Question

Show that T1-space X is regular iff for each point' a' element of X and each open set U containing 'a' ,there is an open set W containing 'a' whose closure is contained in U

Answer 1

$⇒)$ given $x\in X$, and a neighbourhood $U$ of $x$, there is an open set $O\subset X$ such that $x ∈ O ⊂ N$. Consider the point $x$ and the closed set $X − O$, which does not contain $x$. By regularity, there are open sets $N$ and $V$ such that $x ∈ N, X − O ⊂ V$ and $N ∩ V = ∅$. Thus $x ∈ N ⊂ X − V ⊂ O ⊂ U$, so $X − V=W$ is a closed neighbourhood of $x$ contained in the given neighbourhood $U$ of $x$.

$⇐)$ given $x ∈ X$ and the closed set $C ⊂ X − {x}$ , since $X − C$ is open and contains $x$, there is a closed neighbourhood $W$ of $x$ so that $W ⊂ X − C$. Let $V = X − W$. Then $V$ is open and $C ⊂ V$. Since $W$ is a neighbourhood of $x$, there is an open set $U$ such that $x ∈ U ⊂ W$. Then $U ∩ V ⊂ W ∩ (X − W) = ∅$, so $X$ is regular.