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Open sets and metric spaces

  1. Jun 23, 2010 #1
    I'm reading Analysis on Manifolds by Munkres and in the section Review of Topology Munkres states the following theorem without proof:

    Let X be a metric space; let Y be a subspace. A subset A of Y is open in Y if and only if it has the form A = U ∩ Y where U is open in X.

    All he has defined is a metric space, subspace, B(y;ε) = {x|d(x,y) < ε} which is the ε neighborhood of y. And he defined open sets to be: A subset U of X is said to be open in X if for each y ∈ U there is a corresponding ε > 0 such that B(y;ε) is contained in U.

    I have never taken Topology or even read about it, so I wrote a proof for it which I'm not sure is correct. Here it is:

    Since A is open then for x0 ∈ A there exists an ε > 0 such that B(x0;ε) is in A. Further, B(x0;ε) is open (I've proved this already). Now taking the union of all such ε neighborhoods of x's in A also produces and open set (I've also proved this). Therefore, letting U = U B(x; ε) proves this direction as clearly, A = U B(x;ε) ∩ Y.

    (⇐)

    A = U ∩ Y

    This means that any x ∈ A is also in U and Y. Therefore, since x ∈ U, which is open, there exists B(x; ε) in U. But I want the open ball to be in A so letting ε' = ε and then B'(x;ε') = B(x;ε)∩Y. However since B(x;ε) is in U, then B'(x;ε') is in U∩Y which is A.

    Am I on the right track? Btw, what if I have a metric X, and a closed set Y in X and then I choose U to be open and not entirely in Y but entirely in X. I define A to be U∩Y, but then A isn't open since it contains part of the boundary of Y. What am I not understanding? And if X is a metric space and Y a subspace, if A is open in Y, does it necessarily imply that A is open in X also?

    I'd appreciate the help!
     
  2. jcsd
  3. Jun 25, 2010 #2
    It may be intended as a definition. Given a subset Y of a metric space X there are two routes to defining a topology on it.

    (i) metric space X -> topological space X -> topological subspace Y.
    (ii) metric space X -> metric subspace Y -> topological subspace Y.

    In (i) the open sets for the topological subspace Y are defined as the sets U ∩ Y where U is open in the topological space X.

    The final topology on Y is the same via either route which is what you're proving. The proof looks OK (but would be clearer if you made a separate notation such as BY(y;ε)=B(y;ε)∩Y for the open spheres in Y).

    The open spheres in the metric space X are not necessarily open spheres in the metric subspace Y nor vice versa. Also if B(x;ε) means the open sphere in X with centre x and radius ε, then B(y;ε)∩Y is an open sphere in Y if y∈Y, but B(x;ε)∩Y is not necessarily an open sphere in Y.
     
    Last edited: Jun 25, 2010
  4. Jun 25, 2010 #3
    I just read your penultimate paragraph more carefully. What you seem to be misunderstanding is that open sets in the topological space X are not necessarily open sets in the topological space Y nor vice versa. The example you give is open in Y but not in X. (Which also answers your last question).
     
  5. Jun 25, 2010 #4
    Thanks for your help. I was reading up on topologies on a set X a bit and I feel like I understand how it is possible for something to be open in Y but not in X (if Y is a subspace of X). So open sets are relative to their universe so to speak. However, it seems like the balls B(y;ε) = {x|d(x,y) < ε} are not really dependent on the set they're in. As the proof that B(y;ε) is open (Munkres asks the reader to prove it) follows from the triangle inequality which is part of the definition of a metric and so, they're always open. So why is it that B(y;ε) seems to be special? Is there a reason why we prove sets are open by using these open balls?

    Sorry if these are just dumb questions...
     
    Last edited: Jun 25, 2010
  6. Jun 25, 2010 #5
    Consider [itex]\mathbb{R}[/itex] as a metric space (with [itex]d_{\mathbb{R}}(x,y)=|x-y|[/itex] for [itex]x,y\in\mathbb{R}[/itex]). Then [itex]\mathbb{Q}[/itex] is a metric subspace (with metric [itex]d_{\mathbb{Q}}(p,q)=d_{\mathbb{R}}|_{\mathbb{Q}\times\mathbb{Q}}(p,q)=|p-q|[/itex]).

    But [itex]\frac{\sqrt{2}}{2}\in B_\mathbb{R}(0;1)[/itex] and [itex]\frac{\sqrt{2}}{2}\notin \mathbb{Q}[/itex], so [itex]B_\mathbb{R}(0;1)[/itex] is not an open ball in [itex]\mathbb{Q}[/itex].

    Similarly [itex]B_\mathbb{Q}(0;1)[/itex] can't be [itex]B_\mathbb{R}(x,r)[/itex] for any [itex]x\in \mathbb{R}[/itex] and [itex]r\in \mathbb{R}^+[/itex], because it would have to be
    [tex]B_\mathbb{R}(\frac{\text{sup }B_\mathbb{Q}(0;1)+\text{inf }B_\mathbb{Q}(0;1)}{2};\frac{\text{sup }B_\mathbb{Q}(0;1)-\text{inf }B_\mathbb{Q}(0;1)}{2})=B_\mathbb{R}(0;1)[/tex]

    but [itex]\frac{\sqrt{2}}{2}\in B_\mathbb{R}(0;1)[/itex] and [itex]\frac{\sqrt{2}}{2}\notin B_\mathbb{Q}(0;1)[/itex].

    Further [itex]B_\mathbb{R}(\sqrt{2};1)\cap\mathbb{Q}[/itex] is also not an open ball in [itex]\mathbb{Q}[/itex].

    We don't so much prove sets are open using open balls as define open sets in metric spaces using open balls, though of course to prove any particular set in a metric space is open we have to show the definition is satisfied. Open balls don't exist in a topological space where the topology is not derived from a metric.
     
    Last edited: Jun 25, 2010
  7. Jun 25, 2010 #6
    I don't understand what you mean. Could you please explain?

    Just another question, I was trying to understand what a topology on a set X is and if I take X = R and have [tex] \tau [/tex] contains all subsets of X then this would be a topology for X, right? Topology on Munkres defines an open set in the following way, a subset U of X with topology [tex] \tau [/tex] is open in X if U is in [tex] \tau [/tex]. So with my example above, then sets which we normally call closed (like [a,b]) would then be called open? This is all because when we normally say (a,b) is open and [a,b] is closed we're using the [tex] \tau [/tex] which only has open sets in the "normal" sense?
     
  8. Jun 25, 2010 #7
    You pretty much explained it yourself in the second paragraph.

    If [itex]S[/itex] is a set of cabbages then [itex]\mathcal{P}(S)=\{O:O\subset S\}[/itex] is a topology on [itex]S[/itex]. It is unnecessary to define a distance function [itex]d_S(b,c)[/itex] between cabbages [itex]b,c\in S[/itex] in order to define the topology [itex]\mathcal{P}(S)[/itex]. If no such distance function is defined then the normal definition of [itex]B_S(c;r)[/itex] as [itex]\{b\in S:d_S(b,c)<r\}[/itex] becomes meaningless.

    If [itex]S[/itex] is any set and [itex]B\subset\mathcal{P}(S)[/itex], then [itex]B[/itex] can be used to generate a topology on [itex]S[/itex] in the same way that the set of open balls generates a topology in a metric space. The open balls are meaningful only in metric spaces.
     
    Last edited: Jun 25, 2010
  9. Jun 25, 2010 #8
    Ahh I get it now. Thanks a lot for your help. I really appreciate it :)
     
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