Results 1 to 3 of 3

Thread: compact subset of R

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    2

    compact subset of R

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

    I know that the set $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is bounded because $\displaystyle A$ is bounded, so I guess I just need to prove that $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,743
    Thanks
    2814
    Awards
    1
    Use the notation: $\displaystyle B = \left\{ {x \in A:0 \leqslant f(x) \leqslant 1} \right\}$.
    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 $\displaystyle B$ is closed.
    If $\displaystyle B$ were not closed then some limit point, $\displaystyle x_0$, of $\displaystyle B$ and $\displaystyle x_0 \notin B$.
    Then there is a sequence of distinct points from $\displaystyle B$ which converges to $\displaystyle x_0$.
    Remembering that $\displaystyle f$ is continuous, there is a contradiction there.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by junipers232 View Post
    Suppose that $\displaystyle A$ is a compact subset of $\displaystyle \mathbb{R}$ and $\displaystyle f: A \rightarrow \mathbb{R}$ is continuous. Prove that the set $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is a compact subset of $\displaystyle \mathbb{R}$.

    I know that the set $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is bounded because $\displaystyle A$ is bounded, so I guess I just need to prove that $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ 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 $\displaystyle \mathbb{R}$, $\displaystyle f^{-1}(C)$ is closed in A.

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

Similar Math Help Forum Discussions

  1. Closed subset of sequentially compact set is sequentially compact.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Sep 13th 2011, 01:25 PM
  2. inf and sup of a compact subset of R
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Feb 28th 2010, 08:04 PM
  3. Examples of non-compact space with infinite subset
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Jan 17th 2010, 10:44 AM
  4. Continuous function on a compact subset of R
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Mar 5th 2009, 01:12 PM
  5. Compact subset problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 19th 2007, 03:22 PM

Search Tags


/mathhelpforum @mathhelpforum