# Continuous Mappings of Dense subsets

• Dec 4th 2010, 01:50 PM
matt.qmar
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 $f(X)$.

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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?
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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!!
• Dec 4th 2010, 02:19 PM
Drexel28
Quote:

Originally Posted by matt.qmar

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.
• Dec 4th 2010, 02:20 PM
Plato
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 $f(X)$.