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Math Help - Continuous Mappings of Dense subsets

  1. #1
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    Continuous Mappings of Dense subsets

    Hello,

    This is a question from Baby Rudin, Exercise 4.4 I was wondering if anyone could verify my reasoning:

    The question is, if f is a continuous function from a metric space X to a metrix space Y, and a set E \subset X is dense in X, then show f(E) in dense in f(X).

    So I have the solution:

    Since E is dense in X, then for any p \in X, we can construct a sequence of points \{p_n\} where the p_n \in E and \{p_n\} \rightarrow p.

    Since f is continuous, we know f(p_n) \rightarrow f(p), and since f(p_n) \in f(E) and f(p) \in f(X), f(E) is dense in f(X).

    -----

    So my qualms with that solution are
    a) The first statment,

    Since E is dense in X, then for any p  \in X, we can construct a sequence of points \{p_n\} where the p_n \in E and \{p_n\} \rightarrow p.

    Is that a true statement for all dense subsets E of a metric space X?? it seems like a serious claim to make, I have never heard or seen a theorem that says anything like that and I can't find anything in Rudin.

    b) The last statement,

    Since f(p_n) \in f(E) and f(p) \in f(X), f(E) is dense in f(X).

    How does f(p_n) converging to f(p) show us that f(E) is dense in f(X)? Is it because f(p) is arbitrary? is the theorem (which I don't know about) an if and only if theorem?
    --
    Essentially I am asking if there is a theorem (which I can't seem to find anywhere) which states:
    E is dense in X, if and only if for any p  \in X, we can construct a sequence of points \{p_n\} where the p_n \in E and \{p_n\} \rightarrow p.

    Thanks, any help appreciated!!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by matt.qmar View Post

    Essentially I am asking if there is a theorem (which I can't seem to find anywhere) which states:
    E is dense in X, if and only if for any p  \in X, we can construct a sequence of points \{p_n\} where the p_n \in E and \{p_n\} \rightarrow p.
    A subspace D of a metric space X is dense if \overline{D}=X. For metric spaces, saying x\in \overline{D} is equivalent to saying there exists a sequence \{x_n\} with x_n\to x. Combine these two for the desired result.


    Remark: Personally, I would give the following proof:

    By continuity we know that for any G\subseteq X that f\left(\overline{G}\right)\subseteq \overline{f\left(G\right)}. Taking G=E gives f(X)=f\left(\overline{E}\right)\subseteq\overline{  f(E)} from where the conclusion follows.
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  3. #3
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    I will leave someone else to comment on a sequence proof.
    Suppose that p\in f(X) that \mathcal{O} is open in Y and contains p.

    Because f is continuous, f^{-1}(\mathcal{O} ) is open in X and is not empty.
    But E is dense in X, thus \left( {\exists t} \right)\left[ {t \in \left( {E \cap f^{ - 1} (\mathcal{O})} \right)} \right]

    So f(t)\in f(E)\cap\mathcal{O}. Or f(E) is dense in f(X).

    That will put to rest your doubts about sequences.
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