Proof of Characterization of the Closure in a Metric Space by Open Balls

Condition 2 is equivalent to condition 3. By the definition of the open ball, for a real number $\varepsilon>0$ the set $B_d(x,\varepsilon)$ consists of the points $y\in X$ with $d(x,y)<\varepsilon$. Hence a point belongs to $B_d(x,\varepsilon)\cap A$ if and only if it is a point $a\in A$ with $d(x,a)<\varepsilon$; here every $a\in A$ lies in $X$ because $A\subseteq X$. So for each real $\varepsilon>0$ the statement $B_d(x,\varepsilon)\cap A\neq\varnothing$ holds if and only if there is $a\in A$ with $d(x,a)<\varepsilon$, and the two conditions are equivalent.

Condition 1 implies condition 2. Assume $x\in\operatorname{cl}_X(A)$ and let $\varepsilon>0$ be a real number. By Open Ball in a Metric Space is Open the set $B_d(x,\varepsilon)$ is open in $(X,d)$, that is, $B_d(x,\varepsilon)\in\mathcal{T}_d$. By condition 2 of the definition of a metric we have $d(x,x)=0$, and $0<\varepsilon$, so $d(x,x)<\varepsilon$ and therefore $x\in B_d(x,\varepsilon)$. Applying the definition of the closure to the open set $B_d(x,\varepsilon)$, which contains $x$, gives $B_d(x,\varepsilon)\cap A\neq\varnothing$.

Condition 2 implies condition 1. Assume condition 2 and let $U\in\mathcal{T}_d$ with $x\in U$. Since $U$ is open in $(X,d)$, there is a real number $r>0$ with $B_d(x,r)\subseteq U$. Applying condition 2 with $\varepsilon=r$ produces a point $y\in B_d(x,r)\cap A$, and then $y\in U\cap A$, so $U\cap A\neq\varnothing$. As $U$ was an arbitrary member of $\mathcal{T}_d$ containing $x$, and $x\in X$, this gives $x\in\operatorname{cl}_X(A)$.