# Prove that every $T_1$-space is $T_0$-space

Exercise: A topological space is said to be $$T_0$$-space if for each pair of distinct points $$a,b$$ in $$X$$, either there exists an open set containing $$a$$ and not $$b$$, or there exists an open set containing $$b$$ and not $$a$$.A topological space $$(X,\tau)$$ is said to be $$T_1$$-space if every singleton set $$\{x\}$$ is closed in $$(X,\tau)$$

Prove that every $$T_1$$-space is $$T_0$$-space.

I attempted the following proof:

If we have two topological spaces $$T_1=(X,\tau_1)$$ and $$T_0=(X,\tau_0)$$. If $$a,b\in X$$, by the definition of $$T_1$$-space: $$B=X\setminus{a}$$ is an open set such that $$a\notin B$$ and $$b\in B$$. In analogous way $$A=X\setminus{b}$$ is an open set such that $$b\notin A$$ and $$a\in A$$. This proves that the open sets of $$\tau_0$$are the open sets of $$\tau_1$$, then $$\tau_0\subset\tau_1$$.

Question:

Is the proof right? Can it be said "open sets of $$\tau_0$$are the open sets of $$\tau_1$$, then $$\tau_0\subset\tau_0$$"?

• Why do you have two different topological spaces in your proof? Sep 28, 2018 at 17:04
• @bitesizebo Because the question refers to two different spaces and the underlying topologies define two different topological spaces regardless of the fact $X$ is the same set fro both. Sep 28, 2018 at 17:07
• Your proof shows that the space you are calling $T_1$ is by definition a $T_0$-space. It does nothing to relate the open sets of the topological space $T_1$ to the open sets of the topological space $T_0$. Sep 28, 2018 at 17:09
• @BadamBaplan $A,B$ as defined, are sets of $\tau_0$. Sep 28, 2018 at 17:10
• Right, so where you're getting confused is that $T_0$ and $T_1$ are not distinct spaces. They're properties of topological spaces, like being Hausdorff. You're given that the topological space $(X, \tau)$ satisfies the $T_1$ property. You need to prove that the same space also satisfies the $T_0$ property. Sep 28, 2018 at 17:11

If every singleton is closed, then given $$a \neq b \in X$$, the set $$X \, \backslash \, \{ a \}$$ is an open subset of $$X$$ containing $$b$$ and not $$a$$.

• That is what I meant in my proof. Sep 28, 2018 at 17:09

Say that $$(X,\tau)$$ is $$T_1$$ space.

$$\iff$$ $$\{x\}$$ is a closed subset of $$X$$ for all $$x\in X$$

$$\iff$$ $$\forall y\in \{x\}^c$$ $$\exists U_y \in \tau: y\in U_y\subset\{x\}^c$$ for all $$x\in X$$

$$\Rightarrow$$ For all $$y\neq x$$ in $$X$$ there is an open subset $$U$$ of $$X$$ such that $$y\in U$$, $$x\notin U$$

$$\Rightarrow$$ $$(X,\tau)$$ is $$T_0$$ space.