Proof of Continuity Between Metric Spaces is Equivalent to Sequential Continuity

Throughout, $d_{X}$ and $d_{Y}$ are symmetric by Metric Space §symmetric, and for a sequence $(y_{n})$ in $X$ with $y_{n}\in A$ for every $n\in\mathbb{N}$, $(f(y_{n}))$ is the sequence $n\mapsto f(y_{n})$ in $Y$ given by Sets and Maps: Ordinary Notation §maps. The clauses sequential and criterion are proved for an arbitrary $x\in A$.

Clause sequential. Assume $f$ is continuous at $x$ relative to $A$, and let $(y_{n})$ be a sequence in $X$ with $y_{n}\in A$ for every $n\in\mathbb{N}$ that converges to $x$ in $(X,d_{X})$. Let $\varepsilon>0$ be real. By Continuous Map Between Metric Spaces §continuous-at there is a real $\delta>0$ such that every $y\in A$ with $d_{X}(x,y)<\delta$ satisfies $d_{Y}(f(y),f(x))<\varepsilon$. Then, $\delta$ being chosen, Convergent Sequence in a Metric Space §converges applied with $\delta$ in place of $\varepsilon$ gives $N\in\mathbb{N}$ with $d_{X}(y_{n},x)<\delta$ for every $n\in\mathbb{N}$ with $n\ge N$. For such $n$, $d_{X}(x,y_{n})<\delta$ and $y_{n}\in A$, so $d_{Y}(f(y_{n}),f(x))<\varepsilon$. As $\varepsilon$ was arbitrary, $(f(y_{n}))$ converges to $f(x)$ in $(Y,d_{Y})$ by Convergent Sequence in a Metric Space §converges.

Clause criterion. We prove the contrapositive. Assume $f$ is not continuous at $x$ relative to $A$. Negating that definition, there is a real $\varepsilon_{0}>0$ such that for every real $\delta>0$ some $y\in A$ satisfies $d_{X}(x,y)<\delta$ while $d_{Y}(f(y),f(x))<\varepsilon_{0}$ fails; fix such an $\varepsilon_{0}$.

For $n\in\mathbb{N}$, read in $\mathbb{R}$ as in The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §identification, $0<n$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §naturals, so $n\neq0$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization and $0<n^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal. Let

$$A_{n}=\{y\in A:d_{X}(x,y)<n^{-1}\text{ and }d_{Y}(f(y),f(x))<\varepsilon_{0}\text{ fails}\},$$

a subset of $A$ by Sets and Maps: Ordinary Notation §set-builder, which has at least one element by the choice of $\varepsilon_{0}$ applied with $\delta=n^{-1}$. The family $(A_{n})_{n\in\mathbb{N}}$ is the map $n\mapsto A_{n}$ from $\mathbb{N}$ to $\mathcal{P}(A)$, each $A_{n}$ lying in $\mathcal{P}(A)$ by The Union Set and the Power Set of a Set §power, given by Sets and Maps: Ordinary Notation §maps. By the assumed axiom of dependent choice, Countable Choice Follows from Dependent Choice §countable-choice with $b=A$ gives a map $\mathbb{N}\to A$, written $(y_{n})_{n\in\mathbb{N}}$, with $y_{n}\in A_{n}$ for every $n\in\mathbb{N}$; this is the only use of choice in the proof. Its range is a subset of $A\subseteq X$, so it is a map $\mathbb{N}\to X$ by Functions, Values of a Function, and Functions from One Class to Another §map, that is, a sequence in $X$ with $y_{n}\in A$ for every $n\in\mathbb{N}$.

The sequence $(y_{n})$ converges to $x$ in $(X,d_{X})$. Let $\varepsilon>0$ be real. Then $\varepsilon\neq0$ and $0<\varepsilon^{-1}$ as above, and The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §archimedean gives $N\in\mathbb{N}$ with $\varepsilon^{-1}<N$. By Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal, $N^{-1}<(\varepsilon^{-1})^{-1}$, and $(\varepsilon^{-1})^{-1}=\varepsilon$ by Rules of Arithmetic and Order in an Ordered Field §reciprocals; so $N^{-1}<\varepsilon$. Let $n\in\mathbb{N}$ with $n\ge N$. The inequality $N\le n$ holds in $\mathbb{R}$ as well by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §agreement, and $0<N$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §naturals, so $n^{-1}\le N^{-1}$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §reciprocal-order. As $y_{n}\in A_{n}$, $d_{X}(y_{n},x)=d_{X}(x,y_{n})<n^{-1}$. From $d_{X}(y_{n},x)<n^{-1}\le N^{-1}$, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed gives $d_{X}(y_{n},x)<N^{-1}$, and with $N^{-1}<\varepsilon$, Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive gives $d_{X}(y_{n},x)<\varepsilon$. Hence $(y_{n})$ converges to $x$ by Convergent Sequence in a Metric Space §converges.

The sequence $(f(y_{n}))$ does not converge to $f(x)$ in $(Y,d_{Y})$. If it did, Convergent Sequence in a Metric Space §converges applied with $\varepsilon_{0}$ would give $N\in\mathbb{N}$ with $d_{Y}(f(y_{n}),f(x))<\varepsilon_{0}$ for every $n\ge N$, in particular for $n=N$, contradicting $y_{N}\in A_{N}$. This proves the contrapositive, and with it the clause criterion.

Clause on-subset. By Continuous Map Between Metric Spaces §continuous, $f$ is continuous on $A$ if and only if it is continuous at $x'$ relative to $A$ for every $x'\in A$. The clause sequential, applied at each $x'\in A$ in place of $x$, gives the forward implication, and the clause criterion, applied likewise, gives the converse.