I'm trying to understand a pretty simple theorem in my intro to Topology notes:
Let $X$ be a topological space and $A \subset X$ . Suppose the topology is given by a basis $B'$; then $x$ is in the closure of $A$ iff every basis element $B$ containing $x$ intersects $A$.
Before this theorem in my notes the author states:
$x \in $ cl($A$) iff $A$ intersects every open set $U$ containing $x$.
First I have a preliminary question, is the basis of a topology necessarily contained in the topology? Ex.: if not, the interval (in this case our topology with open sets in $(0,1)$)
$(0,1)$ could have as a basis all open intervals in $\Bbb R$ (?)
Secondly, my notes state for the forward direction:
If every open set containing $x$ intersects $A$, then every basis element containing $x$ intersects $A$. (In my notes "intersect" means the intersection isn't empty). I think this statement is supposed to be obvious, but it is not obvious to me at all. Any insights appreciated. Also I am new to this topic so if anything I've written doesn't make sense I will elaborate or clean it up immediately.
Best Answer
Every basis element is open. If $x \in \mathrm{cl}(A)$ then every open set containing $x$ intersects $A$, so in particular every basis element containing $x$ intersects $A$.
On the other hand, suppose every basis element containing $x$ intersects $A$. If $O$ is an open set containing $x$, there exists a basis element $B$ with $x \in B \subset O$. Since $B$ intersects $A$, so does $O$. Thus every open set $O$ containing $x$ intersects $A$, so that $x \in \mathrm{cl}(A)$.