Prove that $(A \cap B)^\circ = A^\circ \cap B^\circ$ Let $(X,d)$ be a metric space and let $A, B \subset X$  
How can I show that  $(A \cap B)^\circ = A^\circ \cap B^\circ$ ?
Please just tell me a Hint. ($A^\circ$ and $B^\circ$ are sets of interior points of A and B)
 A: Use $x \in A^\circ \iff \exists r >0 : B(x,r) := \{ y\in X \mid d(x,y) < r \} \subset A$.
A: One definition of $A^\circ$ is the largest open set contained in $A$.
Using this definition, you can note that $A^\circ \cap B^\circ$ is an open set contained in $A \cap B$.
The other direction is similar but note that $A^\circ \cap B^\circ$
is an open set contained in both $A$ and $B$.
A: In general one proves "A= B", for A and B sets, by showing that $A\subset B$ and $B\subset A$.  And one shows "$A\subset B$" by starting "if $x\in A$" and using the properties of A and B to conclude "then $x\in B$".
Here, to show $(A\cap B)^o= A^o\cap B^o$, start "if $x\in (A\cap B)^o$" then x is an interior point of $A\cap B$.  That is, there exist $\epsilon> 0$ such that all points, y, such that $d(x, y)< \epsilon$ are in $A\cap B$.  Such a y is in A so "there exist $\epsilon> 0$ such that if $d(x, y)< \epsilon$ then $y\in A$".  That is, x is in $A^o$.  The same argument, using B instead of A, shows that x is in $B^o$.  That is, $(A\cap B)^o\subset A^o\cap B^o.
Now go the other way: If "$x\in A^o\cap B^o$" then $x\in A^o$ and $x\in B^o$.  Since $x\in A_0$, there exist $\epsilon_1$ such that the set $\{y| d(x,y)< \epsilon_1\}$ is a subset of A.  Since $x\in B_0$, there exist $\epsilon_2$ such that the set $\{y| d(x,y)< \epsilon_2\}$ is a subset of B.  Take $\epsilon$ equal t the smaller of $\epsilon_1$ and $\epsilon_2$.  Then $\{y|d(x,y)< \epsilon\}$ is a subset of both A and B so is a subset of $(A\cap B)$.  That means that $x\in (A\cap B)^o$.
Note, by the way, that this can be immediately extended to any finite collection of sets because any finite set of positive numbers contains a smallest positive number.  It does not extend to an infinite collection of sets because an infinite set of positive numbers does not necessarily contain a smallest positive number.
A: All you need to prove is that:


*

*$V^{\circ}$ is an open set

*$V^{\circ}=V$ if and only if $V$ is open

*If $V \subset W$ then $V^{\circ} \subset W^{\circ}$
which follows directly from the definition of interior of a set. Once this is done the demonstration is as follows:

First note that $A^{\circ} \subset A$ and $B^{\circ} \subset B$, then
  $A^{\circ} \cap B^{\circ} \subset A \cap B$. 
As $A^{\circ}$ and $B^{\circ}$ are both open their intersection is
  also open, i.e, $A^{\circ} \cap B^{\circ}$ is open and by (2) we get
$(A^{\circ} \cap B^{\circ})^{\circ}=A^{\circ} \cap B^{\circ}$ .
By (3) we have that  $A^{\circ} \cap B^{\circ} \subset A \cap B$ implies that
$(A^{\circ} \cap B^{\circ})^{\circ} \subset (A \cap B) ^{\circ}$
but we have seen that $(A^{\circ} \cap B^{\circ})^{\circ}=A^{\circ} \cap B^{\circ}$ . So, 
$A^{\circ} \cap B^{\circ} \subset( A \cap B)^{\circ} $. 
Conversely,using (3) we have that $A\cap B \subset A$ implies $(A\cap B)^{\circ} \subset
 A^{\circ}$ and similarly $(A\cap B)^{\circ} \subset B^{\circ}$.
Therefore we have that 
$(A\cap B)^{\circ} \subset A^{\circ} \cap B^{\circ}$
So, we can conclude that 
$(A\cap B)^{\circ} =A^{\circ} \cap B^{\circ}$.

