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Math Help - If f: R^n --> R^m is continuous everywhere and S is bounded, then f(S) is bounded

  1. #1
    Senior Member Pinkk's Avatar
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    If f: R^n --> R^m is continuous everywhere and S is bounded, then f(S) is bounded

    I have a general idea on how to go about it; if \{x_{n}\} is a sequence in S, then it has a subsequence \{x_{n_{k}}\}that converges to some x\in \mathbb{R}^{n}, and since f is continuous there \{f(x_{n_{k}})\} is a convergent sequence in f(S), which means that sequence is bounded. Now I am not sure if I can conclude that f(S) is bounded. If I can't, how would I go about proving that if a function is continuous everywhere and a set S, subset of \mathbb{R}^{n}, is bounded, then f(S) is bounded? Any help would be appreciated, thanks.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Pinkk View Post
    If f: R^n --> R^m is continuous everywhere and S is bounded, then f(S) is bounded

    I have a general idea on how to go about it; if \{x_{n}\} is a sequence in S, then it has a subsequence \{x_{n_{k}}\}that converges to some x\in \mathbb{R}^{n}, and since f is continuous there \{f(x_{n_{k}})\} is a convergent sequence in f(S), which means that sequence is bounded. Now I am not sure if I can conclude that f(S) is bounded. If I can't, how would I go about proving that if a function is continuous everywhere and a set S, subset of \mathbb{R}^{n}, is bounded, then f(S) is bounded? Any help would be appreciated, thanks.
    If S is bounded then S is a subset of a compact set K, and the image f(S) of S under the continuous map f is a subset of the compact set f(K), and thus bounded.
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  3. #3
    Junior Member
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    Quote Originally Posted by Pinkk View Post
    I have a general idea on how to go about it; if \{x_{n}\} is a sequence in S, then it has a subsequence \{x_{n_{k}}\}that converges to some x\in \mathbb{R}^{n}, and since f is continuous there \{f(x_{n_{k}})\} is a convergent sequence in f(S), which means that sequence is bounded. Now I am not sure if I can conclude that f(S) is bounded. If I can't, how would I go about proving that if a function is continuous everywhere and a set S, subset of \mathbb{R}^{n}, is bounded, then f(S) is bounded? Any help would be appreciated, thanks.
    Assume that f(S) is not bounded.
    Then there is a sequence f(x_n) in f(S) which has no convergent subsequence.
    According to your argument above this is impossible. Therefore f(S) must be bounded.
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