Proving that if a set is both open and closed then it is equal to the real numbers Prove that if $A$ is both open and closed then $A = \mathbb{R}$ also as one suggested let $A \neq \emptyset$
You may use what ever definition of open and closed you would like, just avoid going into metric spaces, haven't covered that topic yet. My question is well essentially how to prove this statement but considering I like to do things myself, I was wondering if anyone had any particular suggestions to help me solve this.
attempted proof: Let $a\in A$ then since $A$ is open there exists an open interval $N(a,\epsilon)$ such that $$a\in N(a,\epsilon)\subset A$$ Then $a$ must be an interior point of $A$, but since $A$ is also closed then $a$ must also be an accumulation or limit point of $A$ as well. 
I am going to stop here because I am not sure if this is the right approach here, any suggestions would be greatly appreciated. 
 A: Suppose a set $A$ of real numbers is both open and closed, and that it is neither $\varnothing$ nor $\mathbb R$.  Since $A\ne\varnothing$, there exists some $a\in A$.  Since $A\ne\mathbb R$, there exists some $b\not\in A$.
Case 1: $a<b$.
Let $c=\sup (A\cap [a,b])$.  This exists since $A\cap[a,b]$ is non-empty (since it contains $a$) and bounded above (since $b$ is an upper bound).   Either $c\in A$ or $c\not\in A$.  If $c\in A$ then $c\ne b\not\in A$ so $a\le c<b$. Since $A$ is open, some open interval centered at $c\in A$ is a subset of $A$.  But that open interval contains numbers between $c$ and $b$, contradicting the fact that $c$ is an upper bound of $A\cap [a,b]$.  Thus we rule out the possibility that $c\in A$ and conclude $c\not\in A$.  Since no number smaller than $c$ can be an upper bound of $A\cap[a,b]$, every interval $(c-\varepsilon,c)$ contains some member of $A$. That means $c$ is an accumulation point of $A$. Since $A$ is closed, that means $c\in A$, and now we have a contradiction.
Case 2: $b<a$.
In this case, let $c=\inf(A\cap [b,a])$ and then proceed as above (but with $(c,c+\varepsilon)$ where $(c-\varepsilon,c)$ appeared above).
In either case, the assumption that $A$ is both open and closed and equal to neither $\varnothing$ nor $\mathbb R$ leads to a contradiction.
Summary: The idea is to find a point $c$ at the boundary between $A$ and $\mathbb R\setminus A$ and show that since $A$ is both open and closed it must both include and exclude that point.
A: Use that X is path connected implies that it's connected and that a space is connected iff the only sets that are both open and close in X are X itself and the empty set 
