Proving the Containment of Sint in S: A Simple Task or a Mind-Bending Challenge?

[trap101]

Prove that S^int$\subseteq$ S

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

So I said:

for all $x$ let x$\in$S^int

==> there exists a B(r, x) s.t with any point a in S^int |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 S^int also apply to S. Help

 

[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.)

 

[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?

 

[trap101]

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?

That's exactly it, I do know what S^int 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?

 

[Dick]

That's exactly it, I do know what S^int 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?

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.

 

[trap101]

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.

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?

 

[Dick]

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?

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