A map between metric spaces preserving convergent sequences is continuous

Pugh, "Mathematical Analysis", exercise 2.17:

Assume $f : M \to N$ is a map from one metric space to another which satisfies the following condition: for every convergent sequence $(a_n) \subset M$, the sequence $(f(a_n)) \subset N$ converges. Prove that $f$ is continuous.

(I know that if $f$ commutes with limits then the answer is affirmative, but I can't prove it with that hypothesis).

Solution 1:

Suppose that the desired property is satisfied. Let $(a_n) \to a$ be a convergent sequence. By hypothesis, we know that that $f(a_n) \to L$ for some $L \in \mathbb{R}$. Assume for contradiction that $L \neq f(a)$. Now consider the sequence

$(b_n)= \begin{cases} \ a_n \textrm{ if $n$ is even} \\ \ a \textrm{ otherwise} \\ \end{cases} $

Clearly $b_n$ is convergent, but $f(b_n)$ is not, (by our assumption, take $\epsilon=|L-f(a)|/2$ ) so we have a contradiction.

Therefore, "convergence preserving" implies sequential continuity.

See Munkres 21.3 for a proof of the equivalence of continuity and Sequential continuity in metric spaces. (or https://proofwiki.org/wiki/Sequential_Continuity_is_Equivalent_to_Continuity_in_Metric_Space)

Solution 2:

I assume the statement is supposed to hold for every sequence which converges. If that's true specialize to sequences which are constant, like $a_n = a$ for every $n$, to get what you consider missing. If then $(b_k)$ is an arbitrary sequence converging to $a$ replace $(b_k)$ by $(b_0, a, b_1, a, b_2, a,\dots)$ to find the (unique) limit of $f$ applied to the sequence.