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Math Help - Quotient space

  1. #1
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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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  2. #2
    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 \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?
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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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  4. #4
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    Re: Quotient space

    Quote Originally Posted by Drexel28 View Post
    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.
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