TITLE: How to "convert" from net to sequence in a first countable space
QUESTION [7 upvotes]: 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?

REPLY [5 votes]: 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).