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Thread: metric spaces

  1. #1
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    metric spaces

    Let (X,d) and (Y,p) be metric spaces, and assume Y is complete. Let f: X→Y and x∈X.
    Show that f has a limit at x if and only if given any ε>0 there exists δ>0 such that whenever y, z ∈B_δ(x)\{x},p(f(y),f(z)) < ε.



    can anyone please help me to solve this question??? ,,,
    Last edited by jin_nzzang; May 2nd 2009 at 04:55 AM.
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  2. #2
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    Quote Originally Posted by jin_nzzang View Post
    Let (X,d) and (Y,p) be metric spaces, and assume Y is complete. Let f: X→Y and x∈X.
    Show that f has a limit at x if and only if given any ε>0 there exists δ>0 such that whenever y, z ∈B_δ(x)\{x},p(f(y),f(z)) < ε.
    Let $\displaystyle (x_n)$ be a sequence in X that converges to x. Then $\displaystyle (x_n)$ is a Cauchy sequence in X. Use the given ε-δ condition to conclude that $\displaystyle (f(x_n))$ is a Cauchy sequence in Y. The completeness of Y tells you that $\displaystyle (f(x_n))$ converges to some element w in Y.

    It remains to show that for every sequence that converges to X, its image under f converges to the same limit in Y. So suppose that $\displaystyle x_n\to x$ with $\displaystyle f(x_n)\to w$, as above, and also that $\displaystyle y_n\to x$ with $\displaystyle f(y_n)\to v$. Then $\displaystyle d(x_n,y_n)\to0$, and therefore $\displaystyle p(f(x_n),f(y_n))\to0$, from which it follows that $\displaystyle p(w,v)=0$ and hence v=w.

    Thus for every sequence that converges to x, its image under f converges to w. Therefore $\displaystyle \textstyle\lim_{y\to x}f(y)$ exists and is equal to w.
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