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The text of the problem is the following:

$X = [0,1] \times \{0,1\} $ with the topology induced by the euclidean topology, is an Hausdorff space. Let's define an equivalence relation: $(x_1,y_1)R(x_2,y_2)$ iff:

1)$(x_1,y_1)=(x_2,y_2);$

2)$x_1 = y_1 $ and $x_2 = 0, \ y_2 = 1$ or viceversa.

Show that $X/R$ is not Hausdorff.

Firstly it isn't to clear to me what viceversa does it mean. I interpreted it in the following way: $x_2=y_2$ and $x_1 =0, y_1 = 1$.

Now I will try to explain my thoughts. The quotient space $X/R$ can be thought as a segment of the following form: $[0,1)$. Let's define the set $S = \{(0,0),(1,0),(0,1),(1,1)\}$ and the projection:

$\pi(x,y) \to x \ if (x,y) \notin S$

$\pi(x,y) \to \{0\} \ if (x,y) \in S$

$\forall z_1,z_2 \in X/R $, without loss of generality we can take $z_1 < z_2$. Now I will define two disjoint and open sets (in the quotient topology), containing $z_1$ and $z_2$.

$A_1 = [0,(z_1+z_2)/2) \ \bigcup \ ((z_2+1)/2,1);$

$A_2= ((z_1+z_2)/2,(z_2+1)/2)$.

So the quotient seems to be Hausdorff. Where is my mistake?

Any help is appreciated. Thanks in advance.

Luigi
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    Vice versa most certainly means $x_1 = y_1$, $x_2 = 1$, $y_2 = 0$. – Arno Dec 30 '22 at 13:17
  • @Arno
    Firstly thanks for your help. I also thought about that possibility. In that case the quotient topology can be interpreted as a closed segment $[0,1]$, with the obvious projection, and for every $z_1,z_2$ I can find two open disjoint subsets:

    $z_1 \in A_1 = [0,(z_1+z_2)/2)$ and $z_2 \in A_2 = ((z_1+z_2)/2,1] $.

     – Luigi Dec 30 '22 at 13:23
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    Maybe this is a meta problem: find out which of the possible interpretations of "vice versa" gives a non-Hausdorff quotient topology :-/ – Lee Mosher Dec 30 '22 at 14:56
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    $x_1=y_1$ over $X$ can be satisfied by two points only. And so I don't see a way to interpret (2) so that infinitely many pairs of points satisfy it. But then $X/R$ is Hausdorff, because $R$ is closed in $X\times X$. There's something really missing here. – freakish Dec 30 '22 at 16:14
  • @freakish Thanks for your contribution. I don't understand your argument about infinitely many pairs of points, and why R would be closed. Could you please explain a little more? – Luigi Dec 30 '22 at 16:37
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    @alfa let $\Delta={(v,v)\ |\ v\in X}\subseteq X\times X$. Clearly $\Delta\subseteq R$, like with any equivalence relation. Note that $\Delta$ represents your condition (1). For Hausdorff spaces $\Delta$ is a closed subset of $X\times X$. Now if $R=\Delta\cup F$ where $F$ is finite (in your case $F$ represents condition (2)) then $R$ is closed in $X\times X$ as a union of two closed subsets. And that is equivalent to $X/R$ being Hausdorff. – freakish Dec 30 '22 at 21:35
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    And so if $R=\Delta\cup F$ (with $F$ disjoint from $\Delta$) then the only way for $X/R$ to be non-Hausdorff is when $F$ is infinite. That's necessary, but not sufficient of course. But I don't see a way to interpret (2) so that it produces infinitely many pairs in $R$. Where did you get that problem from? Some book? – freakish Dec 30 '22 at 21:52
  • @freakish in our university they give us problems sheet where i found this problem. You are using the fact that the projection is an open map to state the equivalence between $R$ being closed and $X/R $ being Hausdorff? – Luigi Dec 30 '22 at 22:37
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    @alfa Ahh, sorry, the equivalence is when $X$ is compact Hausdorff. Regardless of whether the quotient map is open or not. See this: https://math.stackexchange.com/questions/1093373/question-about-quotient-of-a-compact-hausdorff-space and apply it to the quotient map $\pi:X\to X/R$. – freakish Dec 30 '22 at 22:46
  • @freakish does not my interpretation or the one suggested by Arno produces infinite many pairs in R? And I don't understand why $ \Delta $ is finite? Doesn't it have the same cardinality of $[0,1]$? – Luigi Dec 30 '22 at 22:48
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    @alfa I didn't say $\Delta$ is finite. I said it is closed. I also said that $F$ is finite (and thus closed), because the condition (2) can only be satisfied by finitely many pairs. Here $F$ is the set of all pairs satisfying (2). And these are only $(0,0)R(0,1)$ and $(1,1)R(0,1)$, right? Plus "vice versa", i.e. $(0,1)R(0,0)$ and $(0,1)R(1,1)$. Or maybe $(0,0)R(1,0)$, and so on, like Arno suggests. Either way it is finitely many pairs, because only $0$ and $1$ can participate in this definition (because $x_1=y_1$ implies $x_1=y_1=0$ or $x_1=y_1=1$). And so there's something deeply wrong here. – freakish Dec 30 '22 at 22:50
  • @freakish Sorry for my mistake, confusing closed and finite. I would say that all the pairs that have the same element in the first coordinate and a "different" second coordinate satisfy this condition. Such that $(s,1)R(s,0) \ \forall \ s \in [0,1]$ – Luigi Dec 30 '22 at 23:04
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    @alfa yeah, but that is Hausdorff as well. For different reasons. This just glues two intervals together, in fact $X/R$ is homeomorphic to $[0,1]$. – freakish Dec 30 '22 at 23:08
  • @Arno so following my interpretation $F$ would be infinite, but $X/R$ would be Hausdorff anyway, because it would be homeomorphic to $[0,1]$. – Luigi Dec 30 '22 at 23:19
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    @alfa yes, that's correct. As I said earlier: for $X/R$ to be non-Hausdorff we need $F$ to be infinite. But it is not a sufficient condition, as your example shows. – freakish Dec 31 '22 at 07:25
  • @alfa all in all: I think you have to ask someone from your university about this problem. And present him with all the findings we've talked about here. Because as it is, I don't see a way to fix it. – freakish Dec 31 '22 at 11:47

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