Results 1 to 3 of 3

Thread: Compactness

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    22

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    a) $\displaystyle M=\{(x,y)|0<x<1 , 0 \leq y \leq 1\}$
    you can prove why

    b) Taken an open cover of $\displaystyle K_x$ 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 $\displaystyle a \in A$ then the corresponding open set in B is $\displaystyle a \times \mathbb{R}$. This is an open set since it is a product of two open sets in $\displaystyle \mathbb{R}^2$, 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 $\displaystyle K_x$ 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) $\displaystyle \Delta = \{(x,x)\in \mathbb{R}^2|x\in \mathbb{R}\}$ 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 $\displaystyle \Delta$ is closed in $\displaystyle \mathbb{R}^2$ but its projection is $\displaystyle \mathbb{R}$ which is open.


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

    Hope that helps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by Gamma View Post
    c) $\displaystyle \Delta = \{(x,x)\in \mathbb{R}^2|x\in \mathbb{R}\}$ 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 $\displaystyle \Delta$ is closed in $\displaystyle \mathbb{R}^2$ but its projection is $\displaystyle \mathbb{R}$ which is open.


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

    To find a closed set in $\displaystyle \mathbb{R}^2$ whose projection is not closed, you need to look at something like $\displaystyle \{(x,1/x):x>0\}$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compactness
    Posted in the Differential Geometry Forum
    Replies: 11
    Last Post: Mar 11th 2010, 11:17 AM
  2. Compactness
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Jan 19th 2010, 01:12 AM
  3. compactness
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 26th 2009, 11:26 AM
  4. Topological Compactness and Compactness of a Set
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Sep 22nd 2009, 12:16 AM
  5. Compactness
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Aug 16th 2007, 06:00 AM

Search Tags


/mathhelpforum @mathhelpforum