Let $f$ be continuous function on $X$ and let $E$ be a dense subset of $X$. Then if $y \in f(E)'$, i.e. if $y$ is limit point of $f(E)$, show that there exists $x \in X$ such that $y = f(x)$.
I can't solve this problem. Any help or hint will be appreciated.
This is similar to the last question and is also false. Let $f:(0,1)\rightarrow [0,1]$ be defined by $f(x) = x$. Let $E = (0,1)$. Every topological space is dense in itself, so $E$ is a dense subset of $X = (0,1)$. $1$ is a limit point of $f(E) = (0,1)\subset [0,1]$, but there is no $x\in X = (0,1)$ so that $x= f(x) = 1$.
edit: This appears to be part of a proof attempt at question 4.4 of Rudin's analysis book. The above example should still show that the claim you are trying to prove is not correct. To be clear, $X = (0,1)$ and $Y = [0,1]$. These are metric spaces with the metric $d(x,y) = |x - y|$. $f(x) = x$ is a continuous function between these metric spaces.
In this problem, you are trying to show that $f(E)$ is a dense subset of $f(X)$. This is not the same as showing that $\overline{f(E)} = f(X)$.
You want to prove that if $y$ is in $f(X)$, then $y$ is a limit point of the set $f(E)$. You don't want to show that every limit point of $f(E)$ is a point in $f(X)$.
In symbols, you want to show $f(E)\subset f(X) \subset \overline{f(E)}$, not $f(X) = \overline{f(E)}$.