Proof. Let \(x,y\in X\), where \(x\neq y\). Let \(\epsilon =\dfrac {1}{2}d(x,y)\). Then the open balls \(B(x,\epsilon )\) and \(B(y,\epsilon )\) of radius \(\epsilon \) centered at \(x\) and \(y\) are open sets. If
\(B(x,\epsilon )\cap B(y,\epsilon )\neq \emptyset \), then there is a \(z\in X\) satisfying \(d(x,z)<\epsilon \) and \(d(y,z)<\epsilon \). From the triangle inequality, this implies that
\(d(x,y)\leq d(x,z)+d(z,y)<\epsilon +\epsilon =2\epsilon \). This contradicts the fact that \(\epsilon =\dfrac {1}{2}d(x,y)\). Thus \(B(x,\epsilon )\cap B(y,\epsilon )=\emptyset \). Since \(x\in B(x,\epsilon )\) and \(y\in B(y,\epsilon )\), it follows that the metric space \(X\) is a Hausdorff space.

□

Proof.

i).

Let \(X\) be a Hausdorff space and \(Y\) be a subspace of \(X\). Let \(x,y\in Y\), with \(x\neq y\). Then \(x,y\in X\), since \(Y\subseteq X\). Since \(X\) is a Hausdorff space, there exists open sets U and V such that \(x\in U\), \(y\in V\) and \(U\cap V=\emptyset \). Now \(U\cap Y\) and \(V\cap Y\) are both open sets in Y. Moreover, \(x\in U\cap Y\) and \(y\in V\cap Y\). Also \((U\cap Y)\cap (V\cap Y)=(U\cap V)\cap Y\). Since \(U\cap V=\emptyset \), it follows that \((U\cap Y)\cap (V\cap Y)=\emptyset \). Thus Y is a Hausdorff space.


 

ii).

Let \(X\) and \(Y\) be Hausdorff spaces and let \((a,b),(c,d)\in X\times Y\), with \((a,b)\neq (c,d)\). Then \(a,c\in X\) and \(b,d\in Y\). Since X and Y are Hausdorff spaces, there exists open sets \(U,V\subseteq X\) and \(P,Q\subseteq Y\) such that \(a\in U\), \(c\in V\), \(b\in P\) and \(d\in Q\), with \(U\cap V=\emptyset \) and \(P\cap Q=\emptyset \).
Now, we have \((a,b)\in U\times P\) and \((c,d)\in V\times Q\). Clearly, \(U\times P\) and \(V\times Q\) are open sets in \(X\times Y\). Since \(U\cap V=\emptyset \) and \(P\cap Q=\emptyset \), we have that \((U\times P)\cap (V\times Q)=(U\cap V)\times (P\cap Q)=\emptyset \). This proves that \(X\times Y\) is Hausdorff.

iii).

Let \(x,y\in X\) with \(x\neq y\). Consider \(f(x)\) and \(f(y)\) in \(Y\). Since \(f\) is injective, \(f(x)\neq f(y)\). Since \(Y\) is Hausdorff there exists open sets \(U\) and \(V\) in \(Y\) such that \(f(x)\in U\), \(f(y)\in V\) and \(U\cap V=\emptyset \).
Moreover, \(x\in f^{-1}(U)\) and \(y\in f^{-1}(V)\). Since \(f\) continuous, \(f^{-1}(U)\) and \(f^{-1}(V)\) are open in \(X\). Thus we have that \(x\in f^{-1}(U)\), \(y\in f^{-1}(V)\) and \(f^{-1}(U)\cap f^{-1}(V)=\emptyset \). This proves that \(X\) is a Hausdorff space.