Define Formally a set $T$ which includes all sets

Question

How may I define formally a set $T$ which includes all sets of sets which their members' size is finite?

For example $\{\{1\},\{2\},\{3\},\{4\}, \cdots\}$ is an element of $T$ since each set size inside of it is finite and equal to $1$.

However, the set $\{\{1,2,3,4, \cdots\}\}$ isn't in $T$, because its only member has infinitely many elements.

Note: By formally I mean this is how we define a set which includes all even numbers: $T=\{x \in R\mid x\text{ is even}\}$.

Another example from wikipedia:

$$F=\{n\mid n{\text{ is an integer, and }}0\leq n\leq 19\}.$$

Please Note, The last edit is wrong, I am talking about set that contains all sets of sets... (Rolled back the edit) — Clark, Nov 06 '20 at 17:24
Don’t know what you mean by “formally”, but within most standard set theories, such a collection would not be a set. The collection of all singletons is not a set: see here. — Arturo Magidin, Nov 06 '20 at 17:26
You cannot “define” such a set, because one can prove that there is no such set, at least in the two standard set theories (Zermelo-Fraenkel and Goedel-Bernays-von Neumann) — Arturo Magidin, Nov 06 '20 at 17:28
In other words: there is no “set of all finite sets”. There isn’t even a “set of all singletons”. — Arturo Magidin, Nov 06 '20 at 17:29
I already linked to a question that proves there is no set of all singletons. Whether it makes sense is immaterial. To paraphrase Neil DeGrasse Tyson, the nice thing about mathematics is that it doesn’t really care whether something makes sense or not. If you can prove it is not so, then it is not so. — Arturo Magidin, Nov 06 '20 at 17:30
Does this answer your question? Why does the set of all singleton sets not exist? — Arturo Magidin, Nov 06 '20 at 17:31
What you can do is define the set of all finite subsets of a particular set. All your examples are finite subsets of $\omega$ (or of $\mathbb{N}$, if you prefer). That you can do, but that is a lot more narrow than what your first paragraph describes. — Arturo Magidin, Nov 06 '20 at 17:33

score 1 · Answer 1 · answered Nov 06 '20 at 17:38

Basically, like this:

$$\{A\ |\ \forall a\in A, |a|<\aleph_0 \}$$

The condition $|a|<\aleph_0$ means "the cardinality of $a$ is less than $\aleph_0$", $\aleph_0$ being the smallest infinite cardinal number. You could replace this with many other equivalent ways of saying "$a$ is finite", such as "there is no injection from $a$ into itself".

I think, given the examples you've listed, this is basically what you mean by "formal". Of course, strictly speaking in formal set theory this is not correct, but I think those sorts of concerns might be a bit beyond the scope of the question as you intended it.

The reason it isn't correct is basically because we haven't specified the domain of the variable $A$ on the left hand side of the $|$ sign. For example, the following would be perfectly fine in formal set theory:

$$\{A\in\mathcal P(\mathbb N) \mid \forall a\in A, a \text{ is finite} \}$$

...where again, we replace "$a$ is finite" via some appropriate formalization. However, this only gives you the set of all finite sets of positive integers.

Define *Formally* a set $T$ which includes all sets

1 Answers1

Define Formally a set $T$ which includes all sets