Continuity Between Metric Spaces is Equivalent to Sequential Continuity

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces, let $A\subseteq X$, let $d_{A}$ be the restriction of $d_{X}$ to $A$, which is a metric on $A$ by claim 1 of The Restriction of a Metric to a Subset Induces the Subspace Topology, and let $\mathbb{N}$ be the set of natural numbers. Let $\mathbb{R}$ be the set of real numbers with the order $\le$ of its ordered field structure, where $a<b$ means that $a\le b$ and $a\ne b$.

Let $f:A\to Y$ and let $x\in A$. Below, a sequence $(y_{j})_{j\in\mathbb{N}}$ in $A$ is said to converge to $x$ in $A$ if it converges to $x$ in the metric space $(A,d_{A})$, and the sequence $\bigl(f(y_{j})\bigr)_{j\in\mathbb{N}}$ in $Y$ is a sequence in $Y$ whose convergence is understood in $(Y,d_{Y})$.

Then the following hold.

1. (Continuity implies sequential continuity)¶ Suppose $f$ is continuous at $x$ relative to $A$. Then for every sequence $(y_{j})_{j\in\mathbb{N}}$ in $A$ converging to $x$ in $A$, the sequence $\bigl(f(y_{j})\bigr)_{j\in\mathbb{N}}$ converges to $f(x)$ in $(Y,d_{Y})$.

2. (Sequential continuity implies continuity)¶ Suppose that for every sequence $(y_{j})_{j\in\mathbb{N}}$ in $A$ converging to $x$ in $A$ the sequence $\bigl(f(y_{j})\bigr)_{j\in\mathbb{N}}$ converges to $f(x)$ in $(Y,d_{Y})$. Then $f$ is continuous at $x$ relative to $A$.

3. (On a subset)¶ $f$ is continuous on $A$, that is continuous at every point of $A$ relative to $A$, if and only if for every $x'\in A$ and every sequence $(y_{j})_{j\in\mathbb{N}}$ in $A$ converging to $x'$ in $A$ the sequence $\bigl(f(y_{j})\bigr)_{j\in\mathbb{N}}$ converges to $f(x')$ in $(Y,d_{Y})$.