out a way to show (without choice) that the sequence behaves essentially as the particular sequence that we have defined above, i.e., from a limit point we can extract a nontrivial ultrafilter.
Just one more remark on the Tychonov theorem: Tychonov for Hausdorff spaces is not equivalent to full AC. You need to consider spaces that are not Hausdorff to get an equivalent of AC. The argument above shows that from the compactness of $[0,1]^{2^\omega}$ one gets a free ultrafilter on $\omega$, whose existence cannot be proved without AC.
Hence we cannot prove the compactness of this particular product of spaces without AC (same with $[0,1]$ replaced by $\{0,1\}$).