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Math Help - Uniformly Cont. Functions

  1. #1
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    Uniformly Cont. Functions

    Consider the following statements:

    1.) If f : A \rightarrow \mathbb{R} is uniformly continuous, then f(A) is bounded.

    2.) If f : A \rightarrow \mathbb{R} is uniformly continuous, and also if A is bounded, then f(A) is bounded.

    Either prove, or disprove, the statements above.
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  2. #2
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    Quote Originally Posted by alikation0 View Post
    Consider the following statements:
    1.) If f : A \rightarrow \mathbb{R} is uniformly continuous, then f(A) is bounded.
    2.) If f : A \rightarrow \mathbb{R} is uniformly continuous, and also if A is bounded, then f(A) is bounded.
    a) Is obviously false. Consider f:\mathbb{R} \mapsto \mathbb{R}\;,\;f(x) = x.

    Hints for b)
    Uniformly continuous functions map Cauchy sequences into Cauchy sequences.
    If E is closed the result in immediate.
    You may try proof by contradiction.
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  3. #3
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    Quote Originally Posted by Plato View Post
    a) Is obviously false. Consider f:\mathbb{R} \mapsto \mathbb{R}\;,\;f(x) = x.

    Hints for b)
    Uniformly continuous functions map Cauchy sequences into Cauchy sequences.
    If E is closed the result in immediate.
    You may try proof by contradiction.
    That hint for b helped a lot! Thanks a lot, I figured that out. But, for a) (AKA 1), I still don't see how that disproves the statement. Mind elaborating?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by alikation0 View Post
    That hint for b helped a lot! Thanks a lot, I figured that out. But, for a) (AKA 1), I still don't see how that disproves the statement. Mind elaborating?
    the real-valued function f(x) = x is uniformly continuous, but if we take the domain of that function to be the reals, the range is unbounded
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