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Math Help - Compactness

  1. #1
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    Compactness

    For a set M ⊂ R^2, we define its shadow (projection) on the x-axis to be the set in R defined as

    M_x={ x: there is some y so that (x,y) ∈ M }

    a) Find a set M, so that M_x is open (in R) but M is not open (in R^2).

    b) Suppose K is a compact set in R^2. Prove that K_x is also compact in R.

    Hint: Either prove …first "if A is open in R, then A R is open in R^2", then start with an open cover C of K_x and create a cover for K.
    Or: start with a sequence (x_n) in K_x and use that K is sequentially compact to show that (x_n) has a cluster point in K_x.

    c) Find a closed set N in R^2 such that N_x is not closed (in R) .



    If anyone helps me, I will be thankful.
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  2. #2
    Super Member Gamma's Avatar
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    a) M=\{(x,y)|0<x<1 ,  0 \leq y \leq 1\}
    you can prove why

    b) Taken an open cover of K_x call it A. Look at B where B consists of the inverse of the projection map of each open set in A. For instance if a \in A then the corresponding open set in B is a \times \mathbb{R}. This is an open set since it is a product of two open sets in \mathbb{R}^2, so we notice B is an open cover of K, a compact set, so it has a finite subcover. Then look at the projections of each of these sets and it is a finite subcover of K_x and thus its compact.

    I dunno how to do the fancy script A, so I apologize for the awkward notation of a representing a set and A being a collection of sets...

    c) \Delta = \{(x,x)\in \mathbb{R}^2|x\in \mathbb{R}\} This is the diagonal and it is probably a theorem you have done that is pretty easy to show from definitions that hausdorff iff diagonal is closed so \Delta is closed in \mathbb{R}^2 but its projection is \mathbb{R} which is open.


    If you dont believe me that \Delta is closed, just look at \mathbb{R}^2-\Delta and see why it's open.

    Hope that helps.
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  3. #3
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    Opalg's Avatar
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    Quote Originally Posted by Gamma View Post
    c) \Delta = \{(x,x)\in \mathbb{R}^2|x\in \mathbb{R}\} This is the diagonal and it is probably a theorem you have done that is pretty easy to show from definitions that hausdorff iff diagonal is closed so \Delta is closed in \mathbb{R}^2 but its projection is \mathbb{R} which is open.


    If you dont believe me that \Delta is closed, just look at \mathbb{R}^2-\Delta and see why it's open.
    \Delta is certainly closed, but its projection is \mathbb{R}, which is closed as well as open.

    To find a closed set in \mathbb{R}^2 whose projection is not closed, you need to look at something like \{(x,1/x):x>0\}.
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