Claim: Open Ball is an Open Set
Let \((X,d)\) be a metric space. Then every open ball is open.
Proof. Fix \(x\in X\), \(r>0\), and let \(U=B_r(x)\).
If \(u\in U\), set \(\varepsilon=r-d(u,x)>0\). For any \(y\in B_\varepsilon(u)\),
the triangle inequality gives \(d(y,x)\le d(y,u)+d(u,x)<\varepsilon+d(u,x)=r\);
hence \(y\in B_r(x)\). Thus about each \(u\in U\) there is a ball contained in \(U\),
so \(U\) is open.
Proposition: Properties of the Induced Topology
Let \((X,d)\) be a metric space and \(\mathcal T\) its family of open sets. Then \(\emptyset,X\in\mathcal T\); \(\mathcal T\) is closed under arbitrary unions and finite intersections.
- \(\emptyset,X\in\mathcal T\). Trivial and vacuous.
- Arbitrary unions. If \(x\in\bigcup_i U_i\), pick \(U_j\ni x\). Since \(U_j\) is open, some ball about \(x\) lies in \(U_j\subseteq\bigcup_i U_i\).
- Finite intersections. If \(x\in U\cap V\), pick radii \(r,s\) with \(B_r(x)\subseteq U\), \(B_s(x)\subseteq V\). Then \(B_{\min\{r,s\}}(x)\subseteq U\cap V\).
Proposition: Metric Spaces are Hausdorff
If \(x\neq y\), let \(\delta=d(x,y)\), and set \(U=B_{\delta/2}(x)\), \(V=B_{\delta/2}(y)\). These are disjoint by the triangle inequality.
Definition: Interior of a Set
For \(E\subseteq X\), the interior is \(E^\circ=\{x\in E:\exists r>0\ \text{with}\ B_r(x)\subseteq E\}\), i.e., the largest open set contained in \(E\).