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Kae Araki
Frequent poster

Post Number: 120
 Posted on Wednesday, 27 July, 2011 - 03:09 pm:

{x}, as a topological space, with the only possible topology.
The definition says if for all distinct x and y in X, there exist 2 disjoint open sets containing x and y respectively, X is called a Hausdorff space. Seems to me that {x} looks like Hausdorff as we cannot find two points that cannot be "separated". But the condition of the definition does not even apply, since there is only ONE point in it and you cannot find "distinct x and y".
So is {x} Hausdorff?
Yatir Halevi
Veteran poster

Post Number: 1305
 Posted on Wednesday, 27 July, 2011 - 03:55 pm:

But it does apply
Look at it logically:
$\forall x \forall y(x\neq y\rightarrow\dots$
And remember that $(P\rightarrow Q) \Leftrightarrow ((\neg P)\vee Q)$

Yatir
Zhen Lin
Frequent poster

Post Number: 141
 Posted on Wednesday, 27 July, 2011 - 05:05 pm:

In other words, the topological space $1$ with one point (and note that it is uniquely specified up to homeomorphism!) is vacuously Hausdorff.

One definition of a Hausdorff topological space which makes this fact non-vacuous is the following: A Hausdorff topological space is a topological space $X$ with the property that the diagonal map $\Delta : X \to X \times X$, where $X \times X$ is equipped with the product topology, has a closed image in $X \times X$.

Of course, it's still obvious under that definition that $1$ is Hausdorff. But at least there's no need to fiddle with vacuous truths.
Billy Woods
Veteran poster

Post Number: 950
 Posted on Wednesday, 27 July, 2011 - 07:15 pm:

On a more fundamental level, I think you're misunderstanding the definition. There is no need to find distinct x and y. The topology is Hausdorff if, whenever x and y are distinct, the topology can separate them. That is, write down every distinct pair x and y. If any pair can't be separated in the topology, the topology is not Hausdorff; otherwise, it is. It's pretty easy to write down every pair of distinct points in this space...