1. ## Separated

If $\displaystyle A,B\subset X$ are both open and disjoint, then why are they separated? (that is $\displaystyle A\cap\overline{B}=\emptyset=\overline{A}\cap B$)

2. Originally Posted by bram kierkels
If $\displaystyle A,B\subset X$ are both open and disjoint, then why are they separated? (that is $\displaystyle A\cap\overline{B}=\emptyset=\overline{A}\cap B$)
Think about it this way. $\displaystyle \text{cl }B=B\cup D\left(B\right)$ ($\displaystyle D$ is the derived set, the set of limit points) and so $\displaystyle A\cap\text{cl }B=\left(A\cap B\right)\cup\left(A\cap D(B)\right)=A\cap D(B)$. Thus, if $\displaystyle A\cap\text{cl }B\ne \varnothing$ then $\displaystyle A$ must contain a limit point of $\displaystyle B$, but why is that stupid?

3. Since a sequence $\displaystyle \{x_j\}_{j=1}^\infty$ in $\displaystyle A$ does not enter $\displaystyle B$, the sequence cannot converge to any point of $\displaystyle B$. Thanks

4. Originally Posted by bram kierkels
Since a sequence $\displaystyle \{x_j\}_{j=1}^\infty$ in $\displaystyle A$ does not enter $\displaystyle B$, the sequence cannot converge to any point of $\displaystyle B$. Thanks
Well, the idea's right. No neighborhood of a limit point of $\displaystyle B$ can be contained entirely within $\displaystyle A$ and thus can't be a point of $\displaystyle A$.

Note though, I've seen you work in general topological spaces, and the definition you are implying of the closure in terms of sequences is only valid in first countable spaces.