Proof of Sequential Characterization of Closed Subsets of a Metric Space

Throughout, $\mathbb{R}$ denotes the real numbers, an ordered field with order $\le$ and additive identity $0$; for real $s,t$ we write $s<t$ to mean $s\le t$ and $s\ne t$. For $x\in X$ and a real number $r>0$, $B_d(x,r)$ denotes the open ball in $(X,d)$ with center $x$ and radius $r$, and $X\setminus A$ denotes the complement of $A$ relative to $X$. By the definition of a closed subset, $A$ is closed in $(X,\mathcal{T}_d)$ precisely when $X\setminus A\in\mathcal{T}_d$, that is, precisely when $X\setminus A$ is open in $(X,d)$.

Necessity. Assume $X\setminus A$ is open in $(X,d)$. Let $(x_m)_{m\in\mathbb{N}}$ be a sequence in $X$ with $x_m\in A$ for every $m\in\mathbb{N}$, and let $x\in X$ be such that $(x_m)_{m\in\mathbb{N}}$ converges to $x$ in $(X,d)$. Suppose, for contradiction, that $x\notin A$. Then $x\in X\setminus A$, so by the definition of openness there is a real number $r>0$ with

Applying the definition of convergence with $\varepsilon=r$ produces $N\in\mathbb{N}$ such that $d(x_m,x)<r$ for every $m\in\mathbb{N}$ with $m\ge N$. The order on $\mathbb{N}$ is reflexive by Properties of the Order on the Natural Numbers, so $N\ge N$ and therefore $d(x_N,x)<r$. Condition 3 in the definition of a metric gives $d(x,x_N)=d(x_N,x)$, hence $d(x,x_N)<r$ and so $x_N\in B_d(x,r)$. Consequently $x_N\in X\setminus A$, that is, $x_N\notin A$, contradicting $x_N\in A$. Therefore $x\in A$, as required.

Sufficiency. We prove the contrapositive: assuming $A$ is not closed, we exhibit a sequence with all terms in $A$ that converges in $(X,d)$ to a point outside $A$.

So assume $X\setminus A$ is not open in $(X,d)$. Negating the definition of openness, there is a point $x\in X\setminus A$ such that for every real number $r>0$ the inclusion $B_d(x,r)\subseteq X\setminus A$ fails, that is, there exists $y\in B_d(x,r)$ with $y\notin X\setminus A$. Since $y\in X$ and $y\notin X\setminus A$, such a $y$ lies in $A$. Hence

By Existence of a Sequence of Positive Real Numbers with Limit Zero there is a sequence $(h_k)_{k\in\mathbb{N}}$ of real numbers with $0<h_k$ for every $k\in\mathbb{N}$ and with limit $0$. For each $k\in\mathbb{N}$ put

which is a subset of $X$, nonempty by the previous paragraph applied to $r=h_k$. Thus $(A_k)_{k\in\mathbb{N}}$ is a family of subsets of $X$ with every member nonempty, so Axiom of Countable Choice yields a sequence $(y_k)_{k\in\mathbb{N}}$ in $X$ with $y_k\in A_k$ for every $k\in\mathbb{N}$. In particular, for every $k\in\mathbb{N}$ we have $y_k\in A$ and, by the definition of the open ball, $d(x,y_k)<h_k$.

We claim $(y_k)_{k\in\mathbb{N}}$ converges to $x$ in $(X,d)$. Let $\varepsilon$ be a real number with $\varepsilon>0$. Since $(h_k)_{k\in\mathbb{N}}$ has limit $0$, there is $N\in\mathbb{N}$ such that $|h_k-0|<\varepsilon$ for every $k\in\mathbb{N}$ with $k\ge N$, the absolute value being that of the ordered field $\mathbb{R}$. By Additive Cancellation and Elementary Additive Identities in a Field we have $h_k-0=h_k$, so $|h_k|<\varepsilon$ for every such $k$. Statement 3 of Properties of the Absolute Value in an Ordered Field gives $h_k\le|h_k|$, and mixed transitivity, statement 2 of Elementary Order Arithmetic in an Ordered Field, then gives $h_k<\varepsilon$.

Fix $k\in\mathbb{N}$ with $k\ge N$. Condition 3 in the definition of a metric gives $d(y_k,x)=d(x,y_k)$, so $d(y_k,x)<h_k$. Since $h_k<\varepsilon$ we have in particular $h_k\le\varepsilon$, and mixed transitivity again gives $d(y_k,x)<\varepsilon$. As $\varepsilon>0$ was arbitrary, $(y_k)_{k\in\mathbb{N}}$ converges to $x$ in $(X,d)$.

Every term of $(y_k)_{k\in\mathbb{N}}$ lies in $A$, the sequence converges to $x$, and $x\in X\setminus A$ so $x\notin A$. Hence the stated sequential condition fails, which proves the contrapositive and completes the proof.