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Proof check: continuous functions (General topology)

  1. Oct 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Let ##A \subset X##; let ##f:A \mapsto Y## be continuous; let ##Y## be Hausdorff. Show that if ##f## may be extended to a continuous function ##g: \overline{A} \mapsto Y##, then ##g## is uniquely determined by ##f##.


    2. Relevant equations



    3. The attempt at a solution

    Just looking for some verification/feedback/comments on my proof. Not sure if the way I handled it was correct, or the best way to approach the problem. Any help would be greatly appreciated!

    Proof:
    Let ##g## and ##h## both be extensions of the continuous function ##f##. We wish to show that they are equal everywhere. We need not check for points in ##A## since both agree with ##f## everywhere on ##A##, so it is sufficient to consider only limit points of ##A##. First we show that if ##x \in A'##, but ##x \not \in A##, then the set ##\{x\}## is not open in ##\overline{A}##. For since ##x## is a limit point of ##A##, every open set containing ##x## intersects ##A##, so that this open set must contain more than one element.
    Now, suppose ##g(x) \not = h(x)##. By hypothesis, there exists open sets ##U## containing ##g(x)## and ##V## containing ##h(x)## such that the two sets are disjoint. Since ##g## and ##h## are both continuous functions, ##g^{-1}(U)## and ##h^{-1}(V)## must be open in ##\overline{A}##. However, note that ##g^{-1}(U) \cap h^{-1}(V) = \{x\}##, which is a contradiction, since finite intersections of open sets must be open, and we have already shown that ##\{x\}## is not open. Hence, ##g(x) = h(x)## for all ##x \in A'##, so that ##g = h## on ##\overline{A}##.
     
  2. jcsd
  3. Oct 27, 2012 #2
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  4. Oct 27, 2012 #3
    Hi hedipaldi,

    Thanks for the response! Sorry, I forgot to explicitly mention this in the proof, but I believe that since the range $Y$ is a Hausdorff space by hypothesis, this implies that for two distinct points in $Y$, there exists disjoint open neighborhoods containing each point. This ensures that the pre-image of both of these open neighborhoods are also disjoint, except for the single point $x$.
     
  5. Oct 27, 2012 #4
    Yes,but there may be other accumulated points of A common to the two pre images.
     
  6. Oct 27, 2012 #5
    Hmm i see what you're saying. You mean their intersection on ##A## is empty, but they may still intersect at other points of ##A'## right?

    How about this then? I can modify the argument to show that any subset of ##A' - A## cannot be open, where the contradiction will still hold since ##g^{-1}(U) \cap h^{-1}(V) \subset A' - A##. And any nonempty subset of ##A' - A## cannot be open since its elements are all limit points of ##A## and any open set containing them must intersect ##A##. Does this rectify the problem?

    However, I'm confused on the part of your proof where you say "there exists ##a_1,a_2 \in A, a_1 \in g^{-1}(U), a_2 \in h^{-1}(V)##, so ##g(a_1) = g(a_2)## is in ##U \cap V##, contradiction to ##U \cap V##." What do you mean by ##g(a_1) = g(a_2)##? How did we know that ##g(a_2) \in U##?
     
  7. Oct 27, 2012 #6
    since g(a1)=h(a2) and g(a1) is in U while h(a2) is in V,we have an element in the intersection of U and V.
    I,mistakenly wrote g insted of h.
     
  8. Oct 27, 2012 #7
    your last argument for the proof seems to be right.
     
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