(*) Let $f$ be continous mapping of a metric space $X$ into metric space $Y$, and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$.
Following up from my previous asked question: Dense set equivalent definitions , I am aware that the two mentioned definitions in the previous question are not equivalent since the definition of limit point in Rudin is: $1)$ x is a limit point of V iff every open neighbouhood of x contains a point of V not equal to x iff there is a sequence of points $x_n$ in V such that $x_n→x$ and $x_n≠x$. So I am now just trying to prove the mentioned statment (*) by this definition of dense set : Given a metric space $X$ and $E \subset X$; $E$ is dense in $X$ iff every point of $X$ is a limit point of $E$ or $E = X$ or both of these are true. Where the definition of limit point is that of $(1)$. So the problem that I am facing now that I can't seem to resolve is that what if we have a situation in $f(X)$ where we have an isolated point $y_1$ in $f(X)$ and a limit point $y_2$ of $f(X)$ that is a member of $f(X)$ as well. Now according to the definition of dense set that I have, we have to show in such a scenario that $f(E)$= $f(X)$. However I cannot find a way to show that $y_2$ is contained in $f(E)$. So a help would be appreciated here
Take $y\in f(X)$. Then $y=f(x)$ for some $x\in X$. Take $\varepsilon>0$. Then $f^{-1}\bigl(D(y,\varepsilon)\bigr)$ is an open subset of $X$ to which $x$ belongs; in particular, it is not empty. Since it is open and not empty, it contains some $e\in E$. But then $f(e)\in D(y,\varepsilon)$