R^n topological quesion-Calculus

[WannaBe22]

Homework Statement

Let K be a closed&bounded set in R^n which isn't empty.
Prove that K isn't open.

Homework Equations

No topology! I can't use the fact that the only sets in R^n which are closed and open are the empty set and R^n... Only the definitions of open sets and closed sets (closed sets=sets which satisfy that the limit of every converged sequence is an element of the set...)

The Attempt at a Solution

I have no idea how to do it...

Hope you'll be able to help me

Thanks

 

[Dick]

Can you prove it in R^1? Then use the fact any two points in R^n can be joined by a line segment.

 

[WannaBe22]

Actually I have no idea how to prove it on R... I've tried doing it by contradiction but without any success... Hope you'll be able to give me some further guidance


Thanks !

 

[Dick]

Actually I have no idea how to prove it on R... I've tried doing it by contradiction but without any success... Hope you'll be able to give me some further guidance


Thanks !

How did you try to do it by contradiction? What went wrong?

 

[WannaBe22]

I've tried this way:
Let K be an open,closed, boundless set. We'll try to get a contradiction from the assumption that K isn't empty. Let a be an element in K. We know that there is a disk around a which is contained completely in K. So B(a,r) is contained in K. On the other hand, K is also closed... This is where I got stuck... I've know idea how to continue...

I've also tried to take an element in a closed,bounded,not empty set and to show that there is an element inside this set which isn't part of a neighbourhood contained in the set... I can't understand how to continue from this point, because there is a possibility that $$\partial K = \emptyset$$ ...

Hope you'll be able to help me

Thanks!

 

[Office_Shredder]

You can prove the boundary exists. Pick a point in K. Draw a line in a random direction. Eventually the line has to leave the set, because otherwise K would be unbounded. Find the point farthest down the line, and that has to be in the boundary.

This is just an outline, there's a lot of details to prove here

 

[Dick]

I've tried this way:
Let K be an open,closed, boundless set. We'll try to get a contradiction from the assumption that K isn't empty. Let a be an element in K. We know that there is a disk around a which is contained completely in K. So B(a,r) is contained in K. On the other hand, K is also closed... This is where I got stuck... I've know idea how to continue...

I've also tried to take an element in a closed,bounded,not empty set and to show that there is an element inside this set which isn't part of a neighbourhood contained in the set... I can't understand how to continue from this point, because there is a possibility that $$\partial K = \emptyset$$ ...

Hope you'll be able to help me

Thanks!

Try proving it in R first. Take K to be bounded and closed. To show K isn't open, show the complement K^C=R-K isn't closed. Find a convergent sequence of points in K^C whose limit isn't in K^C.

 

[WannaBe22]

Well...Let K=[a,b] be a closed set in R. Let $$b+ \frac{1}{n}$$ be a converged sequence in $$K^c$$. Then the limit of this sequence isn't an element of the set itself, which means K isn't open... If K is a union of finite amount of closed segments or an intersection of closed segments- we can prove it the same way... But in order to finish this proof, we also need to prove that the only closed sets in R are the closed segments or a finite union of such segments... Which makes it more difficult...

Anyway- when taking a closed and bounded set in $$R^n$$ , we can say that it must have at least two points (if it has only one point - this point has no open set contained in the set... We can even show that the set must be infinite but it isn't that necessary) ... So, we can draw a line between the two points we have chosen (if the two points were a and b, the line is ta+(1-t)b ) . K is also open, by our assumption, so there are other points on this line which are in K. From this point, I've no idea how to continue... In R, we could continue because we knew what are the points in $$\partial K$$...Here we can't make such a guess...

Hope you'll be able to give me some further guidance

Thanks

 

[WannaBe22]

Actually, we can take a closed and bounded set K and a point a in it... We can take a line segment from 0 to this point (mark it as D) and denote: $$b_i= D \cap K$$ and then build a sequence in a similar way we did in R... Is this correct?

 

[Dick]

For the K in R case, you don't know it's a interval. But you do know it's bounded. If it's bounded it has a least upper bound. Can you show that since K is closed, that least upper bound is in K? Use that instead of b and your proof is fine. Now for the R^n case pick a point a in K and a point b in K^C and look at your line a*t+b*(1-t). If you intersect K and K^C with that line then haven't you reduced the problem to the case of the two sets being in R? That's I think what you meant to say.

 

[WannaBe22]

Thanks a lot !