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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).

1 Answers 1