Sequential Continuity is Equivalent to Continuity in Metric Space

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Theorem

Let $\left({X, d}\right)$ and $\left({Y, e}\right)$ be metric spaces.

Let $f: X \to Y$ be a mapping.

Let $x \in X$.


Then $f$ is continuous at $x$ if and only if $f$ is sequentially continuous at $x$.


Corollary

$f$ is continuous on $X$ if and only if $f$ is sequentially continuous on $X$.


Proof

We have that a Continuous Mapping is Sequentially Continuous.

To prove the converse, by the Rule of Transposition we may prove the contrapositive:

If $f$ is not continuous at $x$, then $f$ is not sequentially continuous at $x$.

We suppose therefore that there exists $\epsilon_0 > 0$ such that for all $\delta > 0$ there exists $y \in X$ such that $d \left({x, y}\right) < \delta$ and $e \left({f \left({x}\right), f \left({y}\right)}\right) \ge \epsilon_0$.

For $n \ge 1$, define $\delta_n = \dfrac 1 n$.

For $n \ge 1$, we may choose $y_n \in X$ such that $d \left({x, y_n}\right) < \delta_n$ and $e \left({f \left({x}\right), f \left({y_n}\right)}\right) \ge \epsilon_0$.

Therefore, by definition the sequence $\left\langle{y_n}\right\rangle_{n \ge 1}$ converges to $x$.

However, by definition the sequence $\left\langle{f \left({y_n}\right)}\right\rangle_{n \ge 1}$ does not converge to $f \left({x}\right)$.

That is, $f$ is not sequentially continuous at $x$.

$\blacksquare$