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Math Help - Prove that the Image of a Continuous Function on a Dense Set is Dense

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    Prove that the Image of a Continuous Function on a Dense Set is Dense

    I've been working on this for days and have made no progress, so any advice would be appreciated.

    Definition: Let R be an ordered field, and Q \subset R. Q is dense in R iff for any x < y in R there is a q \in Q such that x < q < y.

    Prove that, for any function f continuous on a metric space X and E \subset X a dense subset, then f(E) is dense.

    So far what I have is hardly more than setting the problem up. Let f(x), f(y) \in f(X). Thus x \neq y (for now, lets assume x < y), and by density of E, \exists e \in E such that x < e < y . I know that I will be basically done if I can show that d(f(x), f(e)) < d(f(x), f(y)) and d(f(y), f(e)) < d(f(x), f(y)). I also know that the inverse map must also be continuous. But that's all I got.

    Thanks in advance.
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    The function needs to be surjective (consider the exponential map). The definition you have is equivalent to the definition of every ball containing an element of E (which is much easier to work with). So suppose that f:X\rightarrow Y is continuous and surjective, and E is dense in X. Let B be an open ball in Y, then f^{-1}(B) is open in X (continuity), thus f^{-1}(B) \cap E \neq \emptyset and hence B \cap f(E)\neq \emptyset (surjection).
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    Guy
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    OP: Are you sure that's the appropriate definition of a dense set? You only define it for ordered fields, but the statement of the question talks about dense subsets of metric spaces. You also don't state what set f(E) is supposed to be dense in. Isn't the usual definition:

    A set Q is dense in X if the closure of Q in X, denoted \bar{Q}, is equal to X.

    If the question is asking to show that f(E) is dense in f(X), try to show y \in f(X) implies y \in \overline{f(E)}.

    If that's the case, let y \in f(X) be given, find x such that f(x) = y. If x \in E you are done. If x \notin E then x is a limit point of E, so use this to show that f(x) is a limit point of f(E).
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    Hmm... The question doesn't mention surjectivity. Maybe Rudin made a mistake.

    In response to Guy, I'm sort of piecing together the definition. Rudin states much earlier in the text that, for two real numbers, there is a rational between them. He then says that this means the rationals are dense in the reals--so this is all the definition I have to work with. But as for what f(E) is supposed dense in, it's f(X).

    Thank you for the input, fellas.
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    ... In case I'm not annoying yet with this question, I just e-mailed my prof about what definition I should use for "dense" and he gave the one you did, Guy, though it's nowhere in my book. Which I must say, this prof's lack of clarity in his assignments is starting to... frustrate me. Anyway. With that definition in mind I might be able to answer the problem myself, and if not I can probably figure it out from the responses you guys gave. Thanks again.
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    If you're referring to #4 on p.98 of Rudin's Principles of Mathematical Analysis, the problem requires showing f(E) is dense in the image f(X), and the definition of dense referred to is on p.32:

    E is dense in X if every point of X is a limit point of E, or a point of E (or both).
    So to show that f(E) is dense in f(X), pick a point y \in f(X) and show it's either a point of f(E) or a limit point of f(E).

    Since y \in f(X), there is a point x \in X so that f(x) = y. Now if we assume y \not\in f(E), then x \not\in E. But E is dense in X, so this forces x to be a limit point of E.

    Now we want to show that y is actually a limit point of f(E) and we're done. So take an arbitrary open neighborhood of y, call it U and show it intersects f(E). Since f is continuous, f^{-1}(U) is an open set in X (see theorem 4.8 on p. 86).

    Do you see how this preimage set must intersect E, and how sending it back to the image will give you what you want (that U intersects f(E))?
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