# Quotient space of $\mathbb{R}^2$

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 $$l_2$$.

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 $$A_1 \cap A_2 \neq \emptyset$$.

• One of the neighbourhoods contain both $l_1$ and $l_2$? The separation should not be that bad. – edm Sep 25 '18 at 15:47
• @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. – Sou Sep 25 '18 at 15:50
• 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. – edm Sep 25 '18 at 15:52
• @edm Ok. I'll edit that part. – 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.