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I'm currently working through some exercises in Amann & Escher's Analysis, where I'm asked to prove the following proposition for a set-valued function:

The following hold for the set-valued functions induced from $f$

$A_\alpha \subseteq X, \alpha \in A \Rightarrow f\left( \bigcup_{\alpha \in A} A_\alpha \right) = \bigcup_{\alpha \in A} f(A_\alpha)$

And the definition of Set-Valued Function is

Let $ f : X \to Y $ be a function. Then, using the above definitions, we have two "induced" set valued functions, $f : \mathcal{P}(X) \to \mathcal{P}(Y), \quad A \mapsto f(A) \quad \text{and} \quad f^{-1} : \mathcal{P}(Y) \to \mathcal{P}(X), \quad B \mapsto f^{-1}(B). $ Using the same symbol $f$ for two different functions leads to no confusion since the intent is always clear from context.

$\mathcal{P}$ is Power Set. Now I have a counterexample $$ X = \{1, 2\}, Y=\{1, 2\}, \quad F：X\rightarrow Y $$ Then the power set of $X$ and $Y$ is $$ P(Y)=P(X) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}. $$ Then we can define a set-valued function $$ f: P(X) \to P(Y) $$ such that: $$ f(\{1\}) = \{1\}, \quad f(\{2\}) = \{2\}, \quad f(\emptyset) = \emptyset, \quad f(\{1,2\}) = \emptyset. $$

Let $$ A_1 = \emptyset, \quad A_2 = \{1\}, \quad A_3 = \{2\}, \quad A_4 = \{1, 2\}. $$

Then $$ f(A_1 \cup A_2 \cup A_3 \cup A_4) = f(\{1,2\}) = \emptyset, $$ but $$ f(A_1) \cup f(A_2) \cup f(A_3) \cup f(A_4)=\emptyset \cup \{1\} \cup \{2\} \cup \emptyset= \{1,2\}. $$ In this case, $f\left( \bigcup_{\alpha \in A} A_\alpha \right) = \bigcup_{\alpha \in A} f(A_\alpha)$ does not hold. I don't know whether the mistake is in the book, or there's something wrong with my example, or whether some stronger condition is needed for this so-called 'union-preserving property' to hold.

Asaf Karagila
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Teng Wei
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    It isn't the definition of "set-valued function" it's the definition of "set-valued function induced from $f$". – Sassatelli Giulio Jun 09 '25 at 07:21
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    @SassatelliGiulio This is an answer to the question, not just a comment. Comments, per the Help Center, are temporary notes which are meant to either request clarification, or to suggest changes which might improve a post. See How do comments work for more information. Therefore, I suggest to post your answer as an answer. This brings extra visibility to the answer und puts the question off the unanswered list. – Anne Bauval Jun 09 '25 at 07:27
  • @AnneBauval It isn't an answer to the question, but if you believe so feel free to turn it into one. – Sassatelli Giulio Jun 09 '25 at 07:57
  • @SassatelliGiulio What's the difference? – Teng Wei Jun 09 '25 at 07:59
  • @TengWei A set-valued map is simply a function $g:A\to \mathcal P(B)$. Since this is, arguably, always the case in set theory and therefore in mathematics, the term is usually used in contexts where there is an implicit distinction between "objects" and "sets of objects" (for instance, real numbers and subsets of real numbers). – Sassatelli Giulio Jun 09 '25 at 10:04

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Your "counterexample" map $f:\mathcal P(X)\to\mathcal P(Y)$ is not induced by a map $\tilde f:X\to Y$, otherwise by definition you would have $$f(\{1,2\})=\{\tilde f(1),\tilde f(2)\}=f(\{1\})\cup f(\{2\})$$

Christophe Boilley
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  • So when I want to apply the mapping to the set ${1, 2}$, I still have to use the original rule $f$ on each element of ${1, 2}$—that is, 1 and 2—rather than defining a new mapping rule myself. That's also why the book uses the same letter $f$ to denote the function $f: \mathcal{P}(X) \to \mathcal{P}(Y)$. – Teng Wei Jun 09 '25 at 12:18