# Alternative Proof of Theorem 21.3 of Munkres’ Topology

let $$f:X\to Y$$. If the function $$f$$ is continuous, then for every convergent sequence $$x_n \to x$$ in $$X$$. The sequence $$f(x_n)$$ converges to $$f(x)$$. The converse if $$X$$ is metrizable.

My attempt:

Approach(1): Assume towards the contradiction, $$f$$ is not continuous i.e. $$\exists V\in \mathcal{T}_Y$$ such that $$f^{-1}(V)\notin \mathcal{T}_X$$. $$X$$ is metrizable, let say with metric d. That means $$\exists x\in f^{-1}(V)$$ such that $$\forall \delta \gt 0, B_d(x,\delta)\nsubseteq f^{-1}(V)$$. Which implies $$\forall \delta \gt 0,\exists x\neq y\in B_d(x,\delta)$$ such that $$y\notin f^{-1}(V)$$. Taking $$\delta_{n}=\frac{1}{n}; n\in \Bbb{N}$$. So $$\forall n\in \Bbb{N}, \exists x\neq x_n\in B_d(x,\delta_n)$$ and $$x_n\notin f^{-1}(V)$$. Claim: $$\{ x_n\} \to x$$. Proof: let $$U\in \mathcal{N}_x$$. Then $$\exists \epsilon \gt 0$$ such that $$B_d(x,\epsilon)\subseteq U$$. For $$\epsilon \gt 0$$, choose $$N\in \Bbb{N}$$ with $$1\lt \epsilon N$$ such that $$x_n\in B_d(x,1/n)\subseteq B_d(x,1/N)\subseteq B_d(x,\epsilon)\subseteq U, \forall n\geq N$$. Hence $$\{x_n\} \to x$$. But $$f(x_n)\notin V,\forall n\in \Bbb{N}$$. Clearly $$V\in \mathcal{N}_{f(x)}$$, since $$x\in f^{-1}(V)$$ and $$V\in \mathcal{T}_Y$$. So $$\nexists \overline{N}\in \Bbb{N}$$ such that $$f(x_n)\in V, \forall n\geq \overline{N}$$. Which contradicts our initial assumption of $$f(x_n)\to f(x)$$, when $$\{x_n\} \to x$$. Thus $$f$$ is continuous. Is this proof correct?

Approach(2): $$\exists x\in X$$ and $$\exists V\in \mathcal{N}_{f(x)}$$, such that $$\forall U\in \mathcal{N}_x$$ we have $$f(U)\nsubseteq V$$. So $$U\nsubseteq f^{-1}(V)$$. $$\forall \delta \gt 0, B_d(x,\delta)\in \mathcal{N}_x$$. So $$B_d(x,\delta)\nsubseteq f^{-1}(V), \forall \delta \gt 0$$. Rest of the proof is similar to approach(1). Is this proof correct?

Can you explicit tell/explain why direct proof of theorem 21.3 using open set and $$\epsilon - \delta$$ definition of continuity don’t work(it’s impossible by How to prove continuity in terms of convergent sequences directly? post)?

You assume non-continuity, so some open $$O \subseteq Y$$ so that $$f^{-1}[O]$$ is not open, so there is some $$x$$ with $$f(x) \in O$$ so that all balls around $$x$$ "stick out of" $$f^{-1}[O]$$. In fact what you need here in the proof is that $$x$$ has a countable local base of neighhbourhoods $$U_n$$ (we can assume it to be decerasing in $$n$$, as in the metric case, where we use $$U_n = B_d(x,\frac1n)$$ as the local base) so that the positive consequence of non-interior-ness of $$x$$ allows us to pick $$x_n \in U_n \setminus f^{-1}[O]$$ giving us a sequence converging to $$x$$ but the $$f(x_n) \notin O$$ so these do not converge to $$f(x)$$, which then contradicts the given of sequential continuity, so the assumption of non-contuity was wrong.
So yes, your proofs are correct, and follow the schedule I described. The converse can be extended even beyoud first countable spaces, to the realm of so-called sequential spaces (defined by the property that a sequentially closed set $$F$$ is always closed; $$F$$ is ssequentially closed when for all sequences $$x_n$$ with terms in $$F$$ we have $$x_n \to x$$, we can say $$x \in F$$, so it contains the limits of all its sequences). Try to generalise the converse to such spaces $$X$$. All first countable (and thus metric) spaces are sequential, but there are many more.