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Show R^2 is locally compact with non-standard metric

  1. Nov 30, 2005 #1

    benorin

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    The Q: Define the distance between points [itex]\left( x_1 , y_1\right) [/itex] and [itex]\left( x_2 , y_2\right) [/itex] in the plane to be

    [tex]\left| y_1 -y_2\right| \mbox{ if }x_1 = x_2 \mbox{ and } 1+ \left| y_1 -y_2\right| \mbox{ if }x_1 \neq x_2 .[/tex]

    Show that this is indeed a metric, and that the resulting metric space is locally compact. I need help with the second part.

    My A: Write

    [tex]d\left( \left( x_1 , y_1\right) , \left( x_2 , y_2\right) \right) = \delta_{x_1}^{x_2} + \left| y_1 -y_2\right| ,[/tex]

    where

    [tex]\delta_{x_1}^{x_2}=\left\{\begin{array}{cc}0,&\mbox{ if }
    x_{1} = x_{2}\\1, & \mbox{ if } x_{1} \neq x_{2}\end{array}\right.[/tex]

    is the Kronecker delta function. Then [itex]d:\mathbb{R} ^2 \times \mathbb{R} ^2 \rightarrow \mathbb{R}[/itex] is a metiric on [itex]\mathbb{R} ^2[/itex] since the following hold:

    i. d is positive definite since d is obviously positive and

    [tex]\delta_{x_1}^{x_2}=0 \Leftrightarrow x_{1} = x_{2} \mbox{ and } \left| y_1 -y_2\right| = 0 \Leftrightarrow y_{1} = y_{2}[/tex]

    ii. d is symmetric in its variables, that is

    [tex]d\left( \left( x_1 , y_1\right) , \left( x_2 , y_2\right) \right) = \delta_{x_1}^{x_2} + \left| y_1 -y_2\right| = \delta_{x_2}^{x_1} + \left| y_2 -y_1\right| = d\left( \left( x_2 , y_2\right) , \left( x_1 , y_1\right) \right)[/tex]

    iii. d the triangle inequality, that is: if [itex]\left( x_j , y_j\right) \in \mathbb{R} ^2, \mbox{ for } j=1,2,3,[/itex] then

    [tex]d\left( \left( x_1 , y_1\right) , \left( x_2 , y_2\right) \right) \leq d\left( \left( x_1 , y_1\right) , \left( x_3 , y_3\right) \right) + d\left( \left( x_3 , y_3\right) , \left( x_2 , y_2\right) \right) ,[/tex]

    which can be reasoned thus: the triangle inequality in R^2 with the Euclidian metric gives

    [tex]\left| y_1 -y_2\right| \leq \left| y_1 -y_3\right| + \left| y_3 -y_2\right| , \forall y_{1},y_{2},y_{3}\in\mathbb{R}[/tex]

    and

    [tex]\delta_{x_1}^{x_2} \leq \delta_{x_1}^{x_3} + \delta_{x_3}^{x_2} \mbox{ holds } \forall x_{1},x_{2},x_{3}\in\mathbb{R}[/tex]

    for suppose not: then

    [tex]\exists x_{1},x_{2},x_{3}\in\mathbb{R} \mbox{ such that }\delta_{x_1}^{x_2} > \delta_{x_1}^{x_3} + \delta_{x_3}^{x_2} \Leftrightarrow x_1 \neq x_2 \mbox{ and } x_1 = x_3 = x_2 ,[/tex]

    which is a contradiction; adding these inequalities yields the required result, viz. the triangle inequality.

    By i,ii, and iii, d is a metric on [itex]\mathbb{R} ^2[/itex].

    The locally compact part I don't get: a metric space is locally compact iff every point of has a neighborhood with compact closure.

    An open neighborhood of a point, say [itex]\left( x_0 , y_0\right) [/itex], is given by: for some k>0, put

    [tex]\left\{ \left( x , y\right) : d\left( \left( x , y\right) , \left( x_0 , y_0\right) \right) < k \right\}[/tex]

    but what does that look like? How do I grasp what compact means in this metric?

