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 $x \in X$.

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

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 of Theorem
We have that a Continuous Function 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 $d(x,y) < \delta$ and $e\left(f(x),f(y)\right) \geq \epsilon_0$.

For $n \geq 1$, define $\delta_n = 1/n$.

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

Therefore, by definition the sequence $\left( y_n \right)_{n \geq 1}$ converges to $x$.

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

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

Proof of Corollary
This follows immediately from the definitions:
 * 1) A function is sequentially continuous everywhere in $X$ if and only if it is sequentially continuous at each point
 * 2) A function is continuous everywhere in $X$ if and only if it is continuous at each point