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**Drexel28** So what do you think? Since the only elements of $\displaystyle \mathbb{R}$ that get collapsed to the same point as $\displaystyle (0,1)$ are points of $\displaystyle (0,1)$ I think you can conclude that $\displaystyle p^{-1}(Q)=(0,1)$ and so $\displaystyle Q$ is open by definition. Now, suppose that $\displaystyle X-Q$ is open, then you have that $\displaystyle p^{-1}(Q-X)$ is open, but what does this set look like? Now, suppose that you had a closed set $\displaystyle C$ containing $\displaystyle Q$, then $\displaystyle X-C$ is an open set not containing $\displaystyle Q$ and so $\displaystyle p^{-1}(Q-C)$ is not an open set containing $\displaystyle (0,1)$. So, what do the intersetion of all those sets look like?