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If $f$ is entire and $A$ is dense in $\Bbb{C}$, why is $f(A)$ dense in $\Bbb{C}$? I am allowed to use the fact that $f(\Bbb{C})$ is dense in $\Bbb{C}$.

This claim was used in the course of showing that an essential singularity $z_0$ of holomorphic function $h$, is an essential singularity of $f\circ h$ if $f$ is entire, but it was simply claimed as if it were perfectly clear.(Which it might be, due to which I take into account that I might be overlooking something.)

Conceptually, my intuitive approach to it is somehow analogous to "a set A dense in B which is dense in C, is dense in C", but it doesn't seem applicable and I am quite stuck, avoiding a long proof by contradiction similar to that proving that $f(\Bbb{C})$ is dense in $\Bbb{C}$ for entire $f$. Do you have any intuitive perspective on that matter?

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    the [open mapping](https://en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)) so $f(\overline{A}) = \overline{f(A)}$ – 2017-01-23
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    What about a constant function (which is, of course, entire). So, it seems to me that we should assume $f$ in ubbounded (see Liouville's Theorem). – 2017-01-23
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    To add on what szw1710 said, if $f$ is entire and non-constant then $f(\mathbb{C})$ is dense in $\mathbb{C}$, because otherwise there is $a$ such that $\frac{1}{f(z)-a}$ is entire and bounded and hence constant – 2017-01-23
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    I unintentionally excluded it. I will edit. – 2017-01-23
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    @user1952009, I have shown that. What as for the image of dense set under entire function? – 2017-01-23

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This follows from Liousville's Theorem with a little thinking, presumably you've assumed that $f$ is non-constant at some point since the claim isn't true for constant functions. Assume that there exists some $z_0,r$ such that $B_r(z_0)\cap f(A)=\emptyset$. Then $|\frac{1}{f(z)-z_0}|<\frac{1}{r}$. But $g(z)=\frac{1}{f(z)-z_0}$ is then a bounded entire function, and so is constant. It follows that $f(z)$ is constant, contradiction.

You can also use it's strengthening, Picard's Little Theorem to immediately conclude the result.

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    I actually wanted to avoid that approach as I already proved something similar showing that $f(C)$ is dense, and I thought I could use a simpler method. Moreover, I couldn't see how to modify that approach. It seemed pretty meticulous. – 2017-01-23
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    @Meitar I think your proof is overly complicated, as the Liousville proof is actually really straight forward. I edited my answer to include it. – 2017-01-23
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continuity alone will give you that $f[A]$ is dense in $f[\mathbb{C}]$ and as the latter set is dense in $\mathbb{C}$ (this uses entire) $f[A]$ is dense in $\mathbb{C}$ as well.

This only uses basic topology: $f: X \rightarrow Y$ continuous, then for $A \subseteq X$, $f[\overline{A}] \subseteq \overline{f[A]}$.

(or more directly for dense sets: if $O$ is non-empty open in $f[X]$, so is $f^{-1}[O]$ in $X$, so this intersects $A$, and so $O$ intersects $f[A]$.)

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You know $f(\mathbb C)$ is dense in $\mathbb C.$ So let $w\in f(\mathbb C).$ Then $w=f(z)$ for some $z\in \mathbb C.$ Since $A$ is dense in $\mathbb C,$ there exists a sequence $(a_n)$ in $A$ such that $a_n\to z.$ By continuity, $f(a_n) \to f(z) = w.$ Thus $f(A)$ is dense in $f(\mathbb C).$ Hence $\overline {f(A)} \supset f(\mathbb C),$ which implies $\overline {f(A)} \supset \overline {f(\mathbb C)} = \mathbb C.$ Therefore $f(A)$ is dense in $\mathbb C.$

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    Nicely demonstrated. This is what I had in mind, but I couldn't bring it to the right form. – 2017-01-23