Results 1 to 10 of 10

Math Help - metric space

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    154

    metric space

    In a metric space (S,d), let A and B be disjoint closed subsets of S. Prove that \exists disjoint open sets U and V of S such that A \subset U and B \subset V.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2006
    From
    Florida
    Posts
    228
    Let r=\inf_{x\in{A},y\in{B}}d(x,y), then r>0 since they are disjoint and closed (I'll leave it to you to show that this is true). Then for each point x\in{A} create the neighborhood N_{\frac{r}{3}}(x) (neighborhood about x with radius r/3), similarly create N_{\frac{r}{3}}(y) for y\in{B}.

    Finally take U=\bigcup_{x\in{A}}N_{\frac{r}{3}}(x) and V=\bigcup_{y\in{B}}N_{\frac{r}{3}}(y).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by putnam120 View Post
    Let r=\inf_{x\in{A},y\in{B}}d(x,y), then r>0 since they are disjoint and closed
    This is not true: Take the graph of f(x)= \frac{1}{x^2} in \mathbb{R} ^2 it is like two hyperbolas that get arbitrarily close as you approach 0 (they both go to infinity). It is clearly closed (each 'hyperbola' is also closed), and the two 'hyperbolas' are disjoint but \inf  =0. for this to work either A or B must be compact
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2006
    From
    Florida
    Posts
    228
    I don't think that f(x)=\frac{1}{x^2} is closed. There is no point so that f(x)=0. Or is there some other definition of closed that you are using?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by putnam120 View Post
    I don't think that f(x)=\frac{1}{x^2} is closed. There is no point so that f(x)=0. Or is there some other definition of closed that you are using?
    I'm saying the graph of f is (topologically) closed as a subset of \mathbb{R} ^2 (it's easy to check the complement is open).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Kat-M View Post
    In a metric space (S,d), let A and B be disjoint closed subsets of S. Prove that \exists disjoint open sets U and V of S such that A \subset U and B \subset V.
    For each point x\in S, define d(x,A) = \inf\{d(x,s):s\in A\}. Then d(x,A) = 0\ \Longleftrightarrow\ x\in A. (That is because A is closed. In fact, that condition is equivalent to A being closed.) Similarly, define d(x,B) = \inf\{d(x,s):s\in B\}. Again, this will be 0 if and only if x\in B.

    Now let U = \{x\in S: d(x,A)<d(x,B)\} and V = \{x\in S: d(x,A)>d(x,B)\}. All you have to do now is to show that U and V are open.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2008
    Posts
    154

    closed set

    Quote Originally Posted by Opalg View Post
    For each point x\in S, define d(x,A) = \inf\{d(x,s):s\in A\}. Then d(x,A) = 0\ \Longleftrightarrow\ x\in A. (That is because A is closed. In fact, that condition is equivalent to A being closed.)
    so if x \in S is an elemnt of B, d(x,A) > 0 since x \notin A. Is that correct? if so, doesnt putnam120's argument work?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Kat-M View Post
    so if x \in S is an elemnt of B, d(x,A) > 0 since x \notin A. Is that correct? if so, doesnt putnam120's argument work?
    That is correct: if x\in B then d(x,A) > 0. But putnam120's argument doesn't work for the reason pointed out by Jose27. The distance from B to A can be zero, as shown by the example of the two components of the graph of y=1/x^2. But the distance from any individual point in B to A must be strictly positive.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Nov 2006
    From
    Florida
    Posts
    228
    Quote Originally Posted by Kat-M View Post
    so if x \in S is an elemnt of B, d(x,A) > 0 since x \notin A. Is that correct? if so, doesnt putnam120's argument work?
    My argument doesn't work because I took the \inf over all pairs x\in{A} and y\in{B}. But Opalg's argument you look at a particular x\in{S}. To see that d(x,A)>0 if x\notin{A} just use the limit point definition of a closed set.

    I have a question for Jose27: How are you able to see that my original argument would work if one of A or B were compact?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jun 2009
    Posts
    113
    Quote Originally Posted by putnam120 View Post
    My argument doesn't work because I took the \inf over all pairs x\in{A} and y\in{B}. But Opalg's argument you look at a particular x\in{S}. To see that d(x,A)>0 if x\notin{A} just use the limit point definition of a closed set.

    I have a question for Jose27: How are you able to see that my original argument would work if one of A or B were compact?
    Let me get into this nice discussion. If A is compact d(\cdot,B):A\to R^+,  x\mapsto d(x,B) is a continuous function, hence the infimum is attained (it is a minimum I mean). It can't be 0, if this were the case x would belong to B since B is assumed to be closed, contradicting the assumed disjointness.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 8th 2011, 03:16 PM
  2. Is it a metric space?
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 3rd 2011, 07:44 PM
  3. Limit of function from one metric space to another metric space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 17th 2010, 03:04 PM
  4. Sets > Metric Space > Euclidean Space
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 25th 2010, 11:17 PM
  5. Metric Space
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 11th 2009, 04:47 AM

Search Tags


/mathhelpforum @mathhelpforum