How to "convert" from net to sequence in a first countable space

Question

In a first countable space, what's a good way of going from nets to sequences?

Let me explain more clearly what I mean. Suppose $f:X\to Y$ is a topological map and $X$ is first countable. Then I believe the following three conditions are equivalent:

$$ \tag {i}\text{ For every convergent net $x_{\alpha} \to x,\,f(x_{\alpha}) \to f(x)$. }$$ $$ \tag{ii} \text{ For every convergent sequence $x_{i} \to x,\,f(x_{i}) \to f(x)$. }$$ $$\tag{iii} \text{ $f$ is continuous. }$$

The only way I know right now to prove (ii)$\implies$(i) is to show (ii)$\implies$(iii)$\implies$(i). But suppose I wanted to attempt a direct proof. I would start with a net $x_\alpha \to x$, and I'd like to find a way of converting it into a sequence $x_i$, whose convergence guaranteed the convergence of $x_{\alpha}$. It seems like there should be a trick for this.

Similarly, in a first countable space, the following three conditions are equivalent:

$$ \tag {1}\text{ Every net $x_{\alpha}$ in $X$ has a convergent subnet. }$$ $$ \tag{2} \text{ Every sequence $x_{\alpha}$ in $X$ has a convergent subsequence. }$$ $$\tag{3} \text{ $X$ is compact. }$$

In proving (2)$\implies$(1) directly, I'd like to make a similar conversion from net to sequence.

Does anyone know how to do this?

Answer 1

The three items that you have pointed out are not equivalent in the class of first-countable spaces.

• To see that (1) and (2) are not equivalent, not that $[0,1]$ is a compact subset of $\mathbb{R}$, and is therefore sequentially compact (which is your statement (2)). Recall that the least uncountable ordinal $\omega_1$ is linearly ordered by $<$ (and is therefore directed). Fixing an injective mapping $\omega_1 \to [0,1]$, $\{ x_\xi : \xi < \omega_1 \}$, I claim that no subnet of this net is convergent. The basic idea here is that any subnet of a net indexed by $\omega_1$ must itself have a subnet which is indexed by $\omega_1$, so we might as well assume that the original net converges to some $x \in [0,1]$. Then there must be an increasing sequence $\{ \xi_n \}_{n \in \mathbb{N}}$ in $\omega_1$ such that for any $n \in \mathbb{N}$ and any $\xi \geq \xi_n$ we have $| x - x_\xi | < \frac{1}{n}$. Setting $\xi_* = \sup_{n \in \mathbb{N}} \xi_n$ it follows that $x_\xi = x$ for all $\xi \geq \xi_*$ contradicting that we began with an injective net! (Only the fact that $[0,1]$ is first-countable and Hausdorff was really used here.)

• The basic example of a sequentially compact first-countable space which is not compact is the ordinal space $\omega_1 = [0, \omega_1)$ of all countable ordinals with the order topology. Any sequence $\{ \alpha_n \}_{n \in \mathbb{N}}$ in $\omega_1$ will have either a constant subseqeunce or a strictly increasing subsequence. In the former case it is easy to see that it has a convergent subsequence. In the latter case, you take the strictly increasing subsequence $\{ \alpha_{n_k} \}_{k \in \mathbb{N}}$, and see that the supremum $\alpha = \sup_{k \in \mathbb{N}} \alpha_{n_k} < \omega_1$ is the limit of this subsequence. (It is not compact because the family $\{ [ 0 , \alpha ) : \alpha < \omega_1 \}$ is an open cover with no finite (indeed, no countable) subcover.

Answer 2

As an addition to Arthur's answer: the conditions (i),(ii) and (iii) are equivalent for a fixed function $f:X \rightarrow Y$ from a first countable (and probably sequential suffices) space $X$.

To see this: (i) implies (ii) trivially, as all convergent sequences are in particular convergent nets.

(iii) implies (i): this is a standard argument: let $O$ be an open neighbourhood of $f(x)$, then $f^{-1}[O]$ is open by continuity and contains $x$, so it contains a tail of the net $x_\alpha$, and so $f(x_\alpha)$ for those same $\alpha$ are in $O$, as required.

(ii) implies (iii): Let $C$ be closed in $Y$, and we want to show $f^{-1}[C]$ is closed. As in first countable (or sequential) spaces, closed is equivalent to sequentially closed we only need to show: let $(x_n)$ be a sequence in $f^{-1}[C]$, and suppose $x_n \rightarrow x$. Then $f(x_n) \rightarrow f(x)$ by assumption (ii), and the points $f(x_n)$ are in $C$, and $C$ is closed and so in any space $f(x) \in C$, so indeed $x \in f^{-1}[C]$, as we needed to see that $f^{-1}[C]$ is sequentially closed and thus closed.

(1) and (3) are equivalent in all spaces, (2) is the definition of sequentially compact. In first countable spaces, compactness (so (1),(3)) implies sequential compactness, but not reversely (as Arthur showed).