# Why are these two quotient spaces not homeomorphic?

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.

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.

• Thank you, the perfect answer! Jun 28 at 7:54

$$A_1$$ 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.

• 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? Jun 28 at 7:51
• @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 Jun 28 at 8:28