Show that for each $a \in A$ and each $V \in {\mathcal{N}_A}\left( a \right)$ we have that $\overline V \in \mathcal{N}\left( a \right)$

Question

I have the following question:

Let $X$ be a topological space and let $A$ be an open subset of $X$. Let it be given that for any $B \subseteq X$ we have $A \cap \overline B \subseteq \overline {A \cap B}$. Here $\overline{S}$ indicates closure for some set $S$.
Now suppose $A$ is an open subspace of $X$. Use $A \cap \overline B \subseteq \overline {A \cap B}$ to show that for each $a \in A$ and each $V \in {\mathcal{N}_A}\left( a \right)$ we have that $\overline V \in \mathcal{N}\left( a \right)$. Note here ${\mathcal{N}_A}\left( a \right)$ denotes the neighbourhood of $a \in A$ of the subspace $A \subseteq X$.

I think I'm confused because there is so much to use here. I know that $${\mathcal{N}_A}\left( a \right) = \left\{ {A \cap N\;|\;N \in \mathcal{N}\left( a \right)} \right\}$$

So if for each $a \in A$ and each $V \in {\mathcal{N}_A}\left( a \right)$. Then we have $V = A \cap N$ where $N \in \mathcal{N}\left( a \right)$. Now how do I use the given $A \cap \overline B \subseteq \overline {A \cap B}$ ?

William Elliot · Accepted Answer · 2020-05-15T10:51:00.303

1

As V in N$_A$(a), exists A.open U with a in U subset V.
Since U is A.open, exists X.open O with U = O $\cap$ A.
Because A is X.open, U is X.open.
Consequently, a in U subset V subset $\bar V$.
So $\bar V$ in N(a).

edited May 15 '20 at 10:51

answered May 14 '20 at 16:13

William Elliot

17,598

I'm finding it hard to understand your answer, what do all the "." mean? Could you please maybe modify it – pabhp May 14 '20 at 16:21
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A.open means open within the subspace A. @pabhp. – William Elliot May 14 '20 at 20:04

Show that for each $a \in A$ and each $V \in {\mathcal{N}_A}\left( a \right)$ we have that $\overline V \in \mathcal{N}\left( a \right)$

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