Show that $f(E)$ is dense in $f(X)$ if $f$ continuous and $E$ dense in $X$ (Exercice $4$ from Rudin PMA chapter $4$)

Question

Let $f$ be a continuous function defined on $X$ and let $E$ be a set dense in $X$.

We aim to show $f(E)$ is dense in $f(X)$.

We know that: $f(\overline{E})\subseteq \overline{f(E)}$ for every continuous function $f$.

Since $\overline{E} = X$, we have $f(X) \subseteq \overline{f(E)}$.

From there, I fail to prove that $\overline{f(E)} \subseteq f(X)$.

My attempt:

I tried to start with $f(E) \subseteq f(X)$. Hence, $\overline{f(E)} \subseteq \overline{f(X)}$.

I however am not sure that $\overline{f(X)} = f(X)$.

Answer 1

The definition of a subset $U$ being dense in another subset $V$ is $U\subseteq V\subseteq \overline{U}$. So your proof is complete with $f(X) \subseteq \overline{f(E)}$.

The converse you're tring to prove is not true. For example take $X=\Bbb{R}$, $E=X$, $f:\Bbb{R}\rightarrow\Bbb{R}$ defined by $f(x) = \tanh x$. $f(X)$ is the open interval $(-1,1)$ and $\overline{f(E)}=\overline{f(X)}$ is the closed interval $[-1, 1]$.