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$X=[-1,1]\times \mathbb{Z}$ and define an equivalence relation on X via

$(t,n)$~$(t´,n´)$ if and only if $(t,n)$=$(t´,n´)$ or $t=t´\ne 0$.

(You may assume that this is indeed an equivalence relation.) Let $p: X → X/∼$ denote the quotient map and set Y := X/∼.

a) Show that Y is not compact

b) Show that Y is not Hausdorff

c) Show that $p([−1, 1]\times \{0\})$ is compact and calculate its closure.

a) and b) I'm not really sure how to proceed)

But in c) I thought that the set $[−1, 1]\times \{0\}$ is a product of two compact spaces hence compact. And the image of a compact space under a continuous map is compact.

The image $p([−1, 1]\times \{0\})$ is the equivalence class $[(t,0)]$, but I'm quite unsure what it even means to be open in this space X~.

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    What topology do you conisder on $X$? Subspace topology inherited from $\mathbb{R}^2$? – 2017-01-24
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    Nothing is written in the work, so I assume this is the case – 2017-01-24

1 Answers 1

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Notice that $[(0,n)]=\left\{(0,n)\right\}$. Clearly this element is not open in $Y$ as $p^{-1}([(0,n)])=\left\{(0,n)\right\}$ is not open in $X$. This observation is going to be fundamental in understanding this problem.

If we want to find an open cover of $Y$ that doesn't admit a finite subcover, we should focus on the points $[(0,n)]$. Let's find opens in $Y$ that actually contain the element $[(0,n)]$.

Consider $O_n:=p^{-1}(\bigcup_{\alpha\in [-1,0)} [\alpha,0]\cup [0,n]\cup \bigcup_{\beta\in (0,1]}[\beta,0])$. By drawing this, is not difficult to see that this set is open in $X$. Hence $Y_n:=\bigcup_{\alpha\in [-1,0)} [\alpha,0]\cup [0,n]\cup \bigcup_{\beta\in (0,1]}[\beta,0])$ is open in $Y$ and moreover $[0,m]\in Y_n$ if and only if $n=m$.

Notice that $\bigcup_{n\geq 0}O_n=Y$ but it does not admit a finite subcover since we need all $O_n$ to cover the elements $[0,m]$. This shows 1). Try to understand this argument (by drawing it), maybe this helps you to find the other yourself. Good luck.

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    Feel free to ask more details if needed, the notation is quite dense. – 2017-01-24
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    I would love to see an argument for the non Hausdorffness. I tried to take neighborhoods of (0,n) in the quotient, but with distinct n they have different second coordinates and so are disjoint, no? – 2017-04-30
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    @TobiasShuxueLaoshi: It's probably easier to calculate the closure $C$ of $p([-1,1]\times \left\{0\right\})$. Notice that $C\neq p([-1,1]\times \left\{0\right\})$ but $p([-1,1]\times \left\{0\right\})$ is compact. From this it follows that this space is not Hausdorff as in a Hausdorff space compact sets need to be closed. – 2017-04-30