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

• Jun 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 $\displaystyle R$ be an ordered field, and $\displaystyle Q \subset R$. $\displaystyle Q$ is dense in $\displaystyle R$ iff for any $\displaystyle x < y$ in $\displaystyle R$ there is a $\displaystyle q \in Q$ such that $\displaystyle x < q < y$.

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

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

• Jun 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 $\displaystyle f:X\rightarrow Y$ is continuous and surjective, and E is dense in X. Let B be an open ball in Y, then $\displaystyle f^{-1}(B)$ is open in X (continuity), thus $\displaystyle f^{-1}(B) \cap E \neq \emptyset$ and hence $\displaystyle B \cap f(E)\neq \emptyset$ (surjection).
• Jun 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 $\displaystyle f(E)$ is supposed to be dense in. Isn't the usual definition:

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

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

If that's the case, let $\displaystyle y \in f(X)$ be given, find $\displaystyle x$ such that $\displaystyle f(x) = y$. If $\displaystyle x \in E$ you are done. If $\displaystyle x \notin E$ then $\displaystyle x$ is a limit point of $\displaystyle E$, so use this to show that $\displaystyle f(x)$ is a limit point of $\displaystyle f(E)$.
• Jun 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.
• Jun 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.
• Jun 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 $\displaystyle f(E)$ is dense in $\displaystyle f(X)$, pick a point $\displaystyle y \in f(X)$ and show it's either a point of $\displaystyle f(E)$ or a limit point of $\displaystyle f(E)$.

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

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

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