3.6 Hausdorff
Definition 3.6.1 (Hausdorff).label Let $X$ be a topological space, then the following are equivalent:
- (H)
For any $x\neq y\in X$, there exists $U\in\cali{N}(x),V\in\cali{N}(y)$ such that $U\cap V=\emp$.
- (H2)
For any $x\in X$, $\caps{}{U\in\cali{N}(x)}\cl{U}=\curl{x}$.
- (H3)
For any $\scr{F}\suf\cali{P}(X)$ filter converging to $x\in X$, $x$ is the only cluster point of $\scr{F}$.
- (H4)
For any $\scr{F}\suf\cali{P}(X)$, $\scr{F}$ converges to at most one point.
- (H5)
For any index set $I$, the diagonal $\la$ is closed in $X^{I}\define \prods{}{i\in I}X$.
- (H6)
The diagonal $\la$ is closed in $X\ti X$.
If the above holds, then $X$ is a Hausdorff/(T2) space.
Proof. (H)$\implies$(H1): If $x\ne y$ then $\exists U,V$ open neighbourhoods of $x,y$ resp such that $U\cap V=\emp$; hence $y\nin\cl{U}$
(H1)$\implies$(H2): Let $y\ne x$ then there exists a closed nbd $V$ of $x$ such that $y\nin V$, and by hypothesis there exists $M\in\frak{F}$ such that $M\suf V$; thus $M\cap V^{c}=\emp$. But $V^{c}$ is a nbd of $y$; hence $y$ is not a cluster point of $\frak{F}$
(H2)$\implies$(H3): Clear, since every limit point of a filter is also a cluster point.
(H3)$\implies$(H4): Suppose the diagonal $\la$ is not closed then $\exists (x,y)\in \cl{\la}\del \la$ such that $x\ne y$. Since $(x,y)\in\cl{\la}$ every basis element $U\ti V$ of $(x,y)$ meets $\la$ but $(U\ti V)\cap \la\ne\emp\iff U\cap V\ne \emp$. Now consider $\frak{B}=\curl{U\cap V:U\in\cali{N}(x),V\in\cali{N}(y)}$ we see from above that $\frak{B}$ is a non-empty filter base generating some filter $\frak{F}$. It must be that $U\cap V\in \frak{F}$ for fixed $U$ and any $V$ thus $U\in \frak{F}$ so $\cali{N}(x)\suf \frak{F}$ that is to say $\frak{F}\to x$. Similarly $\frak{F}\to y$, a contradiction.
(H4)$\implies$(H5): Obvious
(5)$\implies$(H): If $x\ne y$ then $(x,y)\in X\ti X$ is not in the diagonal $\la$, hence there is a neighbourhood $V$ of $x$ and a neighbourhood $W$ of $y$ in $X$ such that $(V\ti W)\cap \la=\emp$, which means that $V\cap W=\emp$.$\square$
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 $A\inj X$.
$\square$
Proposition 3.6.3.label Let $\curl{X_i}_{i\in I}$ be non-empty topological spaces, then $X=\prods{}{i\in I}X_{i}$ is Hausdorff if and only if $X_{i}$ is Hausdorff for each $i\in I$.
Proof. Suppose $X_{i}$ is Hausdorff for each $i\in I$. Let $x\neq y\in X$ then $\pi_{i}(x)\neq \pi_{i}(y)$ for some $i\in i$ so $X$ is Hausdorff by Proposition 3.6.2(1). Conversely suppose $X$ is Hausdorff. Since the canonical injection $X_{i}\to X$ is a homeomoprhism for each $i\in I$, $X_{i}$ is Hausdorff by Proposition 3.6.2(2).$\square$
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