202303071503

Type :Note Tags : Topology

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Normal Spaces

Note

A topological space $X$ is normal if

1. Every singleton is closed in $X$.
2. for each pair of disjoint closed subsets $A,B\subseteq X$, there is a pair of disjoint open subsets $U,V$ of $X$ such that $A\subset U$ and $B\subseteq V$. We then say that $U,V$ separate $A$ and $B$.

Lemma:

Note

Let $X$ be $T_1$, then $X$ is normal iff for every closed subset $A$ of $X$ and every open subset $U$ containing $A$, there is a open set $V$ containing $A$ and contained in $U$ such that $Cl(V)\subseteq U$.

Proof:

Suppose $X$ is normal, then for any $A$ and every open nbhd $U$ of $A$, take $U^c$ to be the closed set $B$, then there is a separation of $A$, $U^c$ $⟹\exists V,W$ s.t. $A\subseteq V,U^c\subseteq W$, this gives $V\subset Cl(V)\subset W^c\subset U$. We are done.

Conversely, given $A,B$ disjoint and closed, take $U=B^c$ and find $V\subset Cl(V)\subset U$. Then $V$ and $(Cl(V){)}^c$ gives the separation.

NOTE: Subspace and product of normal spaces may not be normal.

Lemma:

Note

Closed subspace of a normal space is normal.

Lemma:

Note

Let $Y$ be a topo space that is not normal. Then there is a topo space $X$ containing $Y$ as a subspace such that $X$ satisfies the second condition in the definition of a normal space.

Proof:

Take $X=Y\cup \{\infty \}$, ${\tau }_X={\tau }_Y\cup \{X\}$. This is normal since any non empty closed subset of $X$ contains $\infty$, hence they are not disjoint, hence it vacuously satisfies the 2nd condition for a normal space.

Lemma:

Note

Every regular 2nd countable space is normal.

Suppose $X$ is the space with $\mathcal{B}$ as the countable basis. Let $C,D$ be the closed disjoint subsets we wish to separate.

Then for any point $x\in C$, construct a separation $U_x,V_x$ of $x,D$. Since $\mathcal{B}$ is a basis, there exists some $U\in \mathcal{B}$ such that $x\in U\subseteq U_x$. Take the union of such $U^{\prime }s$, since $\mathcal{B}$ is countable, there is a countable collection of them, let it be $\{U_n{\}}_{n\in \mathbb{N}}$. Then $U:={\bigcup }_{n\in \mathbb{N}}{U}_n$ is an open set covering $C$, and not intersecting $D$. Similarly get $\{V_n{\}}_{n\in \mathbb{N}}$ such that $V:={\bigcup }_{n\in \mathbb{N}}{V}_n$ is a similar cover for $D$.

The problem is that $U,V$ may intersect. To combat that, consider $U_n^{\prime }:=U_n\setminus (\bigcup Cl(V_n))$ and $V_n^{\prime }:=V_n\setminus (\bigcup Cl(U_n))$.

Check that $C\subset U:={\bigcup }_{n\in \mathbb{N}}{U}_n^{\prime },D\subset V:={\bigcup }_{n\in \mathbb{N}}{V}_n^{\prime }$ and that $U,V$ are disjoint.

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Examples

1. ${\mathbb{R}}_{\ell }$. It is $T_1$, for any element $a$ of $A$, choose $[a,x_a)$ such that it doesn’t intersect $B$, similarly do for all elements of $B$. Then $U:={\bigcup }_{a\in A}[a,x_a)$ and $V:={\bigcup }_{b\in B}[b,x_b)$ forms a separation.
2. ${\mathbb{R}}_{\ell }^2$ is not normal.
   • Take the diagonal, $y=-x$ this is a closed discrete subspace of ${\mathbb{R}}_{\ell }^2$. Then the subset $A:=\{(x,-x)|x\in \mathbb{Q}\}$ is closed in the parent space, similarly $B:=\{(x,-x)|x\notin \mathbb{Q}\}$ is also closed in the parent space.
   • Take a separation $U,V$ of $A,B$.
   • Let $K_n$ be the set of irrationals $x\in \mathbb{R}$ such that $\left[x,x+\frac{1}{n}\right)\times [-x,-x+\frac{1}{n})\subseteq V$.
   • Then $\mathbb{R}$ is a union countably many one point sets ($\mathbb{Q}$) and $K_n^{\prime }s$.
   • By baire category, ${\bar{K}}_n$ has an interval $(a,b)$.
   • Then the parallelogram $(x,-x+ϵ)$ for which $a<x<b;0<ϵ<\frac{1}{n}$ is contained in $V$.
   • Then any rational point on the diagonal in this interval is a limit point of $V$. Contradiction.
3. ${\bar{S}}_{\Omega }\times {\bar{S}}_{\Omega }$ is normal since it is Compact and Hausdorff.

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Related Results

1. Metrizable $⟹$ Normal
2. Regular, Lindelof $⟹$ Normal
3. Regular, 2nd countable $⟹$ Normal
4. Compact, Hausdorff $⟹$ Normal
5. Every well ordered set is normal in the order topology

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Related Problems

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References

Metrizable Spaces Regular Spaces Lindelof Space Second Countability Compactness Hausdorff Property Order Topology Well Ordering S_omega