In the book "Naive Set Theory" of Halmos there is a statement

Show that $E$ is always equal to $\bigcup\{x:x\in \mathcal{P}(E)\}$, but that the result of applying $\mathcal{P}$ and $\bigcup$ to $E$ in the other order is a set that includes $E$ as a subset, typically a proper subset.

I can't understand the second part, i.e. the other way to order $\mathcal{P}$ and $\bigcup$. I've found this question "Applying union and power set in different orders". But the answer there doesn't satisfy me and doesn't seem reasonable. Because to write $\bigcup{E}$ means $E$ is a set of sets. This is because the author says:

The simplest symbol for $U = \{x: x \in X \text{ for some }X \text{ in }\mathcal{C}\}$, called a union of the collection $\mathcal{C}$, is $\bigcup{\mathcal{C}}$.

But from the context it is not clear whether $E$ is a set of sets or not. What author says is only:

We assume nevertheless (in this section only) that all the sets to be mentioned are subsets of one and the same set $E$ and that all complements (unless otherwise specified) are formed relative to that $E$.

Anyway, even if $E$ is a set of sets, then $\bigcup{E} = \{x: x \in X \text{ for some } X \in E\}$. And the power set $\mathcal{P}(\bigcup{E})$ contains all subsets whose elements are elements of those sets that are in $E$. So in such situation $E$ can never be a subset of the power set $\mathcal{P}(\bigcup{E})$, they have different elements.

Please clarify what I am missing. What is the meaning of the "other order of $\mathcal{P}$ and $\bigcup$"? And if I got it correctly, then how is it possible for $E$ to be a subset of the $\mathcal{P}(\bigcup{E})$? Maybe I misunderstand the statement "a set $A$ includes $E$ as a subset"? For me it means $E$ is a subset of $A$ ($E \subset A$).


You're correct that you need $E$ to be a set of sets in order for $\bigcup E$ to make sense. In this case, the author is saying that $\bigcup \mathcal{P}(E) = E$ and that, if $E$ is a set of sets, then $E \subseteq \mathcal{P}(\bigcup E)$ but not necessarily $\mathcal{P}(\bigcup E) \subseteq E$.

In the most commonly used set theoretic foundations, i.e. ZFC or subsystems of ZFC, everything is a set, in which case it always makes sense to talk about $\bigcup E$ no matter what set $E$ is. If I recall correctly, Halmos's book uses (despite its name) works within a subsystem of ZFC, so that $E$ is automatically a set of sets no matter what it is.

  • $\begingroup$ Could you please show how it is ever possible that $E \subset \mathcal{P}(\bigcup {E})$? Because as I said above, even if $E$ is a set of sets, I can't figure out why it should be a subset of the power set of the union of the elements of $E$ $\endgroup$ – Turkhan Badalov Nov 23 '17 at 19:55
  • 1
    $\begingroup$ @TurkhanBadalov Sure. Let $A \in E$. We want to show that $A \subseteq \bigcup E$, so let $a \in A$. Since there is some $X \in E$ such that $a \in X$ (namely, X = A), so by the definition of $\bigcup E$, $a \in \bigcup E$. Thus, $A \subseteq \bigcup E$, and so, $A \in \mathcal{P}(\bigcup E)$. $\endgroup$ – Duncan Ramage Nov 23 '17 at 20:37
  • $\begingroup$ Here's a concrete example so you can see it yourself. Let $E = \{\{a\}, \{b\}\}$. Then, $\bigcup E = \{a, b\}$, and $\mathcal{P}(\bigcup E) = \{ \varnothing, \{a\}, \{b\}, \{a, b\}\}$. As you can see, $E \subseteq \mathcal{P}(\bigcup E)$. This also provides a convenient example that the reverse containment may not be true. P.S., wonderful choice of textbook, it's one of the few I've ever read cover to cover, and really got me interested in set theory. $\endgroup$ – Duncan Ramage Nov 23 '17 at 20:39
  • $\begingroup$ Thank you very much for the proof and example! Yes, this is my first attempt to approach mathematics. And this book helps me a lot. $\endgroup$ – Turkhan Badalov Nov 23 '17 at 21:05

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