# Thread: Quotient space

1. ## Quotient space

Let X be the space obtained from the real line R (with the usual topology) by identifying all points in the open interval (0, 1) to a single point, and let p : R → X be the quotient map. Show that Q := p((0, 1)) is an open set, but not closed in X. Determine the closure of Q in X.

2. ## Re: Quotient space

Originally Posted by kierkegaard
Let X be the space obtained from the real line R (with the usual topology) by identifying all points in the open interval (0, 1) to a single point, and let p : R → X be the quotient map. Show that Q := p((0, 1)) is an open set, but not closed in X. Determine the closure of Q in X.
So what do you think? Since the only elements of $\mathbb{R}$ that get collapsed to the same point as $(0,1)$ are points of $(0,1)$ I think you can conclude that $p^{-1}(Q)=(0,1)$ and so $Q$ is open by definition. Now, suppose that $X-Q$ is open, then you have that $p^{-1}(Q-X)$ is open, but what does this set look like? Now, suppose that you had a closed set $C$ containing $Q$, then $X-C$ is an open set not containing $Q$ and so $p^{-1}(Q-C)$ is not an open set containing $(0,1)$. So, what do the intersetion of all those sets look like?

3. ## Re: Quotient space

Thanks for your help. But, I don't still understand your argument why the quotient map p is not closed. If you clarify a little more that point, I would really appreciate.

4. ## Re: Quotient space

Originally Posted by Drexel28
So what do you think? Since the only elements of $\mathbb{R}$ that get collapsed to the same point as $(0,1)$ are points of $(0,1)$ I think you can conclude that $p^{-1}(Q)=(0,1)$ and so $Q$ is open by definition. Now, suppose that $X-Q$ is open, then you have that $p^{-1}(Q-X)$ is open, but what does this set look like? Now, suppose that you had a closed set $C$ containing $Q$, then $X-C$ is an open set not containing $Q$ and so $p^{-1}(Q-C)$ is not an open set containing $(0,1)$. So, what do the intersetion of all those sets look like?
Sorry, my last post was wrong. I don't understand your last statement. How $X-C$ an open set not containing $Q$ implies that $p^{-1}(Q-C)$ is an open set not containing $(0,1)$? Thanks you.