# Regular spaces and Hausdorff space

I was going through a topology lecture note and came across the concepts of regular spaces (T3 ) and Hausdorff space (T2).

I came across a statement :Every regular space is Hausdorff .

As those notes did not contain any proof for this I tried to do it .Now a regular space is one in which given any point $\ x$ and any closed set $\ V$,which does not contain $\ x$ , there exists two disjoint open subsets $\ U_1$ and $\ U_2$ such that $\ x$ $\ \epsilon$ $\ U_1$ and $\ V$ $\ \epsilon$ $\ U_2$ .

Now for this to be a Hausdorff space , for every two distinct points $\ a$ and $\ b$, we should be able to find two disjoint open sets $\ U_1$ and $\ U_2$, such that $\ a$ $\ \epsilon$ $\ U_1$ and $\ b$ $\ \epsilon$ $\ U_2$. Now take two points in the space $\ a$ and $\ b$ in a regular space . If we can find one closed set which contains only one of the two points then we are done . If the two points belong to two different open sets in the topology then we can guarantee that we can find a closed set that contains only one of them.How to proceed if both the points belong to the same open set in the topology ? At the same time , whether a set is open or not depends on the topology ,I can't construct an open set as per my wish in a way that it contains only one of the two points. What am I missing here ? Thanks in Advance.

Unfortunately, the terminology is not standardized. Some people define "regular" the way you did, and define "$T_3$" to mean "regular and Hausdorff" (or equivalently "regular and $T_1$" or "regular and $T_0$"); others, unfortunately, interchange the meanings of "regular" and "$T_3$", so that for them "regular" includes Hausdorff and "$T_3$" does not.
X={1,2,3,4} Topology on X be { X, $\phi$, {1,2} , {3,4} }.. You can check the given space is not Hausdorff under the given topology, since there exists no disjoint neighbourhoods of 0 and 1. but, it is regular.