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I am having trouble solving the following exercise:

Let $U$ be a bounded domain, and let $f(z)$ be a continous function on $U cup partial U$ that is holomorphic on $U$. Show that $partialleft( f(U)right)subseteq f(partial U)$, that is, the boundary of the open set $f(U)$ is contained in the image under $f(z)$ of the boundary of $U$.

One of the main issues I have with this exercise is that my teacher has defined the open mapping theorem as:

Open mapping theorem: if $Omega subseteq mathbb{C}$ is a connected open set and $f$ is holomorphic and not constant on $Omega$, then $f(Omega)$ is an open set in $mathbb{C}$.

The additional hypothesis that the domain be connected means that we cannot apply the theorem directly. This is my approach:

Suppose $f$ is not constant (if it is the result is clear). Since $stackrel{circ}{U}$ is open, for each $zinstackrel{circ}{U}$ there exists $r_z>0$ such that the open disk of center $z$ and radius $r_z$, $D(z, r_z)$, is contained in $stackrel{circ}{U}$. Then we can write $stackrel{circ}{U}$ as:

$stackrel{circ}{U} = bigcup_{zinstackrel{circ}{U}} D(z, r_z)$

Hence:

$f(stackrel{circ}{U}) = fleft(bigcup_{zinstackrel{circ}{U}} D(z, r_z)right) = bigcup_{zinstackrel{circ}{U}} fleft(D(z, r_z)right)$

Since $f$ is holomorphic in $stackrel{circ}{U}$, disks are open connected sets and the arbitrary union of open sets is open, we can apply the Open mapping theorem to conclude that $f(stackrel{circ}{U})$ is an open set. This means that $f$ maps interior points of $U$ to interior points of $f(U)$.

Let $w$ be a point of $partial f(U)$, then there exists a sequence $w_n subseteq f(U)$ that converges to $w$, for $partial f(U)$ is a compact set (1). By definition of $f(U)$, for all $kinmathbb{N}$ there exists $z_kinstackrel{circ}{U}$ such that $f(z_k) = w_k$. By hypothesis $U$ is bounded, and therefore $z_krightarrow zinpartial U$ and $f(z_k) rightarrow f(z) = winpartial f(U)$, which implies:

$forall w inpartial f(U) : exists zinpartial U mid f(z) = f(w) implies partial f(U) subseteq f(partial U)$.

One of my concerns is that $U$ is an infinite union of open sets, and I am not sure that $f(bigcup_{k=0}^{infty} U_k) = bigcup_{k=0}^{infty} f(U_k)$ still holds in such conditions. My other issue is that I am not sure that $(1)$ is true.

Thank you.

Your proof is absolutely correct.

As Yanko said in his comment, each function $phi : X to Y$ between sets $X,Y$ has the property $phi(bigcup_{alpha in A} M_alpha) = bigcup_{alpha in A} phi(M_alpha)$, where $(M_alpha)_{alpha in A}$ is an arbitrary family of subsets of $X$.

However, to show that a non-constant holomorphic $f : U to mathbb{C}$ defined on arbitrary open $U subset mathbb{C}$ has an open image $f(U)$, it is unnecessary to consider unions. Let $w in f(U)$. Choose $z in U$ such that $f(z) = w$. Since $U$ is open, there exists an open disk $D(z,r) subset U$. Now $V = f(D(z,r))$ is open and we have $w in V subset f(U)$.