Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question

1 Answer 1

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.