Regarding dense subsets, I still think I am on the way of getting the hang of it, and have the following question:
Problem: If $X= \cup X_i$ denotes the union of open sets, then $B$ will be a dense subset of $X$ iff $B$ is dense in each $X_i$.
I have little idea on how to show this, can anyone give me a proof to this problem? It will be extremely helpful. Thank you in advance.
In order to prove that $B\cap X_i$ is dense in $X_i$ for each $i$, let $A$ be a non-empty open subset of $X_i$. Then, since $X_i$ is an open subset of $X$, $A$ is also an open subset of $X$. So, since $B$ is dense, $B\cap A\neq\emptyset$. Since this holds for any non-empty open subset of $X_i$, $B\cap X_i$ is dense in $X_i$.
Now, suppose that $B\cap X_i$ is a dense subset of $X_i$, for each $i$. Let $A$ be a non-empty open subset of $X$. Since $A$ is not empty and $X=\bigcup_iX_i$, $A\cap X_i\neq\emptyset$ for some $i$. And $A\cap X_i$ is an open subset of $X_i$. But then $(A\cap X_i)\cap(B\cap X_i)\neq\emptyset$ since $B\cap X_i$ is dense in $X_i$. In particular, $A\cap B\neq\emptyset$. Since this takes place for each non-empty open subset $A$ of $X$, $B$ is a dense subset of $X$.