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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)?

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The two approaches are basically the same:

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.

Only non-continuity gives us a concrete point to work with to start contradicting sequential continuity. It's just in the nature of the definitions and their logical structure, as it were.

There is some work in the foundations of logic and constructism (where proofs from contradiction are not allowed) to show that continuity and sequential continuity are really logically different. In many branches of constructive maths (like intuitionism) you cannot even define a non-continuous function.

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.

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