# Mathematics Prelims

## February 22, 2009

### The Hausdorff Property

Filed under: Topology — cjohnson @ 7:53 pm

One of the important tools in topology are properties that are invariant between spaces.  These are properties of topological spaces that remain unchanged under certain “operations” that we may perform on the spaces.  We’ve already seen one such property in connectedness, and now we’re going to discuss another one: the Hausdorff property.  We say that a space has the Hausdorff property if for every pair of points in the set, there are disjoint open sets that contain the points.  For example, in the case of the real line with the standard topology, we can pick two distinct points, $x$ and $y$, and find two open intervals, $U$ and $V$, such that $x \in U$ and $y \in V$, but $U \cap V = \emptyset$.  If $x = 1$ and $y = 2$, for instance, we may take $U = (-0.9, 1.1)$ and $V = (1.5, 2.3)$.

Proposition: If $(X, d)$ is a metric space, then the topology induced by the metric has the Hausdorff property.

Proof:  Let $a, b \in X$ and denote the distance between $a$ and $b$ by $\delta = d(a, b)$.  Clearly $B_{\delta / 2}(a)$ and $B_{\delta / 2}(b)$ are disjoint open sets containing $a$ and $b$, respectively.

At first glance, it would seem that a topological space which does not posess the Hausdorff property is a strange or esoteric abstract space, but there is an easy way to extend the real line, with its standard topology, into a non-Hausdorff space.  We simply add a second zero to real line, which we’ll call Z.  The open sets containing Z are the same as the open sets containing zero, except that zero is replaced by Z.  For example, $(-1, 1) \setminus \{0\} \cup \{Z\}$ is such an open interval.  Now, note that this space (called the real line with a double point) is not Hausdorff:  Let $U$ and $V$ be any two open sets with $0 \in U$ and $Z \in V$.  By the way we’ve defined a topology on this space, $V$ must contain an open interval containing Z.  This open interval is precisely the same as some open interval containing zero (with zero replaced by Z).  Any open interval containing zero must have a non-empty intersection with this open set, however.  Since $U$ must contain some open interval containing zero, $U$ and $V$ must have a non-empty intersection.  This means the space can not be Hausdorff.