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.

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    $\begingroup$ Can you give an example of a subnet that is not a subsequence? $\endgroup$ – SmileyCraft Jan 20 '19 at 5:40
  • $\begingroup$ But if you have a subnet, isn't it always possible to find a subnet of the subnet which is a subsequence of the original sequence? It seems to me that this is easily done using the final property of subnets. $\endgroup$ – SmileyCraft Jan 20 '19 at 5:49
  • $\begingroup$ That definitely sounds like a good idea. However, the proof that there exists a convergent subnet is non-constructive and relies on Tychonoff's theorem. So the example of a sequence in $\{0,1\}^{[0,1]}$ with no convergent subsequence is $\{f_n\}$ such that $f_n(x)$ is the $n$'th bit of $x$. I have no idea what would be a convergent subnet. $\endgroup$ – SmileyCraft Jan 20 '19 at 6:03
  • $\begingroup$ No, a subnet need not have a subsubnet that is a subsequence of the original sequence. $\endgroup$ – David C. Ullrich Jan 20 '19 at 18:46

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 $f$.

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  • $\begingroup$ Thank you very much for your help. I was, however, more looking for a solution based on nets. I will try to remember to give you a bounty when I can, because even though this is not the answer I was looking for, you definitely helped me find it. $\endgroup$ – SmileyCraft Jan 20 '19 at 9:27
  • $\begingroup$ Did you know that you can explicitly state when starting a bounty that it is to reward an existing answer, but you still can not give the bounty until 24 hours later? How wonderful. $\endgroup$ – SmileyCraft Jan 22 '19 at 13:03

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 $(f^+,I\cup\{a\})$.

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