I need help with another complex problem in a general topological space:
Show that a set S is open if and only if each point in S is an interior point.
In the other direction to Plato's without all the details. Let $\displaystyle O$ be an open set in a topological space $\displaystyle X$. Suppose that $\displaystyle x\in O$ was not an interior point, then every neighborhood $\displaystyle N$ of $\displaystyle x$ intersects $\displaystyle X-O$. Therefore, we see that $\displaystyle x$ is a limit point of $\displaystyle O'$. But, we have that $\displaystyle x$ is a limit point of $\displaystyle O'$ that is not in $\displaystyle O'$, so $\displaystyle O'$ is not closed and thus $\displaystyle O$ is not open. This, of course, is a contradiction.
This is an indirect way. Of course it assumes you've defined closed set and limit point. If not, you should be able to see a more direct way.
By writing down two definitions the above statement follows quite easily:
Def 1 x is interior point of S <=> there exists ε>0 and $\displaystyle B(x,\epsilon)\subset S$
Def 2 S is open <=> for all ,x: xεS => there exists ε>0 and $\displaystyle B(x,\epsilon)\subset S$.
And using the two definitions we have:
S is open <=> (for all ,x : xεS => there exists ε>0 and $\displaystyle B(x,\epsilon)\subset S$) <=> (for all ,x : xεS => x is interior of S) <=> (each point of S is interior)