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!!!