TITLE: Clarification on the Hausdorff property
QUESTION [5 upvotes]: If $X$ is a Hausdorff space then for points $a,b \in X$ there are disjoint open sets $U$ and $V$ such that $a \in U$ and $b \in V$.  So, take a set of points $\{a_1, \ldots , a_n\}$ and another point $x$.  Then for each $a_i$ there are disjoint open sets $U_i$ and $V$ containing $a_i$ and $x$, respectively.
My question is: is $V$ disjoint from all the $U_i$, or does this only hold pairwise?  Are there conditions we can impose to guarantee an open set separating $x$ from all points in some given finite set?

REPLY [5 votes]: I just want to point out that with the idea in William's proof, above, you can easily prove the following (seemingly more general) property:
If $X$ is Hausdorff and $x_1, \ldots , x_n \in X$ are distinct, then there are open neighbourhoods $V_1 , \ldots , V_n$ about $x_1, \ldots , x_n$, respectively, such that $V_i \cap V_j = \emptyset$ for $i \neq j$.