Subspaces and interiors of metric spaces problem.

[gottfried]

Homework Statement

If S is a subspace of the metric space X prove (int_xA)\capS\subsetint_s(A\capS) where A is an element of ΩX(Open subsets of X)

The Attempt at a Solution

So int_xA=\bigcupB_d(a,r) where d is the metric on X and the a's are elements of A
and I think
int_sA=\bigcupB_d(a,r)\capS

But I'm not sure how to use this fact but it feels as though the answer comes some how from the above condition.

Any pointers?

 

[Fredrik]

When you want to prove a statement of the type $A\subset B$, the proof should usually start with a statement like "Let x be an arbitrary member of A". Then you use what you know about A to show that x is a member of B. Here you might want to use another symbol instead of x, because it looks weird to use a symbol that looks too much like X.

Let A and S be arbitrary subsets of X. Let $y\in \operatorname{int}_X A\cap S$ be arbitrary. Now what can you tell us about y?

 

[gottfried]

y\inint_XA\capS
y\in(int_XA\capS)\capS
It seems right that int_XA\capS = int_SA
y\inint_SA\capS
int_SS=S(Since all points of S have open balls around them in S making each point interior)
y\inint_SA\capint_SS
y\inint_S(A\capS)

Is this the kind of thing you mean?

 

[Fredrik]

In this type of proof, each step is supposed to be an immediate and obvious consequence of the statement in the previous step. Each of the first few steps is often just a clarification of what the statement in the previous step means. What I had in mind for the first step is that you use the definition of $\cap$ to explain what $y\in\operatorname{int}_X A\cap S$ means. Then you use some other definition to explain what the new statement means.

In one of the steps, you will need to think about how the subspace topology is defined.