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Thread: Quotient space

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    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.
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    MHF Contributor Drexel28's Avatar
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    Re: Quotient space

    Quote Originally Posted by kierkegaard View Post
    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 $\displaystyle \mathbb{R}$ 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?
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    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.
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    Re: Quotient space

    Quote Originally Posted by Drexel28 View Post
    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?
    Sorry, my last post was wrong. I don't understand your last statement. How $\displaystyle X-C$ an open set not containing $\displaystyle Q$ implies that $\displaystyle p^{-1}(Q-C)$ is an open set not containing $\displaystyle (0,1)$? Thanks you.
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