Let X be a topological space.
a) If X is Hausdorff and it has an countable base, so X is normal?
b) If X is regular, and if each point of X has a fundamental countable neighborhood system and is separable, then is X normal?
Being normal means that X is $T_1$ and $T_4$.
We say that:
X is $T_4$ if for any A, B $ \subset $ X closed there are U, V $ \subset $ X open disjoint such that A $ \subset $ U and B $ \subset $ V.
X is regular if X is $ T_3 $ and $ T_1 $.
X is $ T_3 $ iff for each x $ \in $ X and each F $ \subset $ X closed such that x $ \notin $ F there are U, V $ \subset $ X open disjoint such that x $ \in $ U and F $ \subset $ V.
If X is $T_3$ is $T_2$ and $T_1$.
If X is $T_2$ is also $T_1$.
if X is Separable it contains a subset A where for any point x in X, any neighborhood of x contains at least one point from A and A is countable. (A space is called separable if it contains a countable, dense subset).
a) and b) are false, right?
In both cases, there are $T_2$ spaces that we want to prove are normal.
Being $T_2$ means that U, V $ \subset $ X open disjoint such that A $ \subset $ U and B $ \subset $ V.But I don't see how to find a closed for any A, B, even with the countable base (first case), as in the case of having a fundamental system of countable neighborhood and being separable (second case).
No to both: the Sorgenfrey plane is a counterexample to the second.
The first has the $\Bbb R_K$ as a counterexample.
Both are treated in Munkres' text book.