Results 1 to 9 of 9

Math Help - Unirormly Continuous Rectangles

  1. #1
    Member
    Joined
    Oct 2010
    Posts
    133

    Unirormly Continuous Rectangles

    Let f:[a,b] x [c,d]\rightarrow R be continuous on [a,b] x [c,d]. Prove that f is uniformly continuous on [a,b] x [c,d].

    I understand the concept of uniform continuity. I'm assuming it can't be all that more complicated for a rectangle.

    My idea is that if we can show that f is uniformly continuous on [a,b] and f is uniformly continuous on [c,d] then f is uniformly continuous on [a,b]x[c,d]???

    Any help would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by zebra2147 View Post
    Let f:[a,b] x [c,d]\rightarrow R be continuous on [a,b] x [c,d]. Prove that f is uniformly continuous on [a,b] x [c,d].

    I understand the concept of uniform continuity. I'm assuming it can't be all that more complicated for a rectangle.

    My idea is that if we can show that f is uniformly continuous on [a,b] and f is uniformly continuous on [c,d] then f is uniformly continuous on [a,b]x[c,d]???

    Any help would be appreciated.
    This really is the result of Heine-Cantor. Try using the Lebesgue Covering Lemma


    EDIT: This is probably overkill, but I don't have time to think about it right now. There's absolutely an easier way that the one I mentioned above (maybe) if you're given that a continuous function on closed intervals is unif. cont. Let me think about it.

    I prove the much more general result I alluded to earlier here (last theorem).
    Last edited by Drexel28; November 4th 2010 at 08:02 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2010
    Posts
    133
    Well, I looked at all the links you gave me and they were all very interesting but I think I'm supposed to use the following lemma for this proof but extend it to rectangles...

    Let f:[0,1]\rightarrow \complex be continuous. Suppose f is uniformly continuous on I_{1}=[0,1/2], on I_{2}=[1/4,3/4], and on I_{3}=[1/2,1]. Then f is uniformly continuous on [0,1].

    That is why I previously stated my idea is to show that if f is uniformly continuous on [a,b] and is uniformly continuous on [c,d] then f is uniformly continuous on [a,b] x  [c,d]??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by zebra2147 View Post
    Well, I looked at all the links you gave me and they were all very interesting but I think I'm supposed to use the following lemma for this proof but extend it to rectangles...

    Let f:[0,1]\rightarrow \complex be continuous. Suppose f is uniformly continuous on I_{1}=[0,1/2], on I_{2}=[1/4,3/4], and on I_{3}=[1/2,1]. Then f is uniformly continuous on [0,1].

    That is why I previously stated my idea is to show that if f is uniformly continuous on [a,b] and is uniformly continuous on [c,d] then f is uniformly continuous on [a,b] x  [c,d]??
    Can you explain that lemma a little? It looks very strange and very arbitrary.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2010
    Posts
    133
    Well, I can try. He didn't cover it in class but the way I see it is that you have a continuous interval, I, and you break it into 3 seperate intervals I_{1}, I_{2}, I_{3} and we say that all three of these intervals overlap so that every point in I is contained in at least on of these subintervals. Then, if all three of these subintervals are uniformly continuous, then the interval I must also be uniformly continuous.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by zebra2147 View Post
    Well, I can try. He didn't cover it in class but the way I see it is that you have a continuous interval, I, and you break it into 3 seperate intervals I_{1}, I_{2}, I_{3} and we say that all three of these intervals overlap so that every point in I is contained in at least on of these subintervals. Then, if all three of these subintervals are uniformly continuous, then the interval I must also be uniformly continuous.
    And they, presumably, need to agree on the overlap, right?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2010
    Posts
    133
    Well, that isn't stated anywhere in his notes but based on what his lemma says, it would appear the overlap is consistent.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by zebra2147 View Post
    Well, that isn't stated anywhere in his notes but based on what his lemma says, it would appear the overlap is consistent.
    So are you allowed to extend this to arbitrary closed oovers. In other words, if X is a space, f:X\to\mathbb{R}, and \left\{X_{\alpha}\right\}_{\alpha\in\mathcal{A}} is a closed cover for X such that f_{\mid X_{\alpha}}:X_{\alpha}\to\mathbb{R} uniformly continuous then f:X\to\mathbb{R} is uniformly continuous?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2010
    Posts
    133
    Well my professor never used the term "closed covers" but I looked up a little bit about it and that seems to be what is happening in his lemma. He also doesn't use discrete space terminology in his notes but what you wrote seems to work from what I gather.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. rectangles
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: May 11th 2010, 05:17 PM
  2. Rectangles
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: August 31st 2009, 02:03 AM
  3. How many rectangles
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: May 23rd 2008, 05:43 PM
  4. How many rectangles
    Posted in the Statistics Forum
    Replies: 1
    Last Post: May 20th 2008, 04:54 AM
  5. Rectangles
    Posted in the Geometry Forum
    Replies: 6
    Last Post: April 24th 2007, 04:35 PM

Search Tags


/mathhelpforum @mathhelpforum