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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 $\tau_0\subset\tau_0$"?

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    $\begingroup$ Why do you have two different topological spaces in your proof? $\endgroup$ – bitesizebo Sep 28 '18 at 17:04
  • $\begingroup$ @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. $\endgroup$ – Pedro Gomes Sep 28 '18 at 17:07
  • $\begingroup$ 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 $T_0$. $\endgroup$ – Badam Baplan Sep 28 '18 at 17:09
  • $\begingroup$ @BadamBaplan $A,B$ as defined, are sets of $\tau_0$. $\endgroup$ – Pedro Gomes Sep 28 '18 at 17:10
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    $\begingroup$ 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. $\endgroup$ – bitesizebo Sep 28 '18 at 17:11
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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.

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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 $a$.

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  • $\begingroup$ That is what I meant in my proof. $\endgroup$ – Pedro Gomes Sep 28 '18 at 17:09

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