$X_H$ compact $\Rightarrow X$ compact

general-topology

$X$ is a topological space with equivalence relation $\sim$ defined as: $x \sim y$ if for any $Y$ that is a Hausdorff space and any continuous map $f:X \rightarrow Y$ we have that $f(x) = f(y)$. $X/\sim$ is called the largest Hausdorff quotient of $X$ if we equip it with the quotient topology and if we consider the canonical projection map $\pi_H: X \rightarrow X_H$.

I was able to prove that if $X$ is compact then $X_H$ has to be compact as well.

I have been told that the converse statement does not hold. Are there any counterexamples that could show that this statement is false?

Best Answer

Consider a closed interval with infinitely many copies of the origin.

Take the interval $([-3,3], \tau)$, with a base $B$ for the usual topology, and add to the set $[-3,3]$ infinitely many copies of $0$: $\{(0, n) : n\in \mathbb{N}\}$. To the base $B$, add in copies of a neighbourhood base of $0$ for each copy of $0$. For example, one new base for this topology is $B$ union the set of all sets of the form $(-r, r) - \{0\}\cup \{(0,n)\}$ (where $r$ is a positive real, and $n$ natural).

It's clear from the way the neighborhoods of the copies of the origin are defined that any continuous map to a Hausdorff space must send all the origins to the same point.

You can get an open cover of the line with infinite origins that has no finite subcover as follows: Let your open cover contain $[-3, -1)$, $(1, 3]$, $(-2,2)$, and every open set of the form $(-2, 2) - \{0\}\cup \{(0,n)\}$.


EDIT: To be very explicit, we create a new topological space $(X, \tau)$. Let $X = [-3,3] \cup \{(0, n) : n\in \mathbb{N}\}$ (you can imagine these points as being `borrowed' from $\mathbb{R}^2$, the point is just that they are new points not usually on the interval $[-3,3]$).

We let $\tau$ contain every open set that belongs to the normal Euclidean topology on $[-3,3]$. We also let $\tau$ contain every set of the form $(-r, r) - \{0\}\cup \{(0,n)\}$, where $r$ is real, and $n$ is natural. AND we let $\tau$ contain any arbitrary union or countable intersection of any kind of set mentioned so far.

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