An interesting problem about functions in set theory

I am working on the following problem:

First prove the following proposition: let $$\mathscr{C}$$ be a set of functions which has the following property: $$\begin{equation*} \left(\forall f, g \in \mathscr{C}\right)\left(f \subseteq g \vee g \subseteq f\right), \end{equation*}$$ then $$\bigcup{\mathscr{C}}$$ is a function. Then describe the domain of $$\bigcup{\mathscr{C}}$$ with respect to functions in $$\mathscr{C}$$.

I have proved that $$\bigcup\mathscr{C}$$ is indeed a function. The proof goes as follows.

To prove that $$\bigcup{\mathscr{C}}$$ is a function, we have to show that for arbitrary $$a, b, c$$, if $$\langle a,b \rangle \in \bigcup{\mathscr{C}}$$ and $$\langle a,c \rangle \in \bigcup{\mathscr{C}}$$, then $$b = c$$. Assume that $$z \in \bigcup{\mathscr{C}}$$, then there has to exist a $$y \in \mathscr{C}$$ such that $$z \in y$$. As it is known that $$\mathscr{C}$$ is a set of functions, we can see that $$y$$ is a function and $$z = \langle a,b\rangle \in y$$. Next we assume that $$\langle a,b \rangle \in \bigcup{\mathscr{C}}$$ and $$\langle a,c \rangle \in \bigcup{\mathscr{C}}$$, and $$b \neq c$$. As a result, there have to exist two different functions $$f,g \in \mathscr{C}$$ such that $$\langle a,b \rangle \in f$$ and $$\langle a,c \rangle \in g$$ while $$\langle a,b \rangle \not\in g$$ and $$\langle a,c \rangle \not\in f$$. On the other hand, we know $$f \subseteq g$$ or $$g \subseteq f$$. If $$f \subseteq g$$, then we require $$\langle a,b \rangle \in g$$, which leads to contradiction. Similarly, if $$g \subseteq f$$, then we require $$\langle a,c \rangle \in f$$, which also leads to a contradiction. As a result, the assumption that $$\langle a,b \rangle \in \bigcup{\mathscr{C}}$$ and $$\langle a,c \rangle \in \bigcup{\mathscr{C}}$$, and $$b \neq c$$ has to be false. We may conclude that $$\langle a,b \rangle \in \bigcup{\mathscr{C}}$$ and $$\langle a,c \rangle \in \bigcup{\mathscr{C}}$$ lead to $$b = c$$, and $$\bigcup{\mathscr{C}}$$ has to be a function.

However, I am stuck on describing the domain of $$\bigcup{\mathscr{C}}$$. It seems that the domain of $$\bigcup{\mathscr{C}}$$ is related to the "largest" element in $$\mathscr{C}$$ in terms of $$\subseteq$$. I cannot find a precise description. Could anyone help me verify my proof and try to propose a description?

Your proof is correct but more complicated than necessary. Suppose that $$\langle a,b\rangle,\langle a,c\rangle\in\bigcup\mathscr{C}$$. Then there are $$f,g\in\mathscr{C}$$ such that $$\langle a,b\rangle\in f$$ and $$\langle a,c\rangle\in g$$. Without loss of generality we may assume that $$f\subseteq g$$, so $$\langle a,b\rangle\in g$$, and since $$g$$ is a function, it follows that $$b=c$$ and hence that $$\bigcup\mathscr{C}$$ is a function.
The domain of $$\bigcup\mathscr{C}$$ is just what you’d expect:
$$\operatorname{dom}\bigcup\mathscr{C}=\bigcup\left\{\operatorname{dom}f:f\in\mathscr{C}\right\}\,.$$
(The proof is very straightforward.) If $$\mathscr{C}$$ has a largest element, $$\bigcup\mathscr{C}$$ will be equal to that element and will of course have the same domain, but there is no reason to expect that $$\mathscr{C}$$ will have a largest element. The union, however, is the least upper bound of $$\mathscr{C}$$, and it does the trick.
• I just have one question: how to prove that $\left\{\mathrm{dom} f: f \in \mathscr{C}\right\}$ is a set? Jan 25, 2021 at 5:44
• I guess technically we should use $\left\{z \in \mathrm{Dom}\left(\bigcup{\mathscr{C}}\right): \exists f\ f \in \mathscr{C} \wedge z \in \mathrm{Dom}\left(f\right)\right\}$, which is clearly a set according to the rule of separation. Jan 25, 2021 at 5:55
• @ZiqiFan: Since $\mathscr{C}$ is a set, and $\operatorname{dom}$ is function-like, you you can get it from Replacement. Alternatively, it’s $$\left\{x\in\bigcup\bigcup\bigcup\mathscr{C}:\exists y\in\bigcup\bigcup\bigcup\mathscr{C}\,(\langle x,y\rangle\in\bigcup\mathscr{C}\right\}\,.$$ Jan 25, 2021 at 6:01