Using the definition of limit points which is as follows:
"Consider $A\subset(X,T)$(a topological space with standard topology), $x\in X$ is a limit point of $A$ if for all $U\in T$ s.t. $x\in U$ we have that the intersection of U and A contains a point other than x itself"
I wanted to work out the limit points of $(0,1)$ (I know they are the set $[0,1]$ from the sequence definition of limit points).
I'm having a little confusion using the definition I mentioned above though , here's my problem.
all $U\in T$ are of the form (a,b), now suppose $x\in (a,b)$ and that $a<0<1<b$. Suppose that $x=-2$ for instance and that $(a,b)=(-3,4)$.
So we have found an open set which x is in , and whose intersection with $(0,1)$ is equal to $(0,1)$, clearly this contains many points other than x.
So my question is why is say $x=-2$ not a limit point (according to this definition )?
Consider the open interval $U=(-2.1,-1.9).$ It contains $-2$ but does not intersect $(0,1)$. This shows that $-2$ is not a limit point of $(0,1)$. $U$ does not contain any point of $(0,1)$, let alone a point different from $-2$.