Continuous Mapping is Sequentially Continuous

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Let $X$ and $Y$ be topological spaces.

Let $x \in X$.

Let $f: X \to Y$ be a mapping that is continuous at $x$.

Then $f$ is sequentially continuous at $x$.


Let $f$ be continuous (everywhere) on $X$.

Then $f$ is sequentially continuous on $X$.


Let $\left\langle{x_n}\right\rangle_{n \ge 1}$ be a sequence in $X$ converging to $x$.

Let $V$ be a neighborhood of $f \left({x}\right)$ in $Y$.

We are required to show that there exists $N \in \N$ such that $f \left({x_n}\right) \in V$ for all $n \ge N$.

By continuity of $f$, choose a neighborhood $U$ of $x$ in $X$ such that $f \left({U}\right) \subseteq V$.

Since $\left\langle{x_n}\right\rangle_{n \ge 1}$ converges, there exists $N \in \N$ such that $x_n \in U$ for all $n \ge N$.

Therefore we must have that $f \left({x_n}\right) \in f \left({U}\right) \subseteq V$ for all $n \ge N$, as required.