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

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June 20th 2010, 10:55 AM
ragnar

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.
June 20th 2010, 11:36 AM
Focus

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).
June 20th 2010, 11:42 AM
Guy

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 $f(E)$ .
June 20th 2010, 12:35 PM
ragnar

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.
June 20th 2010, 12:41 PM
ragnar

... 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.
June 20th 2010, 09:12 PM
gosualite

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:

Quote:

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 $f(E)$ )?