Is there an example where $ A \subseteq\mathcal {P}\bigcup A\ $ is no longer true?

Question

I came up with the following:

Let A = {x}

Then $ \bigcup A = x $

$ \mathcal {P} \bigcup A =$ ?

This is where I get stuck. The definition of power set is the set of all subsets of $A$ but, in this case my $A$ is just an element, which I believe have no subsets, not a set. How do you take the power set of that?

Answer 1

There are $2$ ways of handling your example:

1. In axiomatic set theory (normally using the ZFC axioms), everything in that universe is a set, in particular $x$, so if you want to calculate $\bigcup A$, you have to specify the nature of the set $x$, for example, if $x=\emptyset$, or $x = \{\emptyset, \{\emptyset\}\}$, and so on.

2. There is a concept in set theory for handling objects that are not sets (see Von Neumann–Bernays–Gödel set theory), which are called ur-elements. This objects can belong to a set but are not sets themselves, so the expression $a \in x \,$ doesn't make sense if $x$ is an ur-element, but you can say $x \in A$, if $A$ is a set containing $x$ as a member. Usually in that case, you leave them out of the union (since $x$ has no elements in it), so in your example, $\bigcup A = \bigcup \{x\} = \emptyset$.

If you take approach 2. you have your counterexample. I leave 1. for you.