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?


1 Answer 1


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:


(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.

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    $\begingroup$ Concise and precise. $\endgroup$
    – Ziqi Fan
    Jan 25, 2021 at 5:18
  • $\begingroup$ I just have one question: how to prove that $\left\{\mathrm{dom} f: f \in \mathscr{C}\right\}$ is a set? $\endgroup$
    – Ziqi Fan
    Jan 25, 2021 at 5:44
  • $\begingroup$ 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. $\endgroup$
    – Ziqi Fan
    Jan 25, 2021 at 5:55
  • $\begingroup$ @ZiqiFan: Since $\mathscr{C}$ is a set, and $\operatorname{dom}$ is function-like, you you can get it from Replacement. Alternatively, it’s $$\left\{x\in\bigcup\bigcup\bigcup\mathscr{C}:\exists y\in\bigcup\bigcup\bigcup\mathscr{C}\,(\langle x,y\rangle\in\bigcup\mathscr{C}\right\}\,.$$ $\endgroup$ Jan 25, 2021 at 6:01

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