T_{0} - Space Equivalent Definition

[patric44]

Homework Statement

show that T_{0} - space iff the derived set of every singleton is a union of closed sets.

Relevant Equations

T_{0} - space iff {x}^{'} is a union of closed sets.

Hello everyone,
Concerning the separation axioms in topology. Our topology professor introduced the equivalent definition for a topological space to be a $T_{o}-space$ as:
$$
(X,\tau)\ is\ a\ T_{o}-space\ iff\ \forall\ x\ \in X,\ \{x\}^{\prime}\ is\ a\ union\ of\ closed\ sets.
$$
The direction $\implies$ follows from the result $\bar{\{x\}}\neq\bar{\{y\}}$ for every distinct elements in $T_{o}-space$: Let $z\in \{x\}^{\prime}\implies z\neq x\implies x\notin\bar{\{z\}}$, since
$$
{z}\subseteq{x}^{\prime}\subseteq\bar{\{x\}}\implies \bar{\{z\}}\subseteq\bar{\{x\}}.
$$
Then
$$
\bar{\{z\}}=\bar{\{z\}}-\{x\}\subseteq\bar{\{x\}}-\{x\}=\{x\}^{\prime},
$$
thus $z\in\bar{\{z\}}\subseteq\{x\}^{\prime}$, i.e., $\{x\}^{\prime}$ can be written as a union of closed sets. Unfortunately, the other direction is not that clear. Will appreciate any suggestions.

 

[fishturtle1]

Sorry for the really dumb question, but is there an example where we can't replace the statement with: $(X, \tau)$ is $T_0$ if and only if for all $x \in X$, $\lbrace x \rbrace'$ is empty? (I know $\emptyset$ is closed but...)

Definition: $(X,\tau)$ is $T_0$ if for any distinct points $x,y \in X$, we can find an open set $U$ such that $x \in U$ and $y \notin U$.

So, if $y\neq x$ was a limit point of $\lbrace x \rbrace$, then by definition of $T_0$, there is an open set $U$such that $y \in U$ and $(U - \lbrace y \rbrace) \cap \lbrace x \rbrace = \emptyset$. And this shows $y$ is not a limit point of $\lbrace x \rbrace$.

And for any open set $U$ containing $x$, we have $(U - \lbrace x \rbrace) \cap \lbrace x \rbrace = \emptyset$, which shows $x$ is not a limit point of $\lbrace x \rbrace$.

So $\lbrace x \rbrace$ has no limit points i.e., $\lbrace x \rbrace'= \emptyset$ ?

 

[fishturtle1]

Sorry for the really dumb question, but is there an example where we can't replace the statement with: $(X, \tau)$ is $T_0$ if and only if for all $x \in X$, $\lbrace x \rbrace'$ is empty? (I know $\emptyset$ is closed but...)

Definition: $(X,\tau)$ is $T_0$ if for any distinct points $x,y \in X$, we can find an open set $U$ such that $x \in U$ and $y \notin U$.

So, if $y\neq x$ was a limit point of $\lbrace x \rbrace$, then by definition of $T_0$, there is an open set $U$such that $y \in U$ and $(U - \lbrace y \rbrace) \cap \lbrace x \rbrace = \emptyset$. And this shows $y$ is not a limit point of $\lbrace x \rbrace$.

And for any open set $U$ containing $x$, we have $(U - \lbrace x \rbrace) \cap \lbrace x \rbrace = \emptyset$, which shows $x$ is not a limit point of $\lbrace x \rbrace$.

So $\lbrace x \rbrace$ has no limit points i.e., $\lbrace x \rbrace'= \emptyset$ ?

I think I got the definition wrong. It should be (i think)

$(X,\tau)$ is $T_0$ if for any distinct points $x,y \in X$, we can find an open set $U$ such that $x \in U$ and $y \notin U$ or $x \notin U$ and $y \in U$.

So in post #2, we can't guarantee there is an open set containing $y$ and not containing $x$, which I assumed was possible... sorry. For the problem in OP, I think we can break it into cases:

Proof: $(\Longleftarrow):$ Let $x, y$ be distinct points in $X$ and consider two cases.

Case 1: $x \in \lbrace y \rbrace'$. Then $x \in C$ for some closed set $C \subset \lbrace y \rbrace'$. Since $y$ is not a limit point of $\lbrace y \rbrace$ (because $(X - \lbrace y \rbrace) \cap \lbrace y \rbrace = \emptyset)$, we have that $y \notin C$. So, $y \in X - C$ and $x \notin X - C$ where $X - C$ is open.

Case 2: $x \notin \lbrace y \rbrace'$. Here we show $x \notin \overline{\lbrace y \rbrace}$ and then use a similar argument as in case 1.