Sequential Continuity is Equivalent to Continuity in Metric Space

Theorem

Let $\struct {X, d}$ and $\struct {Y, e}$ 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 $\map d {x, y} < \delta$ and $\map e {\map f x, \map f y} \ge \epsilon_0$.

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

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

Therefore, by definition the sequence $\sequence {y_n}_{n \mathop \ge 1}$ converges to $x$.

However, by definition the sequence $\sequence {\map f {y_n} }_{n \mathop \ge 1}$ does not converge to $\map f x$.

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

$\blacksquare$