# In a topology, if a third open set is formed of intersection of two open sets. How is it Hausdorff?

A very basic question.

If a Topology, $T$ has open sets $O_1$ and $O_2$, and $O_3$ is their intersection or union then how can the neighborhoods of $O_3$ and $O_2$ or $O_3$ and $O_1$ be disjoint, because the neighborhood is a superset of the open set. So how can any topology be ever Hausdorff? "A topological space (X,T) is Hausdorff space if all two distinct points in X have two disjoint neighborhoods." Please correct me. My understanding has gone haywire.

• A space is Hausdorff if disjoint points can be separated by disjoint neighbourhoods. I'm not sure what your problem is here. – Ian Coley Feb 13 '14 at 22:05
• Hausdorffness does not say any two open sets are disjoint. It says for any two fixed points $x$ and $y$, there are disjoint open sets $O_x$ and $O_y$ with $x\in O_x$ and $y\in O_y$. – David Mitra Feb 13 '14 at 22:06
• A point is both in open set $O_1$ and open set $O_3$ (from above). So they can never be disjoint right? And $O_3$ must exist as per the intersection condition of toloplogy. – kosmos Feb 13 '14 at 22:09
• It doesn't say for all pairs of open subsets, it just says there must be at least one pair of open subsets such that the condition is satisfied – Robert Wolfe Feb 13 '14 at 22:12
• Think about the real line. For any two points you can find an infinite number open balls containing both. However, you can also find two disjoint balls that contain only one of them. – Felipe Jacob Feb 13 '14 at 22:14

In a Hausdorff space, given any two distinct points $x$ and $y$, there are disjoint open sets $U$ and $V$ so that $x\in U$ and $y\in V$.

This is not saying that every open set that contains $x$ and every open set that contains $y$ are disjoint.

As mentioned by André Nicolas, $\mathbb{R}$ (in fact, $\mathbb{R}^n$) is a Haudsdorff space in its standard topology, for example.

Here is another simple example. For any set $X$, you can also take its power set to be a topology. This is called a discrete topology and it is Hausdorff. This is because if you take any points $x$ and $y$, $\{x\}$ and $\{y\}$ are open sets, and $\{x\} \cap \{y\} = \{ \}$. (The fact that $\{x,y\}$ is not disjoint from $\{x\}$ (or $\{y\}$) is irrelevant.)

$\{\emptyset, \{1\}, \{2,1\},\{\pi,3,1\},\{2,\pi,3,1\}\}$. I took the liberty of adding the two sets that are required to have a topology. Namely, $\emptyset$ and $\{1\} = \{2,1\}\cap\{\pi,3,1\}$

I assume you mean this to be a topology on $\{2,\pi,3,1\}$, which I'll call $X$. I'll call all the non-empty sets $U$, $V$, and $W$, in the order they appear.

Is this a topology? Well, if it is, then no matter which two points I pick from $X$, I should be able to find two subsets of $X$ in my list that separate the two points.

Is that possible? (Try is before looking at the spoiler.)

First, pick some points.

So I'll pick $1$ and $\pi$.

Then, try to find open sets.

Since $U, V, W$ all contain $1$, it is impossible to separate $1$ and $\pi$.

And so we conclude...

Therefore, $X$ with the topology given is not Hausdorff.

• I understood it clearly. Thanks to you and the large community, I can learn it even if I could not go to a great school! – kosmos Feb 18 '14 at 10:43

I will summarise what's in the comments and add a few details:

The set $\{\{2,1\},\{\pi,3,1\},\{2,\pi,3,1\}\}$ you gave is not a topology since it does not contain some intersections. Even if you took that as a subbasis, and turned it into a topology by adding all unions and intersections, 1, 3 and $\pi$ would be on together on every open set they feature in, so the topology would not be Haussdorf.

Essentially, if two or more points are together on every open set they belong to in the topology, then the topology is not able to "see" both points as independent entities, so for instance, a sequence converging to one of them would be converging to both at the same time. The Haussdorf condition is motivated so that we can always distinguish between two points in the topology.

The way this is defined is, for every two points, there exists some pair of disjoint neighbourhoods such that one neighbourhood contains the first point and the other contains the second. That does not mean every neighbourhood of both points satisfies this condition.

Here's what I think you may have missed: The empty set is open!

In a Hausdorff space, two distinct points have disjoint open neighborhoods. The intersection of those two open neighborhoods is empty, and is open.