Thread: open and closed sets

In this problem you will prove that the only subsets of R that are both open and closed are R and the empty set. Let E be a nonempty subset of R, and assume that E is both open and closed. Since E is nonempty there is an element a in E. Denote the set
Na(E) = {x > 0|(a-x, a + x) contained in E}
(a) Explain why Na(E) is nonempty.
(b) Prove that if x in Na(E) then [a-x, a+x] is contained in E.
(c) Prove that if [a-x, a+x] is contained in E, then there is a y in Na(E) satisfying y > x.
(d) Show that Na(E) is not bounded above (argue by contradiction).
(e) Prove that E = R.


This is what I tried:
a) Well I know that a is both open and closed, so if its closed, it contains its limit points. Since x is a limit point, and a is in E, then by properties of sets that would mean that a+x and a-x would be contained in E, but I really don't know how to show how the interval is contained in E.

b)Think I need to use part a for this one

c)well if [a-x,a+x] is contained in E, and E is also open, that means that there must be an open interval that is also contained in E. So there must be a y in Na(E) so that y>x that would yield the interval (y-a,y+a)

d)Not too sure how to do this

e) Not sure either

Any help would be great!!!

Because is closed and is a limit point of then
Same can be said for .

Because is open Why?
So let then and
The point of all of that is that for any number in there is some number greater in .
Thus can't be bounded above nor below.

Now for part e). If must it follow that
If so how is it possible for to be true?

Thanks for the help, but do you think you can point me in the right direction toward showing how to prove that Na(E) is nonempty. It seems like it would be so easy yet I can't figure out how to do it. I've been trying to use the definiton of limit points but it isn't working.

Originally Posted by mathgirlie

Thanks for the help, but do you think you can point me in the right direction toward showing how to prove that Na(E) is nonempty. It seems like it would be so easy yet I can't figure out how to do it. I've been trying to use the definiton of limit points but it isn't working.

Good grief.
so
But we know that is open so .

thanks, I knew it was easy... must be tired or something