Proof of The Open Subsets of a Metric Space Form a Topology

Every element of $\mathcal{T}_{d}$ is an element of $\mathcal{P}(X)$, so $\mathcal{T}_{d}$ is a subset of $\mathcal{P}(X)$ by Subclasses and Subsets §subclass; and by The Union Set and the Power Set of a Set §power, a set $U$ lies in $\mathcal{T}_{d}$ if and only if $U\subseteq X$ and $U$ is open in $(X,d)$. We verify the three conditions of Topological Space §topology.

Condition empty-whole. The empty set $\emptyset$ has no element, so $\emptyset\subseteq X$, and the condition of Open Subset of a Metric Space §open holds vacuously for it; thus $\emptyset\in\mathcal{T}_{d}$. Next, $X\subseteq X$. For $x\in X$, the real number $1$ satisfies $0<1$ by Rules of Arithmetic and Order in an Ordered Field §squares, and $B_{d}(x,1)$ is a subset of $X$ by Open Ball in a Metric Space §ball; so $X$ is open in $(X,d)$ by Open Subset of a Metric Space §open, and $X\in\mathcal{T}_{d}$.

Condition unions. Let $\mathcal{S}$ be a subset of $\mathcal{T}_{d}$ and put $W=\bigcup\mathcal{S}$, a set by The Union Set and the Power Set of a Set §union, whose elements are the sets $u$ with $u\in V$ for some $V\in\mathcal{S}$, by the same clause. Every $V\in\mathcal{S}$ lies in $\mathcal{T}_{d}$, so $V\subseteq X$; hence $W\subseteq X$. Let $x\in W$, and choose $V\in\mathcal{S}$ with $x\in V$. As $V$ is open in $(X,d)$, Open Subset of a Metric Space §open gives a real number $r>0$ with $B_{d}(x,r)\subseteq V$. Every element of $B_{d}(x,r)$ then lies in $V$, which lies in $\mathcal{S}$, so it lies in $W$; thus $B_{d}(x,r)\subseteq W$. Hence $W$ is open in $(X,d)$ and $W\in\mathcal{T}_{d}$.

Condition intersections. Let $U,V\in\mathcal{T}_{d}$. By The Boolean Operations on Classes, Disjointness, and the Universal Class §operations, the elements of $U\cap V$ are the sets lying in both $U$ and $V$, so $U\cap V\subseteq U\subseteq X$; and $U\cap V$ is a set by Sets and Maps: Ordinary Notation §sets. Let $x\in U\cap V$. As $U$ and $V$ are open in $(X,d)$, Open Subset of a Metric Space §open gives real numbers $r_{1}>0$ and $r_{2}>0$ with $B_{d}(x,r_{1})\subseteq U$ and $B_{d}(x,r_{2})\subseteq V$. The order of $\mathbb{R}$ is a total order, $\mathbb{R}$ being an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals and Ordered Fields §ordered-field; let $r=\min\{r_{1},r_{2}\}$ be the least element of $\{r_{1},r_{2}\}$ given by The Maximum and Minimum of Two Elements of a Total Order §exists, unique by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §least-unique. By totality $r_{1}\le r_{2}$ or $r_{2}\le r_{1}$, so $r=r_{1}$ or $r=r_{2}$ by The Maximum and Minimum of Two Elements of a Total Order §values; in either case $r>0$. Let $y\in B_{d}(x,r)$; then $y\in X$ and $d(x,y)<r$ by Open Ball in a Metric Space §membership. By The Maximum and Minimum of Two Elements of a Total Order §bounds, $r\le r_{1}$ and $r\le r_{2}$, so Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed gives $d(x,y)<r_{1}$ and $d(x,y)<r_{2}$. By Open Ball in a Metric Space §membership, $y\in B_{d}(x,r_{1})\subseteq U$ and $y\in B_{d}(x,r_{2})\subseteq V$, so $y\in U\cap V$. Thus $B_{d}(x,r)\subseteq U\cap V$, so $U\cap V$ is open in $(X,d)$ and $U\cap V\in\mathcal{T}_{d}$.

Hence $\mathcal{T}_{d}$ is a topology on $X$ by Topological Space §topology, and $(X,\mathcal{T}_{d})$ is a topological space by Topological Space §topological-space.