# A question on Hausdorff spaces

Definition: A topological space $X$ is called Hausdorff space if for each $x_1,x_2 \in X$ (they are distinct) we can always find neighborhoods $U_1,U_2$ of $x_1,x_2$ such that $U_1 \cap U_2 = \varnothing$.

If a space be Hausdorff with respect to a topology, is it Hausdorff with respect to others? And if no, why do we define a topological manifold to be Hausdorff without considering the induced topology by the differentiabe atlas ?

• It may or may not be depending on the specific topology you have. Oct 28, 2015 at 7:57
• Consider a two point set in $\Bbb{R}$ equipped with the standard topology. It is Hausdorff . Consider the same two point set equipped with the indiscrete topology
– user210387
Oct 28, 2015 at 7:57

No; for any space $X$ the set $\mathcal{T}=\{\varnothing,X\}$ defines a topology on $X$, which is not Hausdorff if there are (at least) two points in $X$.

• This is the Indiscrete topology. So T is not Hausdorff’,correct?
– user837396
Feb 6, 2021 at 14:15

No it is not. The Hausdorff property really depends on the topology you put on $X$. Take $X=\mathbb{R}$ then for the usual topology $X$ is clearly Hausdorff whereas for the topology given by $\mathcal{T}_0:=\{\emptyset,\mathbb{R}\}$ is not Hausdorff.

More generally take $X$ to be a set with at least two elements. Then the topology on $X$ given by $\mathcal{T}_0:=\{\emptyset,X\}$ is never Hausdorff whereas the topology on $X$ given by $\mathcal{T}_d:=\mathcal{P}(X)$ (all the subsets of $X$) is always Hausdorff.

As you can see the definition invokes open sets around certain points . A set can be open in one topology but might be closed in another

A space is a pair $(X, \tau)$ where $\tau$, satisfies the definition of a topology. $\tau$ is the collection of open sets of the space. The underlying set almost doesn't have to explicitly part of the structure, as it can be recovered from the topology: $X = \bigcup \tau$.

It's a topology that is Hausdorff or not. Sets alone are not "Hausdorff" or "non-Hausdorff" any more than a set (alone) can be "Abelian" or not.

Note that the definition of Hausdorff is a condition on $\tau$:

$\tau$ is Hausdorff $\iff$ for all $x, y \in X$ with $x \neq y$, there are $U, V \in \tau$ such that $x \in U, y \in V$ and $U \cap V = \emptyset$.

Yes, the same set can have different topologies, some Hausdorff and others not. Example: any 2-element set with the discrete topology (Hausdorff), and with the indiscrete topology (not).