# Show that f(E) is dense in f(X) if E is dense in X

1. Jun 6, 2013

1. The problem statement, all variables and given/known data
Let $f$ be a continuous mapping of a metric space $X$ into a metric space $Y$. Let $E$ be a desnse subset of $X$. Prove that $f(E)$ is dense in $f(X)$.

3. The attempt at a solution
Choose any $x\in X$.

Since $x\in X$ it follows that $f(x)\in f(X)$. We need to show that $f(x)$ is a limit point of $f(E)$ or is a point of $f(E)$. If $f(x)\in f(E)$ then we are done so suppose not. Let $\epsilon>0$ be given. $f$ is continuous so $f(p)\in N_{\epsilon}(f(x))$ whenever $p\in N_{\delta}(x)$ for $\delta>0$ and we know that there exists a $p\in N_{\delta}(x)$ such that $p\not=x$ and $p\in E$ since $x$ is a limit point of $E$. $p\in E$ so $f(p)\in f(E)$ and from continuity we saw that $f(p)$ is contained in the neighborhood of $f(x)$, $N_{\epsilon}(f(x))$, whenever $p\in N_{\delta}(x)$. The neighborhood of $f(x)$ and the point $p$ were chosen arbitrarily. We have shown that for any neighborhood of $f(x)$ there exists a point $f(p)$ contained in that neighborhood such that $f(p)\not=f(x)$ and $f(p)\in f(E)$. Hence $f(x)$ is a limit point of $f(E)$. It follows that $f(E)$ is dense in $f(X)$.

Could someone take a look over this proof and correct me on any errors?
Rereading this I think I already see a problem because I haven't shown that $f(x)\not=f(p)$ but this can be easily justified because we assumed that $f(x)\not\in f(E)$. Since $f(p)\in f(E)$ it can't be that $f(x)=f(p)$ or else we contradict ourselves.

Last edited: Jun 6, 2013
2. Jun 6, 2013

### haruspex

Looks ok to me.