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Math Help - Topological Spaces

  1. #1
    Senior Member slevvio's Avatar
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    Topological Spaces

    Let  \mathcal{Z} be the family of sets  U in  \mathbb{R} such that  U is empy or  \mathbb{R} \setminus U is finite.
    Show that  \mathcal{Z} is a topology on  \mathbb{R} . By considering the inverse images of closed sets, show that every polynomial function  f: \mathbb{R} \rightarrow \mathbb{R} is  (\mathcal{Z},\mathcal{Z}) -continuous.

    Hey everyone, I was wondering if anyone could assist me with the 2nd part. If  A is closed in  \mathbb{R} then  \mathbb{R} \setminus A is the empty set, or A is finite, i.e. A is  \mathbb{R} or finite.

    I don't understand though how  f^{-1} (A) is closed and I would really appreciate some advice here, thank you so much. This would show continuity of the polynomial functions.
    Last edited by slevvio; March 4th 2010 at 05:23 AM.
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  2. #2
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    Quote Originally Posted by slevvio View Post
    Let  \mathcal{Z} be the family of sets  U in  \mathbb{R} such that  U is empy or  R \setminus U is finite.
    Show that  \mathcal{Z} is a topology on  \mathbb{R} . By considering the inverse images of closed sets, show that every polynomial function  f: \mathbb{R} \rightarrow \mathbb{R} is  (\mathcal{Z},\mathcal{Z}) -continuous.

    Hey everyone, I was wondering if anyone could assist me with the 2nd part. If  A is closed in  \mathbb{R} then  \mathbb{R} \setminus A is the empty set, or A is finite, i.e. A is  \mathbb{R} or finite.

    I don't understand though how  f^{-1} (A) is closed and I would really appreciate some advice here, thank you so much. This would show continuity of the polynomial functions.

    Hint: if f(x) is a polynomial of degree > 0, how many solutions at most can the equation f(x)=k , for fixed k, have?

    So now, how many elements can the set f^{-1}(A) have, when A is finite?

    Tonio
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  3. #3
    Senior Member slevvio's Avatar
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    it can have at most the degree of the polynomial am I right? eg a polynomial of degree three has 3 complex roots so at most 3 real roots.

    so then the number of elements in  f^{-1} (A) \le deg(f)|A| . Is this correct?

    i.e.  |f^{-1} (A)| \le deg(f)|A| < \infty Hence  |f^{-1} (A)| is finite!

    But what about if A =  \mathbb{R}? Then I guess I am trying to show that  f^{-1}(\mathbb{R}) = \{ x \in \mathbb{R} : f(x) \in \mathbb{R} \} = \mathbb{R} . I guess if f is a polynomial it just moves every point in R to some other point in R. i.e. take any point in R and the polynomial will plop out another point, so it is its own image.

    Thanks for the help
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    Let  \mathcal{Z} be the family of sets  U in  \mathbb{R} such that  U is empy or  \mathbb{R} \setminus U is finite.
    Show that  \mathcal{Z} is a topology on  \mathbb{R} . By considering the inverse images of closed sets, show that every polynomial function  f: \mathbb{R} \rightarrow \mathbb{R} is  (\mathcal{Z},\mathcal{Z}) -continuous.

    Hey everyone, I was wondering if anyone could assist me with the 2nd part. If  A is closed in  \mathbb{R} then  \mathbb{R} \setminus A is the empty set, or A is finite, i.e. A is  \mathbb{R} or finite.

    I don't understand though how  f^{-1} (A) is closed and I would really appreciate some advice here, thank you so much. This would show continuity of the polynomial functions.
    This is called the cofinite topology on \mathbb{R}. Notice that the resulting topological space is T_1. And so let p:\mathbb{R}\mapsto\mathbb{R} be a polynomial, and let F\subseteq\mathbb{R} be closed, then since it is the compliment of an open set we have that F is finite. Let \deg\text{ }p=n. Then clearly for each \xi\in F we have that \text{card }p_n^{-1}(\{\xi\})\leqslant n (otherwise the polynomial p_n(x)-\xi would be a polynomial of degree n with more than n roots! Thus, we see that \text{card }p^{-1}(F)=\text{card }\bigcup_{\xi\in F}p^{-1}(\xi)\leqslant n\cdot\text{card }F. Thus, \text{card }p^{-1}(F) is finite and thus by T_1ness closed.

    The conclusion follows.
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