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Given a metric space (X,d), the set X is open in X. HELP!

  1. Apr 20, 2013 #1
    I must be overlooking something! Given a metric space (E,d), the improper subset E is open in E. How? Here is my understanding:

    1) We call a set S(subset of E) open iff for all x(element of S) there exist epsilon such that an open ball of radi epsilon centered about s is wholly contained in S.

    So, how can E always be open. Here is my counterexample:

    Take E to be a subset of R^2 described by...

    E={(x,y):0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 }

    If now analyze E as a subset of itself, the set E is clearly not open as it contains its boundary. What am I missing here?

    Thanks,

    Topology Newbie
     
  2. jcsd
  3. Apr 20, 2013 #2

    rubi

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    The set E is not open in R^2, but it is open in E with the subspace topology. The balls of that topology might not be circles. For example a ball at (1/2,0) with radius 1/4 is really only the upper half of a circle.
     
  4. Apr 20, 2013 #3

    WannabeNewton

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    ##\partial E## is relative to the larger space you are taking ##E## to be a subset of. If you are looking at ##E\subseteq E##, with the topology on ##E## being the subspace topology inherited from ##\mathbb{R}^{2}##, then ##\partial E = E\setminus (\text{int}E \cup (E\setminus \bar{E})) = \varnothing ## so it has empty topological boundary. Keep in mind that whether or not sets are closed and/or open in a space depends entirely on the topology endowed on that space. If ##A\subseteq X##, where ##X## is a topological space, there is a natural topology induced on ##A## called the subspace topology (alluded to above) given by ##\mathcal{T}_{A} = \left \{ U\subseteq A:U = A\cap V \text{ for some V open in X} \right \}##.
     
  5. Apr 20, 2013 #4
    Wow. i think I see it now! Is this correct:

    If one examines the definition of the open ball, it must be in the context of a particular set... or else it is not sensible. The precise definition of an open ball of radi r centered at p(element of E) in the set E would be the set.....

    {x(ELEMENT OF E): d(p,x) < r}

    So, in my metric space (E,d), if I take E to be an improper subset of itself, it must be that at any x(element of E) an open ball does exist of some radi that is wholly contained in the set E..... this is because of the precise definition of the open ball IN THE CONTEXT OF THE SET E, precisely the part of the set description... all x WHICH ARE ELEMENTS OF E for which the distance between p and x is less than r.

    Do I have the right understanding now?

    Thanks.
     
  6. Apr 20, 2013 #5

    rubi

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    Yes, that's correct. The open balls of a metric space [itex](X,d)[/itex] are defined to be [itex]B_\epsilon(x) = \{y \in X : d(x, y) < \epsilon\}[/itex] and thus if you check your definition of openness, you will find that your set [itex]E[/itex] is open in [itex](E,d)[/itex], but closed in [itex](\mathbb R^2,d)[/itex].
     
  7. Apr 20, 2013 #6

    WannabeNewton

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    I can see what you are trying to say but let's just make it a bit more precise. Take again the case of ##\mathbb{R}^{n}##. We can look at the two metrics ##d_{1}(x,y) = (\sum |x_i - y_i|^2)^{1/2}, d_{2}(x,y) = \text{max}\left \{ |x_1 - y_1|,...,|x_n - y_n| \right \}## on ##\mathbb{R}^{n}##. The open balls of ##d_{1}## are not necessarily the same as the open balls of ##d_{2}## (try to draw the open balls of the respective metrics for the case of ##\mathbb{R}^{2}##). This is what you were trying to say; it isn't the set itself that determines the open balls but rather the metric endowed on that set. In the case of ##A\subseteq \mathbb{R}^{n}##, we don't exactly look at the open balls of ##d_{1}## when trying to see if ##U\subseteq A## is open or not but rather the open balls of ##d_{1}|_{A}## which is the induced metric on ##A##. For example the open balls in ##[0,1]\subset \mathbb{R}## of the metric ##d_{1}|_{[0,1]}## will be sets of the form ##(a,b), [0,a), (a,1]## for appropriate values of ##a,b##.

    It is probably worth noting that even though the open balls of ##d_{1}## and ##d_{2}## need not agree, both metrics induce the same topology on ##\mathbb{R}^{n}##, which can be proven easily. This is a special case of a more general result that all norms on a finite dimensional real or complex vector space ##V## induce the same topology on ##V##. In our special case, both ##d_{1}## and ##d_{2}## are metrics induced by the 2-norm and the max-norm on ##\mathbb{R}^{n}##.
     
    Last edited: Apr 20, 2013
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