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Here is the problem:

Let $(X,\tau) $ be a topological space and $ A \subset X $. Define $$\tau _ A = \{ U \cup (V \cap A) : U , V \in \tau \}.$$ Prove that if $ (X, \tau)$ is T3 and $A$ is a closed subset, then $(X, \tau _A)$ is T3.

So I have to prove 2 things

  1. $(X,\tau_A)$ is T1, and

  2. for each closed subset $K$ and $x \notin K$ there are $U,V \in \tau_A$ such that $x \in U$, $K \subset V$, and $U \cap V = \emptyset$ .

To prove (1) I did:

Let $p,q \in X$ be distinct. Because $(X,\tau)$ is T3 (hence also T1) there exist an open $V \subseteq X$ such that $p \in V$ and $q\notin V$. Note that $V = V \cup ((X \setminus A) \cap A)$ is also open in $(X , \tau _A)$.

But I don't know how to prove (2).

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Probably an important thing to investigate is the connection between the topologies $\tau$ and $\tau_A$.

  • $\tau_A$ is finer than $\tau$. This follows from the fact that $U = U \cup ( \varnothing \cap A )$ for every $U \in \tau$.
  • Since $\tau_A$ is finer than $\tau$, then $\operatorname{cl}_{\tau_A} ( B ) \subseteq \operatorname{cl}_{\tau} ( B )$ for every $B \subseteq X$ (where $\cl_\sigma ( B )$ denotes the closure of $B$ with respect to the topology $\sigma$).

Next, it is probably easier to prove this using the open-neighbourhood characterisation of regularity:

A T1-space $Y$ is regular if (and only if) given any $y \in Y$ and any open neighbourhood $U$ of $Y$ there is an open neighbourhood $W$ of $x$ such that $\operatorname{cl} (W) \subseteq U$.

We can now demonstrate that $\langle X , \tau_A \rangle$ is regular:

Let $x \in X$, and let $U \cup ( V \cap A )$ be a $\tau_A$-open neighbourhood of $x$ (where $U , V \in \tau$). There are two cases:

  • If $x \in U$, then by the regularity of $\langle X , \tau \rangle$ there is a $\tau$-open neighbourhood $W$ of $x$ such that $\operatorname{cl}_\tau ( W ) \subseteq U$. But then $W$ is also a $\tau_A$-open neighbourhood of $x$, and $$\operatorname{cl}_{\tau_A} ( W ) \subseteq \operatorname{cl}_\tau ( W ) \subseteq U \subseteq U \cup ( V \cap A ).$$

  • If $x \in V \cap A$, then $V$ is a $\tau$-open neighbourhood of $x$, so by the regularity of $\langle X , \tau \rangle$ there is a $\tau$-open neighbourhood $W$ of $x$ such that $\operatorname{cl}_\tau ( W ) \subseteq V$. Then $W \cap A$ is a $\tau_A$-open neighbourhood of $x$, and $$ \operatorname{cl}_{\tau_A} ( W \cap A ) \subseteq \operatorname{cl}_\tau ( W \cap A ) \subseteq \operatorname{cl}_\tau ( W ) \cap \operatorname{cl}_\tau ( A ) \subseteq V \cap A \subseteq U \cup ( V \cap A ).$$ (Though a bit hidden, we are using the fact that $A$ is closed, since then $\operatorname{cl}_\tau ( A ) = A$. Were $A$ not closed, the penultimate set-inclusion above may be false.)

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