Lemma 1

(a) Obviously true since $d(x,x)=0<r$

(b) Let $x\in \mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(E)$ then we know $\exists <!--\; FUNCTION\; APPLICATION\; -->r>0$ such that $B(x,r)\subseteq E$, but from (a) we know $x\in B(x,r)$ and hence $x\in E$. Therefore $\mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(E)\subseteq E$.

(c) (i) Assume that the intersection is not empty and let $x\in \mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(E)\subseteq E$ (by (b)) and $x\in \mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)$. Then by definition $\exists <!--\; FUNCTION\; APPLICATION\; -->r>0$ such that $B(x,r)\subseteq E$ and $\exists <!--\; FUNCTION\; APPLICATION\; -->r^{\prime }>0$ such that $B\left(x,r^{\prime }\right)\cap E=\varnothing$. However, $x\in B\left(x,r^{\prime }\right)$ and $x\in E$, a contradiction. Thus $\mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(E)\cap \mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)=\varnothing$

(ii), (iii), and (iv) are all true by definition.

(d)

Hence, $\mathit{\partial E}=\partial (X\setminus E)$.

(e) First note that $X\setminus Ē=X\setminus (\mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(E)\cup \mathit{\partial E})=\mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)$ (by definition, and Lemma (c:iv)).

Let $x\in \mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(X\setminus E)$ then $\exists <!--\; FUNCTION\; APPLICATION\; -->r>0$ such that $B(x,r)\subseteq \mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(X\setminus E)\subseteq X\setminus E$.

Now we will show that $x\in \mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)$ by showing that $\exists <!--\; FUNCTION\; APPLICATION\; -->r^{\prime }>0$ such that $B\left(x,r^{\prime }\right)\cap E=\varnothing$.

Let $r^{\prime }=r$. Then if $y\in B\left(x,r^{\prime }\right)=B(x,r)\subseteq X\setminus E$ we get that $y\notin E$. Hence $B\left(x,r^{\prime }\right)\cap E=\varnothing$. Thus $y\in \mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)$. Hence, $\mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(X\setminus E)\subseteq \mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)$. Let $x\in \mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)$. Then $\exists <!--\; FUNCTION\; APPLICATION\; -->r>0$ such that $B(x,r)\cap E=\varnothing$. But since $x\in B(x,r)$ we must have that $x\notin E$. Thus $x\in X\setminus E$. Hence $\mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)\subseteq \mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(X\setminus E)$.

Thus $\mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)=\mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(X\setminus E)$, but as we noted earlier: $X\setminus Ē=X\setminus (\mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(E)\cup \mathit{\partial E})=\mathrm{Ext}<!--\; FUNCTION\; APPLICATION\; -->(E)$.

Therefore we have $\mathrm{Int}<!--\; FUNCTION\; APPLICATION\; -->(X\setminus E)=X\setminus Ē$.