# Easy Containment question

1. May 27, 2012

### trap101

Prove that Sint$\subseteq$ S

Ok I thought this was going to be easy, but apparently i'm having issues.

So I said:

for all $x$ let x$\in$Sint

==> there exists a B(r, x) s.t with any point a in Sint |X-a| < r.

But now S is the B(r,x) s.t |x-a| = r.....

This is where I'm stuck. I know I have to show that the properties of Sint also apply to S. Help

2. May 27, 2012

### algebrat

you should have said,

If x is in the interior of S, there exists a ball B(x,r) contained in S. Thus, for any a in the entire ambient space, if |a-x|<r, then a is in B(x,r) and thus in S.

The ball is in S, but why do you think S is in the ball?

I think the key point is that x itself is in the ball! Thus x is in... where?! (... as desired.)

3. May 27, 2012

### Dick

I have no idea what you are talking about. But I am pretty sure you don't know what S^int means. This IS be easy. Review the definition of S^int, ok?

4. May 27, 2012

### trap101

That's exactly it, I do know what Sint means: it's the B(r,x) $\subseteq$S. i.e: The ball of radius "r" about the point "x". contained in S, but then I didn't "prove" anything I just stated the definition.

Is there some characteristic about S that I should be using to join them? Or is it just that simple as the definition?

5. May 27, 2012

### Dick

It's super easy. x is in B(r,x). If B(r,x) is contained in S then x is an element of S.

6. May 27, 2012

### trap101

maaaaaaannnn. Is a question like that just made to mess with my head because it is just TOO easy, so you start thinking in circles?

7. May 27, 2012

### Dick

If you overthink something that is easy, then you can start thinking in circles. But some things are just easy.