Proof of Metric Open Sets Form a Topology

Let $\mathcal{T}_d$ be the collection of all subsets of $X$ that are open in the metric space $(X,d)$. We verify the axioms in Topological Space.

First, the empty set belongs to $\mathcal{T}_d$, because the defining condition for openness is vacuous when there is no point $x\in\varnothing$. Also, $X\in\mathcal{T}_d$, because for every $x\in X$ the open ball Open Ball in a Metric Space $B_d(x,1)$ satisfies

So $\varnothing$ and $X$ belong to $\mathcal{T}_d$.

Next, let $A$ be a set, and let $(U_a)_{a\in A}$ be a family of members of $\mathcal{T}_d$. Set

To show that $U\in\mathcal{T}_d$, let $x\in U$. Then there exists $a\in A$ such that $x\in U_a$. Since $U_a$ is open in $(X,d)$, there exists a real number $r>0$ such that

Hence $B_d(x,r)\subseteq U$. Therefore $U$ is open in $(X,d)$, so $U\in\mathcal{T}_d$.

Finally, let $n\in\mathbb{N}$ and let $U_1,\dots,U_n\in\mathcal{T}_d$. Set

If $x\in W$, then $x\in U_i$ for every $i\in\{1,\dots,n\}$. Since each $U_i$ is open in $(X,d)$, for each such $i$ there exists $r_i>0$ such that

Let $r=\min\{r_1,\dots,r_n\}$. Then $r>0$, and if $y\in B_d(x,r)$, one has $d(x,y)<r\le r_i$ for every $i$, so $y\in U_i$ for every $i$. Thus

Hence $W$ is open in $(X,d)$, so $W\in\mathcal{T}_d$.

We have shown that $\mathcal{T}_d$ contains $\varnothing$ and $X$, is closed under arbitrary unions, and is closed under finite intersections. Therefore $(X,\mathcal{T}_d)$ is a topological space.