Let $X$ and $Y$ be topological spaces and $f: X \to Y$ a function.
a) Show that if $f$ is continuous, then for each $ x \in X$ and each sequence $ (x_n) _ {n \in \mathbb N}$ in $X$ such that $ x_n \to x $ we have to $f( x_n) \to f(x) $.
b) Show that the reciprocal is true in the case where $X$ has a fundamental countable neighborhood system.
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Let $U$ be an open neighborhood of $f (x)$, we have that $ f^{- 1} (U)$ is a neighborhood of $x$, so there is $V$ such that $ x_n \in f^{- 1}(U) $ , for $n \ge \mathbb N $, and therefore $f (x_n) \in V$ for $n \ge \mathbb N $.
I know that the exercise is a variation of the Theorem 30.1 in the book Topology by James R. Munkres, 2nd edition, but I am confused, mainly if my answer to item a) is acceptable, because I have the impression that I cannot use what I used in item a) without first proving item b ) what I'm not able to do for now.
Thank in advance for any assistance.
Your proof is almost there, except for the part where you say that $n \geq \mathbb{N}$ which doesn't make much sense. ($\mathbb{N}$ isn't even a number).
Suppose that $U$ is some arbitrary open neighborhood of $f(x)$. Our goal is to show that there is some natural number $N \in \mathbb{N}$ such that for all $n \geq N$, $f(x_n) \in U$.
Since $f$ is continuous, $f^{-1}(U)$ is an open neighborhood of $x$, and since $x_n \rightarrow x$, there is some $N \in \mathbb{N}$ such that for all $n \geq N$, $x_n \in f^{-1}(U) \implies f(x_n) \in U$. In summation for $n \geq N$, $f(x_n) \in U$, which is exactly what we needed.
Part (b) is a tad more complicated. Suppose we have sequential continuity, that is, if $x_n \rightarrow x$, then $f(x_n) \rightarrow f(x)$ for every sequence and point. Suppose that $f$ is not continuous at some point $x \in X$, i.e, there is some $U \subseteq Y$ open such that $f(x) \in U$ but for every open $V \subseteq X$ with $x \in V$, we have that $V \not \subseteq f^{-1}(U)$ which means $f(V) \not \subseteq U$. Now let $\{B_n : n \in \mathbb{N} \}$ be a countable local base at $x$. Notice that we may assume $B_{n+1} \subseteq B_{n}$ for every $n$ (why?). Then by our assumption $f(B_n) \not \subseteq U$ for every $n$, meaning there is some point $x_n \in B_n$ such that $f(x_n) \not \in U$. Notice that the $x_n$ form a sequence such that $x_n \rightarrow x$. By sequential continuity this means $f(x_n) \rightarrow f(x)$. But $U$ is an open neighborhood of $f(x)$ and for every $n$, we have that $f(x_n) \not \in U$, which is a contradiction. So $f$ is continuous at $x \in X$, and $x$ was chosen arbitrarily.