[Math] Image of a convergent sequence under a continuous function converges

Question

Let $f:X \mapsto Y$ be a continuous function between two topological spaces. Let $\{x_i\} \subset X$ be a convergent sequence with limit $x \in X$. Does it follow that $\{f(x_i)\} \subset Y$ converges to $f(x) \in Y$?

This came up in a manifolds topology class, where we are mapping between a subset of metric space and a subset of the manifold with a homeomorphism.

Answer 1

Yes it does. In fact, for first countable spaces (in particular, metric spaces), this is equivalent to $f$ being continuous. More precisely,

Theorem 1. If $X$ and $Y$ are topological spaces and $X$ is first countable, then $f:X\to Y$ is continuous iff whenever a sequence $(x_n)$ in $X$ converges to a limit $x$, then $f(x_n)\to f(x)$.

For general topological spaces, we need a different notion of convergence to get the same equivalence. For instance, using nets one can obtain the analogue:

Theorem 2. If $X$ and $Y$ are topological spaces, then $f:X\to Y$ is continuous iff whenever a net $(x_\alpha)$ in $X$ converges to a limit $x$, then $f(x_\alpha)\to f(x)$.

Since every sequence is a net, the forward direction of Theorem $1$ always holds. It's the converse that fails in general.