Let $A\subseteq P\subseteq X$, where $(X,d)$ is a metric space with distance function $d$. Define the interior of $A$ in $P$, written as $Int_P(A)$, to be the union of all sets $E\subseteq P$ such that $E$ is open in $P$ and $E\subseteq A$. Show that $Int_P(A)=(A\cup P^c)^o\cap P$, where $A^o$ is the set of all interior points of $A$, $A^c$ is the complement of $A$.
We say $a$ is an interior point of $A$ if there exists a neighborhood of $a$ with radius $r>0$ such that $N(a,r)\subseteq A$. We say $A\subseteq P$ is open in $P$ when for all $a\in A$, $\exists r>0$ such that $N_P(a,r)\subseteq A$ where $N_P(a,r)=N(a,r)\cap P$.
We say a neighborhood of $a$ with radius $r>0$ to be the set $N(a,r)=\{x\in X:d(x,a)<r\}$.
Thanks in advance.
Let $O$ be $P$-open and $O \subseteq A$. So $O=O' \cap P$ where $O'$ is open in $X$. Then $O' \subseteq (A \cup P^\complement)$ (if $x \in O'$ it's either in $P$ and then it's in $O$ too (and hence in $A$), or it's in $P^\complement$.) and as the interior is the maximal open subset of a set, $$O' \subseteq (A \cup P^\complement)^\circ$$ and taking intersections with $P$ on both sides we get
$$O= O' \cap P \subseteq (A \cup P^\complement)^\circ \cap P$$ and as the $P$-interior of $A$ is the union of all $P$-open subsets of $A$ we find that the $P$-interior of $A$ (also denoted by $\operatorname{int}_P(A)$) is a subset of $(A \cup P^\complement)^\circ \cap P$.
On the other hand, the set $(A \cup P^\complement)^\circ \cap P$ is the intersection of a set that is open in $X$ with $P$, so $P$-open, and clearly a subset of $A$, as $$(A \cup P^\complement)^\circ \cap P \subseteq (A \cup P^\complement) \cap P = (A \cap P) \cup (P^\complement \cap P) = (A \cap P) \cup \emptyset \subseteq A$$ and again by maximality of the interior
$$(A \cup P^\complement)^\circ \cap P \subseteq \operatorname{int}_P(A)$$ and we have equality.