# Thread: Continuous Mappings of Dense subsets

1. ## 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 $\displaystyle f$ is a continuous function from a metric space $\displaystyle X$ to a metrix space $\displaystyle Y$, and a set $\displaystyle E \subset X$ is dense in X, then show $\displaystyle f(E)$ in dense in $\displaystyle f(X)$.

So I have the solution:

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

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

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

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

Is that a true statement for all dense subsets $\displaystyle E$ of a metric space $\displaystyle X$?? 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 $\displaystyle f(p_n) \in f(E)$ and $\displaystyle f(p) \in f(X)$, $\displaystyle f(E)$ is dense in $\displaystyle f(X)$.

How does $\displaystyle f(p_n)$ converging to $\displaystyle f(p)$ show us that $\displaystyle f(E)$ is dense in $\displaystyle f(X)$? Is it because $\displaystyle f(p)$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:
$\displaystyle E$ is dense in $\displaystyle X$, if and only if for any $\displaystyle p \in X$, we can construct a sequence of points $\displaystyle \{p_n\}$ where the $\displaystyle p_n \in E$ and $\displaystyle \{p_n\} \rightarrow p$.

Thanks, any help appreciated!!

2. Originally Posted by matt.qmar

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

Remark: Personally, I would give the following proof:

By continuity we know that for any $\displaystyle G\subseteq X$ that $\displaystyle f\left(\overline{G}\right)\subseteq \overline{f\left(G\right)}$. Taking $\displaystyle G=E$ gives $\displaystyle f(X)=f\left(\overline{E}\right)\subseteq\overline{ f(E)}$ from where the conclusion follows.

3. I will leave someone else to comment on a sequence proof.
Suppose that $\displaystyle p\in f(X)$ that $\displaystyle \mathcal{O}$ is open in $\displaystyle Y$ and contains $\displaystyle p$.

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

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