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Math Help - Dense subset

  1. #1
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    Dense subset

    Let  X and  Y be metric spaces and let  D be a dense subset of  X . Let  f: X \to Y and  g: X \to Y be continuous functions. Suppose that  f(r) = g(r) for all  r \in D . Prove that  f = g .

    Proof. Let  r \in X \backslash D . Then  r is a limit point of  D by the denseness of  D . Pick a decreasing sequence of positive real numbers  (x_n) . Then in each open ball  B_{x_{n}}(r) , there is a point  q_n \neq r such that  q_n \in D . Form these terms as a sequence  (q_n) . Fix  \epsilon > 0 . Since  x_n \to 0 any open ball centered at  0 contains at least one  x_j for some  j \in \mathbb{N} . So there is some  j such that  x_j \in B_{\epsilon}(0) . Then for  n \geq j ,  q_{n} \in B_{\epsilon}(r) . So  q_n \to r . Then  g(r) = \lim_{n \to \infty} g(q_n) . We know that  q_n \in D for  n \in \mathbb{N} . Thus  f(q_n) = g(q_n) which implies  f(r) = g(r) . Thus  f = g .  \blacksquare


    Is this correct?
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  2. #2
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    Quote Originally Posted by particlejohn View Post
    Let  X and  Y be metric spaces and let  D be a dense subset of  X . Let  f: X \to Y and  g: X \to Y be continuous functions. Suppose that  f(r) = g(r) for all  r \in D . Prove that  f = g .

    First, you must do away with thinking of a general metric space as the real numbers alone. That is, a decreasing sequence of points has no general meaning.

    Now suppose that \left( {\exists x} \right)\left[ {f(x) \ne g(x)} \right] clearly x \in X\backslash D.
    As you noted, by density there is a sequence of distinct points \left( {x_n } \right) \to x\,\&\,x_n \in D.
    \left( {\exists \delta > 0} \right)\left[ {B_\delta \left( {f(x)} \right) \cap B_\delta \left( {g(x)} \right) = \emptyset } \right].
    But f\left( {x_n } \right) = g\left( {x_n } \right)\,\& \,\left( {f\left( {x_n } \right)} \right) \to f(x)\,\& \,\left( {g\left( {x_n } \right)} \right) \to g(x).
    Do you see the contradiction?
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  3. #3
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    I see the contradiction. Thanks. However, in my opinion, proof by contradiction is a "cop out" way of proving things (even though it is correct). Because you can prove anything from contradictory statements. Was my direct way ok?
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