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Connected components of a metric subspace

  1. Oct 29, 2013 #1
    The problem statement, all variables and given/known data.
    Consider the subspace ##U## of the metric space ##(C[0,1],d_∞)## defined as ##U=\{f \in C[0,1] : f(x)≠0 \forall x \in [0,1] \}##. Prove that ##U## is open and find its connected components.

    The attempt at a solution.
    First I've proved that ##U## is open. I want to check if my proof is correct:
    Let ##f \in U##. ##f(x)≠0## and ##f## is continuous, this means ##f(x)>0## or ##f(x)<0##. Suppose ##f(x)>0##. The function ##f## is a continuous function defined on a compact set then there exists ##x_0 \in [0,1]## such that ##f(x)≥f(x_0)>0 \forall x \in [0,1]##. Consider the ball ##B(f,\frac{f(x_0)}{2})##. Let ##g \in B(f,\frac{f(x_0)}{2})##. Then, ##|f(x)-g(x)|≤sup_{x \in [0,1]}|f(x)-g(x)|=d_∞(f(x),g(x))<\frac{f(x_0)}{2} ##.But then ##f(x)-g(x)<\frac{f(x_0)}{2}→0<f(x)-\frac{f(x_0)}{2}<g(x)##. This proves that ##g \in U## for an arbitrary ##g \in B(f,\frac{f(x_0)}{2})→B(f,\frac{f(x_0)}{2}) \subset U##

    The case ##f(x)<0## is analogue. ##f## is a continuous function on a compact set, then there exists ##x_1 \in [0,1]## such that ##f(x)≤f(x_1)<0 \forall x \in [0,1]##. Consider the ball ##B(f,\frac{-f(x_1)}{2})##. Let ##g \in B(f,\frac{-f(x_1)}{2})##. Then ##|f(x)-g(x)|≤sup_{x \in [0,1]}|f(x)-g(x)|=d_∞(f(x),g(x))<\frac{-f(x_1)}{2}##. This means ##\frac{f(x_1)}{2}<f(x)-g(x)→0<\frac{f(x_1)}{2}-f(x)<-g(x)→0>f(x)-\frac{f(x_1)}{2}>g(x)##. This implies ##g \in U## for an arbitrary ##g \in B(f,\frac{f(x_1)}{2})→B(f,\frac{f(x_1)}{2}) \subset U##

    I am stuck on the second part of the problem. If a function f is in the subspace U, then the connected component of f would be the union of all the connected subspaces in U that contain f. My doubt is: What are the connected components of a subspace (in this case, U)?. How could I try to find them?
     
  2. jcsd
  3. Oct 29, 2013 #2

    Dick

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    You've already got the ingredients you need. Define ##U_{+}## to be the set of positive functions and ##U_{-}## to be the set of negative functions. You've proved both of those are open. So? BTW what definition of connectedness are you using?
     
  4. Oct 29, 2013 #3

    pasmith

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    [itex]U[/itex] can be written as a disjoint union of open sets, namely
    [tex]U = \{ f \in U: f(x) > 0\} \cup \{f \in U : f(x) < 0\}[/tex]

    It is possible to show that each of those subsets is path connected, and hence connected.
     
  5. Oct 29, 2013 #4
    If ##(X,d)## is a metric space and ##C \subset X##, then ##C## is connected if:
    1) There aren't any ##U##,##V## non-empty, disjoint and open in ##C## such that ##C=U \cup V##
    2) If ##A \subset C## is non-empty, open and closed in ##C##, then ##A=C##.

    These are the two equivalent definitions I am using.

    So, you've said that I should consider ##U_{+}## and ##U_{-}##. I've already proved they are open and it is easy to see that both are non-empty (##f(x)## constantly ##1## and ##h(x)## constantly ##-1## are in ##U_{+}## and ##U_{-}## respectively) and it's immediate to show that they are disjoint for if ##f \in U_{+} \cap U_{-}## then ##0<f(x)<0 \forall x \in [0,1]##, which is absurd. From here I can conclude that ##U## is not connected. I am pretty sure that ##U_{+}## and ##U_{-}## are the only connected components of ##U##, but I don't know how to prove it. In fact I am not sure of the formal definition of connected components of a subspace. I intuitively think that they are the "maximal" connected subsets of the space (in terms of inclusion). Is that the definition?
     
  6. Oct 29, 2013 #5
    Ok, once I've proved that, should I prove that if ##A \subset U## is connected then ##A \subset U_{+}## or ##A \subset U_{-}##? (using Dick's notation)
     
  7. Oct 29, 2013 #6

    Dick

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    Right you are. Now you just need to show ##U_{+}## and ##U_{-}## are connected.
     
  8. Oct 29, 2013 #7

    Dick

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    One way is to show that they arcwise connected. If ##f_1## and ##f_2## are two functions in ##U_{+}##, then ##t f_1+(1-t) f_2## is in ##U_{+}## for t in [0,1], right? Suppose that there were two sets that disconnect ##U_{+}## but ##f_1## is in one and ##f_2## is in the other?
     
    Last edited: Oct 29, 2013
  9. Oct 29, 2013 #8
    I've just proved that ##U_{+}## and ##U_{-}## are path connected by the same way you did.

    Sorry, I don't know if this was what you were pointing me out but it is what I've came up with:

    Suppose ##A \subset U## is connected and suppose ##A## is not totally included in ##U_{+}## or
    ##U_{-}##. We know that ##U_{+}## and ##U_{-}## disconnect ##U##, so ##A \subset U_{+} \cup U_{-}##. By the previous statements we have ##A \cap U_{+}≠∅## and ##A \cap U_{-}≠ ∅##. I define ##A_{+}=\{f \in A : f \in U_{+}\}## and ##A_{-}=\{h \in A : h \in U_{-}\}##. Then ##A_{+}## and ##A_{-}## disconnect ##A## which is absurd since ##A## is connected. It follows ##A \subset U_{+}## or ##A \subset U_{-}##. This proves ##U_{+}## and ##U_{-}## are the only connected components of ##U##.

    Is this correct? Thanks for your help!
     
  10. Oct 29, 2013 #9

    Dick

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    It's all fine up until "This proves ##U_{+}## and ##U_{-}## are the only connected components of ##U##". It doesn't prove that. Suppose ##U_{+}## could be split into two connected components? The point to what I was telling you is that if ##U_{+}## could be disconnected then the real interval [0,1] could be disconnected. And we know it can't. That's what the arcwise connected bit is about. Maybe there is a another way to do this, but this is the way I usually think about it.
     
    Last edited: Oct 29, 2013
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