7
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $A_1 \cap A_2 \neq \emptyset$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.