quotient space, Hausdorff

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.

• Can you visualize what this space should look like? Say, embedded in some $\mathbb{R}^n$. – G. Chiusole Feb 4 '20 at 16:34
• Yes, I think so. I think about the lines $[-1, -1/2] \cup [1/2, 1]$ and $(- 1/2, 1/2) \times \{ 0, 1 \}$. – lea5619 Feb 4 '20 at 16:41
• Note that the resulting space is connected. – G. Chiusole Feb 4 '20 at 16:44
• Okay, I do not see how this will help to show the statement ... – lea5619 Feb 4 '20 at 22:50

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.