Proof of Continuity Between Metric Spaces is Equivalent to Sequential Continuity

Conventions. For $y,z\in A$ one has $d_{A}(y,z)=d_{X}(y,z)$, and $d_{X}(y,z)=d_{X}(z,y)$ by the symmetry axiom of a metric; both are used without further comment. Elementary order facts are those of Elementary Order Arithmetic in an Ordered Field, whose claim 1 concerns the strict order only, the non-strict law being an axiom of the ordered field $\mathbb{R}$, whose order $\le$ is a total order, so that its reflexivity, antisymmetry, transitivity and totality are axioms of that definition. We write $\iota$ for the canonical map of $\mathbb{N}$ into $\mathbb{R}$ whose properties are recorded in Properties of the Canonical Map from the Natural Numbers to an Ordered Field. The proof of claim 2 selects one point of $A$ for each natural number and thereby uses the axiom of countable choice; it is used nowhere else.

A remark on negated strict inequalities. If $a,b\in\mathbb{R}$ and $a<b$ fails, then $b\le a$. Indeed the order of $\mathbb{R}$ is total, so $a\le b$ or $b\le a$; in the first case $a\ne b$ would give $a<b$, so $a=b$ and $b\le a$ by reflexivity.

Proof of claim 1. Let $(y_{j})_{j\in\mathbb{N}}$ be a sequence in $A$ converging to $x$ in $(A,d_{A})$ and let $\varepsilon\in\mathbb{R}$ be positive. By continuity of $f$ at $x$ relative to $A$ there is a positive $\delta\in\mathbb{R}$ such that every $y\in A$ with $d_{X}(x,y)<\delta$ satisfies $d_{Y}(f(y),f(x))<\varepsilon$. By convergence there is $N\in\mathbb{N}$ such that $d_{A}(y_{j},x)<\delta$ for every $j\in\mathbb{N}$ with $j\ge N$. For such $j$ we have $d_{X}(x,y_{j})=d_{A}(y_{j},x)<\delta$ and therefore $d_{Y}(f(y_{j}),f(x))<\varepsilon$. Since $\varepsilon$ was an arbitrary positive real, $\bigl(f(y_{j})\bigr)_{j\in\mathbb{N}}$ converges to $f(x)$ in $(Y,d_{Y})$.

Proof of claim 2. We prove the contrapositive: if $f$ is not continuous at $x$ relative to $A$, then some sequence in $A$ converges to $x$ in $(A,d_{A})$ while its image sequence does not converge to $f(x)$.

So assume $f$ is not continuous at $x$ relative to $A$. Then there is a positive $\varepsilon_{0}\in\mathbb{R}$ such that for every positive $\delta\in\mathbb{R}$ there is $y\in A$ with $d_{X}(x,y)<\delta$ for which $d_{Y}(f(y),f(x))<\varepsilon_{0}$ fails, hence, by the remark above, with $\varepsilon_{0}\le d_{Y}(f(y),f(x))$.

Let $j\in\mathbb{N}$. By claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field the real number $\iota(j)$ is positive, so its multiplicative inverse $\iota(j)^{-1}$ is positive by claim 7 of Elementary Order Arithmetic in an Ordered Field. Applying the previous paragraph with $\delta=\iota(j)^{-1}$ and choosing one such point for each $j$, the axiom of countable choice yields a sequence $(y_{j})_{j\in\mathbb{N}}$ in $A$ with

The sequence $(y_{j})$ converges to $x$ in $(A,d_{A})$. Let $\varepsilon\in\mathbb{R}$ be positive. By claim 3 of The Archimedean Property of the Real Numbers there is $N\in\mathbb{N}$ with $\iota(N)^{-1}<\varepsilon$. Let $j\in\mathbb{N}$ satisfy $j\ge N$. If $j=N$ then $\iota(j)^{-1}=\iota(N)^{-1}$. If $N<j$ then $\iota(N)<\iota(j)$ by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, hence $\iota(N)\le\iota(j)$; the real number $\iota(N)^{-1}\iota(j)^{-1}$ is positive, being a product of two positive reals by claim 5 of Elementary Order Arithmetic in an Ordered Field, so multiplying the inequality $\iota(N)\le\iota(j)$ by it and using claim 5 of Elementary Arithmetic in an Ordered Field together with the field axioms $\iota(N)\iota(N)^{-1}=1$ and $\iota(j)\iota(j)^{-1}=1$ gives $\iota(j)^{-1}\le\iota(N)^{-1}$. In both cases

so $d_{A}(y_{j},x)<\varepsilon$ by the mixed transitivity of claim 2 of Elementary Order Arithmetic in an Ordered Field. Hence $(y_{j})$ converges to $x$ in $(A,d_{A})$.

The image sequence does not converge to $f(x)$. If it did, then applying the definition of convergence with the positive real $\varepsilon_{0}$ would give some $j\in\mathbb{N}$ with $d_{Y}(f(y_{j}),f(x))<\varepsilon_{0}$; combined with $\varepsilon_{0}\le d_{Y}(f(y_{j}),f(x))$ and the mixed transitivity of claim 2 of Elementary Order Arithmetic in an Ordered Field this gives $\varepsilon_{0}<\varepsilon_{0}$, contradicting the irreflexivity of the strict order. This proves the contrapositive, and with it claim 2.

Proof of claim 3. By the definition of continuity on $A$, $f$ is continuous on $A$ if and only if it is continuous at $x'$ relative to $A$ for every $x'\in A$. Claim 1 applied at each $x'$ gives the forward implication and claim 2 applied at each $x'$ the converse. $\blacksquare$