Proposition 3.6.2.label Let $X$ be a topological space then
- (1)
$X$ is Hausdorff if and only if for any $x\neq y\in X$ there exists a Hausdorff $X'$ and $f\in C(X;X')$ such that $f(x)\neq f(y)$.
- (2)
Hausdorff is hereditary i.e. if $X$ is a Hausdorff space and then $A\suf X$ is Hausdorff with respect to subspace topology.
Proof.
- (1)
Suppose $X$ is Hausdorff then we can take $X'=X$ and $f=\indi{X}$. Conversely since $X'$ is Hausdorff and $x\neq y$ there exists $U'\in \cali{N}(f(x)),V'\in\cali{N}(f(y))$ such that $U'\cap V'=\emp$. By continuity of $f$, $U\define f\inv (U'),V\define f\inv (V')$ are open. Since $U'\cap V'=\emp$ we have $U\cap V=f\inv(U'\cap V')=f\inv(\emp)=\emp$.
- (2)
Apply (1) with the canonical injection $A\inj X$.
$\square$
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