    The delta function above is the discrete metric on R^1 and the absolute value is the Euclidian metric on R^1, and their sum is indeed a metric on the product space R^2. Do I get to keep Heine-Borel? Does Heine-Borel even hold for R^1 with the discrete metric? I don't get the idea of compact sets with H-B, I can tell you "A set is compact if every open cover has a finite subcover," but that topology stuff is so abstract. What does it mean for a set to be compact in terms of a given metric? Is that given by sequential compactness?

    Please help with the second part, and let me know if the first is ok.

    Thanks,
    -Ben

    PS: Please don't answer all the questions in the last paragraph, just the ones that help.
     
  2. jcsd
  3. Nov 30, 2005 #2

    matt grime

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    As is often the case the first hint is: well, just do it. You need to decide what the open sets look like:

    Fix u. Suppose d(u,v)<k, take it into two cases, k<1, k>=1

    if k<1 then obviously the x coords of u and v must agree, so where must v lie?

    if k=>1, split into two subcases, when x_1=x_2 and when it doesn't.

    I think drawing things on paper will help here.

    And you should have proven by now that in a metric space compactness and sequential compactness are the same, not that you need it. I doubt H-B is any use what so ever here; I suspect you can do the rest simply by thinking about what open sets look like.

    For example, take R with the zariski topology: closed sets are sets with a finite number of points. Any set S is compact: let X_r be an open cover of a set, consider X_1, this contains all but a finite number of points of S since X_1 is open and contains all but a finite number of points of R, label the points in S but not in X_1 as s_2, s_3,.., s_r and then pick some X_(r_i) that contains the point s_i and this is a finite refinement of the open cover.
     
    Last edited: Nov 30, 2005
  4. Dec 3, 2005 #3

    benorin

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    proof of compactness

    The closed k-ball about [itex]\left( x_0 , y_0\right) [/itex] is given by

    [tex]\left\{ \left( x , y\right) : d\left( \left( x , y\right) , \left( x_0 , y_0\right) \right) \leq k \right\} = \left\{ \left( x , y\right) : \delta_{x_0}^{x} + \left| y_0 -y\right| \leq k \right\} [/tex]

    which for [itex]0<k<1[/itex] is

    [tex]\left\{ \left( x , y\right) : x=x_{0}\mbox{ and }\left| y_0 -y\right| \leq k \right\} ,[/tex]

    the closed vertical line segment centered about [itex]\left( x_0 , y_0\right) [/itex] of length [itex]2k< 2[/itex].

    and for [itex]1 \leq k <\infty[/itex] is

    [tex]\left\{ \left( x , y\right) : x=x_{0}\mbox{ and }\left| y_0 -y\right| \leq k \right\}\cup \left\{ \left( x , y\right) : x\neq x_{0}\mbox{ and }\left| y_0 -y\right| \leq k-1 \right\} ,[/tex]

    the union of the closed vertical line segment centered about [itex]\left( x_0 , y_0\right) [/itex] of length [itex]2k \geq 2[/itex] and the closed horizontal strip centered about [itex]\left( x_0 , y_0\right) [/itex] of hieght [itex]2(k-1) \geq 0[/itex]extending infinitely in both directions (the degenerate case of k=1 give a horizontal line).

    So how do I prove that these are compact? The open k-balls are the same except with "open" in place of "closed", and recall that closed (open) k-balls are closed (open) in any metric space. I know they're compact, so I am guessing proof by contradiction, but I can't get anywhere.
     
  5. Dec 3, 2005 #4

    benorin

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    Duh! I think it's clear now.

    The case of 0<k<1 is compact (closed and bounded) with the usual Euclidian metric.

    k>1 has a atleast one counter-example.
     
  6. Dec 3, 2005 #5

    AKG

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    You only have to show that each point has some neighbourhood with compact closure. Don't even bother looking at k>1. Looking at 0<k<1, you have to first find out what the closure of the vertical open interval is with respect to this topology, and then whether that closure is compact with respect to this topology. I think you just went straight to talking about the closed ball as the vertical closed interval. This is true, i.e. the closure of the vertical open interval in this topology is the vertical closed interval, but you should prove this. You then have to prove it is compact, and it is pretty simple if you can prove that the closed interval in R is compact under the standard metric, but you can't just say that the set is closed and bounded, thus compact.
     
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