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Math Help - Uniform Continuity definition

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    Uniform Continuity definition

    Let (M,d) and (N,p) be metric spaces, and let f:A \subset M \rightarrow N . Prove that f is an uniformly continuous mapping on A iff for every pair of sequence  \{ x_k \} , \{ y_k \} of A with d(x_k,y_k) \rightarrow 0 , we have p(f(x_k),f(y_k)) \rightarrow 0

    Proof.

    Suppose that f is uniformly continuous, pick sequences  \{ x_k \} , \{ y_k \} of A such that  d( x_k,y_k) \rightarrow 0

    Let  \epsilon > 0 be given, and let  \delta > 0, then  \exists K \in \mathbb {N} such that whenever  k \geq K, we have d(x_k,y_k) < \delta . Since f is uniformly continuous, we then ahve p(f(x_k),f(y_k)) < \epsilon , implies that p(f(x_k),f(y_k)) \rightarrow 0

    Conversely, suppose that math] \{ x_k \} , \{ y_k \} [/tex] are sequences in A with  d( x_k,y_k) \rightarrow , we have p(f(x_k),f(y_k)) \rightarrow 0 .

    Let  \epsilon > 0 be given, pick  \delta > 0, find K \in \mathbb {N} such that whenever  k \geq K, we have d(x_k,y_k) < \delta .

    Now suppose that x_k \rightarrow x_0 and y_k \rightarrow y_0, then d(x_0,y_0) < \delta , but we also have p(f(x_k),f(y_k)) \rightarrow 0 , which means p(f(x_k),f(y_k)) < \epsilon .

    Q.E.D.

    Is this right? Thanks!
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let (M,d) and (N,p) be metric spaces, and let f:A \subset M \rightarrow N . Prove that f is an uniformly continuous mapping on A iff for every pair of sequence  \{ x_k \} , \{ y_k \} of A with d(x_k,y_k) \rightarrow 0
    Clearly if f is uniformly continuous the statement is correct but not your proof.
    From U.C. \varepsilon  > 0 \Rightarrow \left( {\exists \delta  > 0} \right)\left[ {d(x,y) < \delta  \Rightarrow p\left( {f(x),f(y)} \right) < \varepsilon } \right]\quad .
    Suppose that \left( {x_n } \right) \to x\,\& \,\left( {y_n } \right) \to y then <br />
\left( {\exists N} \right)\left[ {n \geqslant N \Rightarrow d\left( {x_n ,x} \right) < \frac{\delta }{3} \wedge d\left( {y_n ,y} \right) < \frac{\delta }{3}} \right].
    Now consider this: If d(x,y) < \frac{\delta }{3} then d\left( {x_n ,y_n } \right) \leqslant d\left( {x_n ,x} \right) + d\left( {x,y} \right) + d\left( {y,y_n } \right) < \delta  \Rightarrow  p\left( {f\left( {x_n } \right),f\left( {y_n } \right)} \right) < \varepsilon<br />
.

    P.S. I was far too tired to think straight last night.

    The converse is easy. If f is not informally continuous then \left( {\exists \lambda  > 0} \right)\left( {\forall \delta  > 0} \right)\left( {\exists x_\delta   \wedge y_\delta  } \right)\left[ {d(x_\delta  ,y_\delta  ) < \delta  \wedge p(f(x_\delta  ),f(y_\delta  )) \geqslant \lambda } \right].
    That is the negation of informally continuous.

    So \left( {\forall n} \right)\left( {\exists x_n  \wedge y_n } \right)\left[ {d(x_n ,y_n ) < \frac{1}{n} \wedge p(f(x_n ),f(y_n )) \geqslant \lambda } \right]
    That is a contradiction, because d(x_n ,y_n ) \to 0.
    Last edited by Plato; October 17th 2008 at 07:13 AM.
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