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Let $X = [-1,1] \times \{0, 1 \} \subset \mathbb{R}^2$ with the induced topology on $\mathbb{R}^2$. Then $X$ is a Hausdorff space as a subspace of a Hausdorff space.

The question is now, if

$Y = X \setminus_{\sim}$ with $\sim$ induced by $(t,0) \sim (t,1)$ $\forall t \in [-1,- 1/2] \cup [1/2, 1]$ is also Hausdorff ?

I think yes, but I do not see how to prove this statement.

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  • $\begingroup$ Can you visualize what this space should look like? Say, embedded in some $\mathbb{R}^n$. $\endgroup$ – G. Chiusole Feb 4 '20 at 16:34
  • $\begingroup$ Yes, I think so. I think about the lines $[-1, -1/2] \cup [1/2, 1]$ and $(- 1/2, 1/2) \times \{ 0, 1 \}$. $\endgroup$ – lea5619 Feb 4 '20 at 16:41
  • $\begingroup$ Note that the resulting space is connected. $\endgroup$ – G. Chiusole Feb 4 '20 at 16:44
  • $\begingroup$ Okay, I do not see how this will help to show the statement ... $\endgroup$ – lea5619 Feb 4 '20 at 22:50
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Let $p,q\in X/\sim$ with $p\neq q$, and $\pi: X \to X/\sim$ the canonical map. Observe that $(X/\sim)\setminus \{ [(-1/2,0)],[1/2,0] \}$ can be cover with the disjoint open sets of $X/\sim$:

  • $X_1 = \{ [(x,y)] \in X/\sim : x \in [-1,-1/2) \}$
  • $X_2 = \{ [(x,y)] \in X/\sim : (x,y) \in (-1/2,1/2)\times\{0,1 \} \}$
  • $X_3 = \{ [(x,y)] \in X/\sim : x \in (1/2,1] \}$

Each $X_i$ is Hausdorff because $\pi^{-1}(X_i)$ is open and Hausdorff, and $\pi^{-1}(p)$ is finite for each $p\in X_i$. So we only need to prove the Hausdorff property for $[(-1/2,0)]$ with respect to al $X/\sim$ and the same for $[1/2,0]$. This easy again because the preimage by $\pi$ of these points are finite.

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