Relationship between continuity and sequential continuity

Question

Let $(X, \mathcal{T})$, $(Y, \mathcal{U})$ be topological spaces and let $f: X \rightarrow Y$.

(a) Suppose $f$ is continuous and $\{x_n\}_{n=1}^{\infty}$ is a sequence in $X$ converging to a point $x$. Show that $\{f(x_n)\}_{n=1}^{\infty}$ converges to $f(x)$

(b) Suppose $(X, \mathcal{T})$ is first-countable, and suppose for each $x \in X$, and each sequence $\{x_n\}_{n=1}^{\infty}$ that converges to $x$, $\{f(x_n)\}_{n=1}^{\infty}$ converges to $f(x)$. Show that $f$ is continuous.

I have a slight idea on how to go about (a)

Consider the set containing $\{f(x_1), f(x_2), \dots, f(x_n), \dots, f(x)\}$. Under the map $f$, the pre-image of this set is an open set containing $\{x_1, x_2, \dots, x_n,\dots, x\}$. However since the set containing $\{x_1, x_2, \dots, x_n,\dots, x\}$ is open, its image must be a open set. Thus, the sequence $\{f(x_n)\}$ must converge to $f(x)$.

But no idea on (b) whatsoever.

Any help is appreciated.

Answer 1

How do you know that $\{ x_n \}$ is an open set? Take for instance the sequence defined by $x_n = 0$ for all $n \ge 1$; this converges obviously to $0$. The set $\{ f(x_n) \} = \{ 0 \}$ is clearly not open!

You need to use the definition of convergence of a sequence. A sequence $\{ x_n \}$ converges to $x$ if for all neighborhoods $U$ of $x$, there exists some $N \in \mathbb{Z}^+$ such that $x_n \in U$ for all $n > N$. Here's how you can proceed:

(a) Let $V$ be a neighborhood of $f (x)$. It is easy to see that $x \in f^{-1} (V)$. By continuity of $f$, $f^{-1} (V) \subseteq X$ is open. Therefore, $f^{-1} (V)$ is a neighborhood of $x$. By hypothesis, there exists some $N_V \in \mathbb{Z}^+$ such that $x_n \in f^{-1} (V) \implies f (x_n) \in V$ for all $n > N_V$. Since $V$ was arbitrary, the sequence $\{ f (x_n) \}$ converges to $f (x)$.

(b) Straightforward application of the Sequence Lemma (see here). Then note that continuity of $f$ is equivalent to the condition that for any subset $A \subseteq X$, $f ({\rm Cl} A) \subseteq {\rm Cl} f (A)$.