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Let $X = \mathbb{R}$ and $x\sim_1 y \iff x,y \in \mathbb{Z}$. Let $A_1 = X / \sim_1$.

Let $Y = [0,1]$ and $x \sim_2 y \iff x,y \in \left\{ \frac{1}{n} \; | \; n \in \mathbb{N} \right\} \cup \{ 0 \} $. Let $A_2 = Y / \sim_2$ .

I am trying to show that these two spaces are not homeomorphic and also geometrically understand why this is so. Geometrically I think of both of these two spaces as a flower with countably many leaves (empty leaves, with just the boundary) or as a Hawaiian earring. I can not see, geometrically, why these two spaces aren't the same.

The second part I have troubles is with how to formally prove they are not homeomorphic. My current idea is to show that $A_1$ is not compact, while $A_2$ is. I know that $A_2$ is compact, since it is a continuous image (by quotient projection) of a compact space $Y$, which is then itself compact. But I do not know how to show that $A_1$ is not compact.

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2 Answers 2

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One open cover of $X$ that does not have a finite subcover is given by taking the open sets $U_n \colon n \in \mathbb{Z}$ whose pre-image in $\mathbb{R}$ is the open interval $(n,n+1)$ (i.e. an open segment of each loop), Then throw in one extra open set that contains the centre of the "petal": its preimage in $\mathbb{R}$ would be something like $\bigcup_{m\in\mathbb{Z}} (m- \varepsilon, m+\varepsilon)$ for some small $\varepsilon$.

This open cover doesn't have a finite subcover because most of each loop is only in one open set so all the open sets are necessary.


The difference between the two spaces $X$ and $Y$ is what's going on at $0$ in $Y$. That point $0$ acts like a limit point of all the loops in the sense that any open set containing $0$ contains infinitely many of the loops.

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  • $\begingroup$ Thank you, the perfect answer! $\endgroup$
    – Matthew
    Jun 28 at 7:54
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$A_1$ is not compact essentially for the same reasons that $\Bbb R$ is not compact. Consider the open cover of $\Bbb R$: $$V_{\kappa}=(\kappa+1/4,\kappa+3/4)\cup\bigcup_{n\in\Bbb Z}(n-1/3,n+1/3)$$For $\kappa\in\Bbb Z$. Now map it into $A_1$, where it remains an open cover. A finite subcover, indexed by some $\{\kappa_k\}_{k=1}^m\subset\Bbb Z$, would supposedly have the preimage of the whole of $\Bbb R$ but this cannot be. E.g. consider that for all natural $N$, for which $N\gt\kappa_k$ for all $k$, will have $(N+1/3,N+2/3)$ not contained in this union.

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  • $\begingroup$ I don't think that this open cover of $\mathbb{R}$ maps to an open cover of $A_1$. Let $q$ be the canonical projection. For example: $q^{-1}(q(-2/3, 2/3)) = (-2/3, 2/3) \cup \mathbb{Z}$ which is not an open set? $\endgroup$
    – Matthew
    Jun 28 at 7:51
  • $\begingroup$ @Matthew You’re right, I stupidly took the preimage of all of them simultaneously and then the integers were contained within, so it appeared open. I’ve posted a fixed solution which should work now $\endgroup$
    – FShrike
    Jun 28 at 8:28

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