Question regarding proof of Tychonoff's theorem

Question

On Wikipedia it states that a space $X$ is compact if and only if every net has a convergent subnet. It then states that a net in the product topology has a limit if and only if each projection has a limit. I understand why both of these facts are true. However it then states this leads to a slick proof of Tychonoff's theorem, and I don't quite see how.

In particular, it seems to me that the first fact implies that every compact space is sequentially compact. Since every sequence is also a net, it has a convergent subnet, which gives a convergent subsequence. This is obviously not true, since $\{0,1\}^\mathbb{R}$ is not sequentially compact, but it is compact by Tychonoff's theorem.

Answer 1

The problem in the reasoning in the OP is that a net can admit no subnet that is a sequence. This is because a subnet needs to be final. For example if the index set of the subnet is $\omega_1$, the first uncountable ordinal, then there exists no final function $h:\mathbb{N}\to\omega_1$, since $\cup h(n)$ is an upper bound on $h(n)$.

To prove Tychonoff's theorem with nets, the main idea is to use Zorn's lemma on partial clusterpoints. We define a partial cluster point of a sequence $\{f_n\}$ in $\Pi X_{\alpha\in A}$ as a tuple $(f,I)$ where $f\in\Pi X_{\alpha\in A}$ and $I\subseteq A$ such that $f$ is a cluster point of $\{f_n|_I\}$. We can then define the partial ordening $(f,I)\leq(g,J)$ iff $I\subseteq J$ and $g|_I=f$.

Then there obviously exists some partial clusterpoint; consider $(f,\emptyset)$. There is a canonical notion of union of clusterpoints which gives us an upper bound for any chain. Hence, by Zorn's lemma there exists a maximal cluster point $(f,I)$. To show that $I=A$, consider a hypothetical element $a\in A\setminus I$. Then the $a$-coordinate of the subnet $\{g_n|_I\}$ of $\{f_n|_I\}$ converging to $f$ must have a cluster point $p$. Then setting $f^+(a)=p$ while $f^+|_I=f$ we find a greater partial cluster point $(f^+,I\cup\{a\})$.

Answer 2

Here is an example of a sequence which has no convergent subsequence, but it has a convergent subnet assuming the axiom of choice.

Let $I=\{0,1\}^\mathbb N$.

The product space $X=\{0,1\}^I$ is a compact space which is not sequentially compact.

For $n\in\mathbb N$ define $f_n:I\to\{0,1\}$ by setting $f_n(i)=i(n)$. Then $\langle f_n:n\in\mathbb N\rangle$ is a sequence in $X$ with no convergent subsequence.

Let $\mathcal U$ be a uniform ultrafilter on $\mathbb N$. (I suppose it can be done without using ultrafilters, but I'm more used to filters than nets.)

Define $f:I\to\{0,1\}$ so that, for each $i\in I$, $\{n\in\mathbb N:i(n)=f(i)\}\in\mathcal U$.

Let $D$ be the collection of all finite subsets of $I$, directed by $\subseteq$.

For $K\in D$, let $h(K)$ be the least $n\in\mathbb N$ such that $i(n)=f(i)$ for all $i\in K$; this defines a monotone final function $h:D\to\mathbb N$.

Define $g_K=f_{h(K)}\in X$; then $\langle g_K:K\in D\rangle$ is a subnet of $\langle f_n:n\in\mathbb N\rangle$ which converges to $f$.