This is an example of from the book Topology of Janich. In the picture $X = \mathbb{R}^2$ with standard topology, and the lines represent the equivalence classes, which are closed $1$ dimensional manifolds. The example is aimed to show: Even if the equivalence classes are closed subsets of the Hausdorff space $\mathbb{R}^2$, the quotient space $X/\sim$ could still be non-Hausdorff. It says that one quotient space in the picture is Hausdorff while the other is not. However I am having hard time finding out which is. Could somebody help me please.


Suppose that the lines $l_1 = \{(-1,y) : y \in \Bbb{R} \}$ and $l_2 = \{(1,y) : y \in \Bbb{R} \}$ are the border lines where $\Bbb{R}^2$ starts to decomposed into curvy lines. Say the quotient space on the left is $X_1$ and quotient space on the right is $X_2$.

You can observe that there are no open subsets in $X_2$ that separate $[l_1]$ and $[l_2]$. You can easily construct disjoint neighbourhoods for $[l_1]$ and $[l_2]$ in $X_1$ by the following : Choose an "$s$"-curve on the region between $l_1$ and $l_2$. The set $\Bbb{R}^2$ without this "$s$"-curve is an open subset consist of two disjoint open subset $U_1$ and $U_2$ in $\Bbb{R}^2$ with $l_1 \subseteq U_1$ and $l_2 \subseteq U_2$. The image of these subsets are disjoint neighbourhoods for $[l_1]$ and $[l_2]$. When you do the same for $X_2$, that is take a complement of a "$c$"-curve between $l_1$ and $l_2$, it will fails, since one of the neighbourhoods contain both $l_1$ and $l_2$. enter image description here

To show that there are no disjoint open subsets in $X_2$ that separate $[l_1]$ and $[l_2]$, you can try as follows : Let $q :\Bbb{R}^2 \to X_2$ be the quotient map. Suppose $A_1$ and $A_2$ are open subsets in $X_2$ such that $A_1$ contain $[l_1]$ and $A_2$ contain $[l_2]$. Since $l_1 \subseteq q^{-1}(A_1)$ and $l_2 \subseteq q^{-1}(A_2)$ are both open in $\Bbb{R}^2$. Choose a point on the line $l_1$, any open ball must contain the tail of a "$c$"-curve. Hence $q^{-1}(A_1)$ contain the tails of all "$c$"-curves above it. By doing the same thing for $l_2$, we obtain a "$c$"-curve that both of its tails contain in $q^{-1}(A_1)$ and $q^{-1}(A_2)$. Therefore $A_1 \cap A_2 \neq \emptyset$.

  • $\begingroup$ One of the neighbourhoods contain both $l_1$ and $l_2$? The separation should not be that bad. $\endgroup$ – edm Sep 25 '18 at 15:47
  • $\begingroup$ @edm What do you mean ? I mean it is not always like that. Only if we take $\Bbb{R}^2 $ minus a curve in between. $\endgroup$ – Sou Sep 25 '18 at 15:50
  • $\begingroup$ I mean, in the last sentence, you seem to say that any neighbourhood of $l_1$ would also contain $l_2$ (and vice versa), but I don't think it is the case. $\endgroup$ – edm Sep 25 '18 at 15:52
  • $\begingroup$ @edm Ok. I'll edit that part. $\endgroup$ – Sou Sep 25 '18 at 15:53

The picture on the right gives a non-Hausdorff quotient space. Specifically, the two vertical lines $l_1$ and $l_2$ next to the cap-shaped lines do not have disjoint open neighbourhood. (from then on, I call cap-shaped lines "cap" lines)

We start at a point $x_1$ on one line $l_1$ of the two lines and see what its neighbourhood contains. Draw an open ball about that point. This ball intersects a certain "cap" line, say $c_1$. Note also that when this ball intersect $c_1$, then this ball also intersects all the "cap" lines above $c_1$. So here, our attempt to construct an arbitrary open neighbourhood $U_1$ (an open set in quotient space) of $l_1$ shows that this neighbourhood contains $c_1$ and all "cap" lines above $c_1$.

By the same argument applied to the other line $l_2$, an open neighbourhood $U_2$ (an open set in the quotient space) of $l_2$ shall contain a certain "cap" line $c_2$ and all the cap lines above $c_2$.

This is how $U_1\cap U_2$ is non-empty: there are "cap" lines which are simultaneously above $c_1$ and $c_2$, and such lines are elements in $U_1\cap U_2$, hence making $U_1\cap U_2$ non-empty.


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