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Math Help - compact subset of R

  1. #1
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    compact subset of R

    Suppose that A is a compact subset of \mathbb{R} and f: A \rightarrow \mathbb{R} is continuous. Prove that the set \{x \in A | 0 \leq f(x) \leq 1 \} is a compact subset of \mathbb{R}.

    I know that the set \{x \in A | 0 \leq f(x) \leq 1 \} is bounded because A is bounded, so I guess I just need to prove that \{x \in A | 0 \leq f(x) \leq 1 \} is closed. I was thinking to take a sequence in the set and show that it converges in the set. This is what is confusing me because I don't know how to do this part. Thanks for any help in advance.
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  2. #2
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    Use the notation: B = \left\{ {x \in A:0 \leqslant f(x) \leqslant 1} \right\}.
    Theorem: Any closed subset of a compact set is compact.
    You should have proved that theorem, if not it is easily proven.
    Thus, you want to prove that the set B is closed.
    If B were not closed then some limit point, x_0, of B and x_0 \notin B.
    Then there is a sequence of distinct points from B which converges to x_0.
    Remembering that f is continuous, there is a contradiction there.
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  3. #3
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    Quote Originally Posted by junipers232 View Post
    Suppose that A is a compact subset of \mathbb{R} and f: A \rightarrow \mathbb{R} is continuous. Prove that the set \{x \in A | 0 \leq f(x) \leq 1 \} is a compact subset of \mathbb{R}.

    I know that the set \{x \in A | 0 \leq f(x) \leq 1 \} is bounded because A is bounded, so I guess I just need to prove that \{x \in A | 0 \leq f(x) \leq 1 \} is closed. I was thinking to take a sequence in the set and show that it converges in the set. This is what is confusing me because I don't know how to do this part. Thanks for any help in advance.
    Lemma 1. Each closed subset of a compact space is compact.

    Since f is continuous, each closed subset C of \mathbb{R}, f^{-1}(C) is closed in A.

    By lemma 1, \{x \in A | 0 \leq f(x) \leq 1 \} is compact. Now, you'll see that \{x \in A | 0 \leq f(x) \leq 1 \} is a compact subset of \mathbb{R}.
    Last edited by aliceinwonderland; March 3rd 2009 at 03:01 PM. Reason: Edit: Plato replied faster. This is basically the same answer with the above.
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