# Prove that this topology yields a $T_3$ space.

Here is the problem:

Let $(X,\tau)$ be a topological space and $A \subset X$. let $\tau _ A = \{ U \cup (V \cap A) \}$ where $U$ and $V$ are open in $X$. I have to prove that if $(X, \tau)$ is $T _3$ and A is a closed subset then $(X, \tau _A)$ is $T_3$.

So I have to prove 2 things

$(\rm i)$ $X$ is $T_1$

$(\rm ii)$ for each closed subset $K$ and $x \notin K \exists U,V \in \tau$ such that$x \in U$ and $K \subset V$ and $U \cap V = \emptyset$

To prove i) I did:

Let p and q different $\in X$ because $(X,\tau)$ is $T_3$ adn so $T_1$ there exist an open V in X such that p $\in V$ and $q\notin V$. And let $W =V \cup (X/A \cap A)$

So W=V and W is an open in X with $\tau _A$ then $p\in W$ and $q\notin W$ so $(X, \tau_A)$ is $T_1$

But I don't know how to prove ii)

Could you help me? Thank you

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$\newcommand{\cl}{\operatorname{cl}}$ Probably an important thing to investigate is the connection between the topologies $\tau$ and $\tau_A$.
• $\tau_A$ is finer than $\tau$. This can easily been seen by noting that if $U \subseteq X$ is $\tau$-open, then $U = U \cup ( \emptyset \cap A )$ is $\tau_A$-open.
• Since $\tau_A$ is finer than $\tau$, then $\cl_{\tau_A} ( B ) \subseteq \cl_{\tau} ( B )$ for every $B \subseteq X$. (Where $\cl_\sigma ( B )$ denotes the $\sigma$-closure of $B$.)
Next, it is probably easier to prove this using the open-neighbourhood characterisation of regularity: A T$_1$-space $Y$ is regular if given any $y \in Y$ and any open neighbourhood $W$ of $Y$ there is an open $V \subseteq Y$ such that $y \in V \subseteq \cl (V) \subseteq W$.
Hint: Suppose $x \in X$ and $W$ is any $\tau_A$-open neighbourhood of $x$. Then there are $\tau$-open sets $U_1 , U_2$ such that $W = U_1 \cup ( U_2 \cap A )$. Handle the cases $x \in U_1$ and $x \in U_2 \cap A$ separately, using the regularity of $\tau$ in each part.