Real Analysis - I open interval contains a point of a set if x is a limit point of that set is in I

Question

I try to prove the statement If $ A \subset R$ is any set, and $I$ is an open interval that contains a limit point of $A$, then $I$ also contain a point of $A$. I try to use the definition of limit point by considering the deleted neighborhood $V'_{\epsilon}(x) = (x-\epsilon, x) \cup (x,x+\epsilon)$ if $x$ is a limit point of $A$. Is there a way or suggestion to go beyond this?

Answer 1

What would happen if $I$ did not contain a point of $A$? Suppose $x\in I$ is the limit point. Since $x$ lies in $I$ and $I$ is open you can find a neighbourhood $(x-\varepsilon,x+\varepsilon)$ contained in $I$. What can you say about $(x-\varepsilon,x+\varepsilon)\cap A?$

Answer 2

If $I$ is an open interval that contains $x \in A'$, then $I$ must, as an open set containing $x$, by definition of a limit point, contain a point of $A$ different from $x$. So you're done: $I$ contains a point of $A$.

Nothing more to say, I think.

Answer 3

Let a be a limit point of A and I an open set with a in I.
U = I - {a} is a deleted nhood of a.
Since a is a limit pt of A, exists x in A $\cap$ U.
This results holds for all topological spaces.