Exercise 2.3.1 (Topological continuity generalizes metric continuity)

Proof. We demonstrate both directions of the equivalence.

$⟹$

Suppose that the topological continuity holds, i.e., for each open neighbourhood $V$ of $f(x)$, there exists an open neighbourhood $U$ of $x$ such that $f(U)\subseteq V$. Let $\{x_n\}$ be an arbitrary sequence converging to $x$, i.e,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->\delta >0\exists <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}\forall <!--\; FUNCTION\; APPLICATION\; -->n\ge N:x_n\in B(x,\delta ).$$

(1)

To demonstrate that $f(x_n)\to x$, pick an arbitrary $𝜖>0$. By the assumption, we can find an open neighbourhood $B(x,\delta )\subseteq U$ of $x$ such that $f(B(x,\delta ))\subseteq B(f(x),𝜖)$. Using 1 ,

$$\exists <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}\forall <!--\; FUNCTION\; APPLICATION\; -->n\ge N:f(x_n)\in f(B(x,\delta ))\subseteq B(f(x),𝜖),$$

which is just another way of saying that $\{f(x_n)\}$ converges to $f(x)$.

$⟸$

Suppose that the metric continuity holds, i.e., whenever $\left\{x_n\right\}$ is a sequence in $X$ such that $x_n\to x$, then $f\left(x_n\right)\to f(x)$. Let $V$ be an arbitrary open neighbourhood of $f(x)$ and suppose for the sake of contradiction that no open neighbourhood $U$ of $x$ satisfies $f(U)\subseteq V$. In particular,

$$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:f\left(B\left(x,\frac{1}{n}\right)\right)⊈B(f(x),𝜖)$$

for some $B(f(x),𝜖)\subseteq V$ (for $𝜖>0$ small enough). By axiom of choice, we can construct a sequence $x_n\in B\left(x,\frac{1}{n}\right)$ and would get $f(x_n)\to ̸f(x)$ by construction - a contradiction.