# Question regarding proof of Tychonoff's theorem

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.

• Can you give an example of a subnet that is not a subsequence? – SmileyCraft Jan 20 '19 at 5:40
• 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. – SmileyCraft Jan 20 '19 at 5:49
• 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. – SmileyCraft Jan 20 '19 at 6:03
• No, a subnet need not have a subsubnet that is a subsequence of the original sequence. – 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 $$f$$.

• 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. – SmileyCraft Jan 20 '19 at 9:27
• 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. – 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 $$(f^+,I\cup\{a\})$$.