Let $(X,\tau)$ be a topologic space and $A \subseteq X$. We can consider the subspace $(A,\tau_A)$.
Let $a \in A$ and $W \in V_A(a)$ if and only if there is a $V \in V(a)$ such that $W=V \cap A$
($V(a)$ is nbd of $a$ and $V_A(a)$ is nbd of $a$ in $(A,\tau_A)$)
My trying: $W \in V_A(a)$ iff $\exists G \in \tau_A $ such that $ a\in G \subseteq W $ and $G=A \cap V$ ($V \in V(a)$ namely $V$ is open in $(X,\tau)$). Thus $A \cap V \subseteq W$ but I couldn’t show $A \cap V \supseteq W$ Is there anyone show it and hint something about other direction?
PS: If I any mistake in trying please correct me. Thanks in advance
Just force it: Let $W \in V_A(a)$, so that indeed $a \in G \subseteq W$ with $G$ open in $A$ so there is some $\hat{G}$ open in $X$ with $\hat{G} \cap A =G$. Define $V = W \cup \hat{G}$. Clearly $V \in V(a)$ and $V \cap A = (W \cap A) \cup (\hat{G} \cap A) = W \cup G = W$ (as $G \subseteq W$ and $W \subseteq A$). So $V \in V(a)$ obeys $V \cap A = W$.
And if $V \in V(a)$ there is some open $G$ with $a \in G \subseteq V$. But then $V \cap A$ contains the open (in $A$) set $G \cap A$ which contains $a$, and so $V \cap A \in V_A(a)$. So that direction is straight forward.