What am I lacking?

Question

Let $f$ be a continuous function on $X$. Let $E$ be any dense subset of $X$. Show that $f(E)$ is dense in $f(X)$.

Below, I have given my solution to the problem. However, I am not satisfied with my solution and feel something important is missing. Any hint is appreciated

solution:- since $E$ is dense in $X$, for any $\epsilon >0$ and $x$ $\in$ $X$ ,$ \exists s\in E $ such that $ 0<|x-s|<\delta $.$f$ is continuous in $X$ so for any $\epsilon >0$, we have $|f(x)-f(s)|< \epsilon $ whenever $|x-s|< \delta$. Then we see for any $\epsilon >0 $ and $f(x) \in X$, we find $f(s) \in f(E).$ This shows $f(E)$ is dense in $f(E)$

Answer 1

The first sentence of the solution makes no sense, since it ends with $\delta$, but you don't say what $\delta$ is.

Let $x\in X$ and ket $\varepsilon>0$. Since $f$ is continuous at $x$, there is a $\delta>0$ such that $\lvert y-x\rvert<\delta\implies\bigl\lvert f(y)-f(x)\bigr\rvert<\varepsilon$. And, since $E$ is dense in $X$, there is a $z\in E$ such that $\lvert z-x\rvert<\delta$. But then $\bigl\lvert f(z)-f(x)\bigr\rvert<\varepsilon$ and $f(z)\in f(E)$. So, every neighborood of any element of $f(X)$ contains an element of $f(E)$. In other words, $f(E)$ is dense in $f(X)$.

Answer 2

Proofs generally ought to be structured by what they want to show. The definition of "$D$ is dense in $Y$" is

For any $y\in Y$ and any $\varepsilon>0$, there is a $d\in D$ such that $|d-y|<\varepsilon$.

You want to show that $f(E)$ is dense in $f(X)$, so a good proof should, in my opinion, look something like this:

Take an arbitrary $k\in f(X)$, and an $\varepsilon>0$. We have [...] and therefore there is an $e\in E$ with $|f(e)- k|<\varepsilon$

Our job is to use the continuity of $f$ and the density of $E\subseteq X$ to fill in the [...]. Here is my way of doing that:

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Take an arbitrary $k\in f(X)$, and an $\varepsilon>0$. By definition of $f(X)$, there is an $x\in X$ with $f(x) = k$. By continuity of $f$, there is a $\delta>0$ such that for any $x_0\in X$ with $|x_0 - x|<\delta$, we have $|f(x_0) - f(x)|<\varepsilon$.

By density of $E$, there is an $e\in E$ with $|e-x|<\delta$. This implies $$ |f(e) - f(x)| = |f(e) - k|<\varepsilon $$