Subnets vs. Subsequences

Question

Can someone give me an example of a compact topological space that is not first-countable such that there is a sequence $(x_n)_{n \in \mathbb{N}}$, with the property that for every subsequence $(x_{n_p})_{p \in \mathbb{N}}$ of $(x_n)_{n \in \mathbb{N}}$, the subsequence wouldn't converge?

In our topology course we learned that in a compact space, every net has a convergent subnet. Now a sequence is of course also a net, so it has to have a convergent subnet. But generally a subnet of a sequence need not be a sequence; but could it still be possible in compact space to choose as subnet of a sequence another sequence? It is of course important that this space should not be first countable, because if it were we wouldn't have to deal with the cumbersome nets. Compactness and first-countable would imply that every sequence contains a convergent subsequence, so my question would be meaningless.

Answer 1

Take $X = \{0,1\}^{[0,1]}$, thought of as the set of all functions from the unit interval $[0,1]$ to the two-point set $\{0,1\}$. Equip $X$ with the product topology; $X$ is compact by Tychonoff's theorem. In the product topology, a sequence $f_n$ converges iff it converges pointwise, i.e. if the $\{0,1\}$-valued sequence $f_n(x)$ converges (i.e. is eventually constant) for every $x \in [0,1]$.

Define $f_n : [0,1] \to \{0,1\}$ to be the function such that $f_n(x)$ is the $n$th bit in the binary expansion of $x$. (If $x$ has more than one binary expansion, take $f_n(x) = 0$ for all $n$; it doesn't really matter here.) Then given any subsequence $f_{n_m}$, you should be able to produce an $x \in [0,1]$ such that $f_{n_m}(x)$ does not converge. Hence $\{f_n\}$ has no convergent subsequence.

Answer 2

My favorite example is $\beta\omega$, in which the sequence $\langle n:n\in\omega\rangle$ has no convergent subsequence. To see this, let $A$ be any infinite subset of $\omega$, and suppose that $\langle n:n\in A\rangle\to p$ for some $p\in\beta\omega$. Clearly $p\in\beta\omega\setminus\omega$, so $p$ is a free ultrafilter on $\omega$. Partition $A$ into two infinite subsets $A_0$ and $A_1$. Since $p$ is an ultrafilter, exactly one of $A_0$ and $A_1$ belongs to $p$, say $A_0$. But then $A_0\cup\{q\in\beta\omega\setminus\omega:A_0\in q\}$ is an open nbhd of $p$ that misses infinitely terms of $\langle n:n\in A\rangle$, so $\langle n:n\in A\rangle$ that does not converge to $p$ after all.