Let $K=\{\frac{1}{n}\mid n \ge1, n \in \mathbb{Z}\}$ and let X the real numbers with the K-topology. Let $Y=X/K$ with the quotient topology.

Question

Let $K=\{\frac{1}{n}\mid n \ge1, n \in \mathbb{Z}\}$ and let X the real numbers with the $K$-topology. Let $Y=X/K$ with the quotient topology. Prove that $Y$ is a $T_1$-space but not $T_2$

I want to prove that $\{[x]\}$ is a closed subset of $Y$. For this, I think I should show that $p^{-1}(Y-\{[x]\})=p^{-1}(Y) -p^{-1}(\{[x]\})$ is open in $X,$ bu I don't know how to apprach this. Could someone help me please? Thanks in advance.

Answer 1

I was actually looking at just this example today. It is a space that provides a counterexample to the following statement (which I very much wanted to be true):

Let $X$ be a Hausdorff space and $\sim$ an equivalence relation on $X$. Then $X/\sim$ is Hausdorff if and only if $\sim$ is closed as a subset of $X\times X$.

This statement appeared (and was subsequently deleted) in:

Let $Y=X/ \sim$ be the topology quotient. Determine if $Y$ is a Hausdorff space.

In your case $X=\mathbb{R}$ and $\sim$ is characterized by $x\sim y$ if $x,y\in K$. Note that $K$ is closed in the $K$ topology (this is most easily seen by passing to a subspace such as $(-1,2)$ for example). Then we note that $X$ is in fact Hausdorff but $not$ $regular$.

It is Hausdorff simply because it is finer than the usual topology on $\mathbb{R}$. It is not regular because you can't separate $0$ and $K$. Why? Assume otherwise and let $U,V$ be basic disjoint open sets of $\mathbb{R}_{K}$ containing $x$ and $K$, respectively. There are two kinds of basic open sets that contain $0$. Specifically, open intervals containing $0$ and the complement of $K$ in open intervals containing $0$. Certainly every ordinary open interval around $0$ will intersect $K$, let alone an open set around $K$. Then $U$ must be of the second variety. That is, if $(a,b)$ is such that $a<0<b<\frac{1}{n}$ (WLOG $\frac{1}{n}$ is the least element of $K$ such that $b<\frac{1}{n}$). Then

$$U=(a,b)\setminus K=(a,0]\cup\left(\frac{1}{n+1},b\right)\cup\left(\bigcup_{m\geq n+1}\left(\frac{1}{m+1},\frac{1}{m}\right)\right)$$

Certainly an odd set to be sure, but it is clear that if $V$ is any open set around $K$ then it will contain an interval around $\frac{1}{n+1}$ which will intersect $U$, so $\mathbb{R}_{K}$ is not regular.

Why did I go through this rigamarole? If you can't separate $0$ from $K$ in $\mathbb{R}_{K}$ then $you$ $won't$ $be$ $able$ $to$ $separate$ $0$ $from$ $the$ $point$ $made$ $by$ $identifying$ $the$ $points$ $of$ $K$ $in$ $\mathbb{R}_{K}/K$. Thus $\mathbb{R}_{K}/K$ is not Hausdorff ($T_{2}$).

Now let's address your actual question. Having established that the quotient space is not Hausdorff, why does $\mathbb{R}_{K}/K$ satisfy the $T_{1}$ axiom? In general, for quotients there is the following fact (shamelessly quoted with minor edits from Munkres):

There is a simple condition for $X/\sim$ to satisfy the $T_{1}$ axiom; one simply requires the each element of the partion $X/\sim$ be a closed subset of $X$.

This is quite easy in our case. There are only two kinds of equivalences classes. Individual points, which are closed in $\mathbb{R}_{K}$ because it is a Hausdorff space. The other is $K$ itself which, as mentioned, is closed. Thus we have that $\mathbb{R}_{K}/K$ is $T_{1}$.

Hope this helps.

Stay tuned for when I inevitably find errors in this solution and frantically edit it.