Solution: Open & Closed Subsets in R: Na(E) Nonempty, [a-x,a+x] ⊆ E, E = R

[dancergirlie]

Homework Statement

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. Denote the set
Na(E) = {x > 0|(a-x, a+x) \subsetE}

(a) Explain why Na(E) is nonempty.
(b) Prove that if x \in Na(E) then [a-x, a+x] \subset E.
(c) Prove that if [a-x; a+x] \subset 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.

Homework Equations

The Attempt at a Solution

a) I don't know if I can just say for (a-x)<(a+x) there has to be an y\inR so that (a-x)<y<(a+x).

b)
I was thinking, since x is in Na(E) that would mean that x\in E and since a and x are both in E, by definition of a set, that would mean that
a-x \in E, and
a+x \in E
and since (a-x, a+x) is contained in E and now (a-x) and (a+x) are in E, that would mean:
[a-x,a+x] \subset E

The rest I'm not too sure about, any help would be great!

 

[VeeEight]

For (e) you can take B = R-E and it is also open and closed. What can you say about A union B and A intersection B? Remembering that the Reals are a connected space, what can you say about B?

Also it seems that you haven't used the open and closed properties. Remember the definition of an open set, involving neighborhoods, and that the complement of an open set is closed and vice versa.

 

[dancergirlie]

how would I incorporate the properties and which ones would it work best on? I can't really see how they would work on part a