Does there exist a set containing infinite elements, whose elements themselves are sets containing infinite elements?

Question

Does there exist a set containing infinite elements, whose elements themselves are sets containing infinite elements?

I think the answer is no, there is a famous paradox for it but I'm forgetting.

Answer 1

There are many sets with this property. One example is $A=\{S\subseteq\mathbb{N}\mid |S|=\infty\}$, i.e. the set of infinite subsets of the naturals.

Answer 2

For each $i\in\mathbb{N}$, set $S_i=\{i,i+1, i+2, \ldots\}$. Each $S_i$ is itself a subset of $\mathbb{N}$.

Then, define $$S=\{S_1, S_2, S_3, \ldots\}$$

Now, $S$ is a set with infinitely many elements, each of which is itself a set with infinitely many elements.

Answer 3

Take the set of all subsets of the natural numbers that are infinite.

I.e. $$S := \{T \in \mathcal{P}(\mathbb{N}) \mid |T| = + \infty\}$$

The key that this is possible is the axiom in ZF that says that the powerset is a well defined set. Hence, this set makes certainly sense since we use the so called 'set builder notation'

Answer 4

There is no paradox here. Indeed, under the usual axioms of set theory (ZFC) there are lots of such sets.

What ZFC disallows is sets which contain themselves, and Russell's paradox (I suspect this might be what you're vaguely remembering) shows that we can't simultaneously have basic set formation axioms and a set of all sets. But there's no problem with infinite sets of infinite sets.

In fact, according to ZFC the "universe of sets" is built entirely from sets of sets of ...! Specifically, ZFC proves that every set $x$ occurs somewhere in the "tower" of sets $V_\alpha$, where

• $V_0=\emptyset$

• $V_{\beta+1}=\mathcal{P}(V_\beta)$ (here "$\mathcal{P}(X)$" is the powerset of $X$), and

• $V_\alpha=\bigcup_{\beta<\alpha} V_\beta$ for $\alpha$ a limit.

Here $\alpha$ is an ordinal. If $\alpha$ is a finite ordinal, $V_\alpha$ will be finite; but once we go into the infinite ordinals we get all sorts of infinite sets, and infinite sets of infinite sets, and etc. So in fact this "sets-of-sets" stuff, which may feel paradoxical at first, is how ZFC interprets the entire mathematical universe!

Answer 5

In standard axiomatic set theory, the answer is certainly yes. For example, you could take the set $$ \{\{0, 1, 2, 3, \ldots\}, \{1, 2, 3,4,\ldots\}, \{2, 3, 4, 5,\ldots\},\ldots\}. $$ In more formal notation you might write this $$ \{ X \subseteq \mathbb N \mid \exists n \in \mathbb N(\forall m \in \mathbb N(m \in X \leftrightarrow m \geq n)) \}. $$

Answer 6

Example:

$$S=\{n\mathbb{Z}\mid n\in \mathbb{Z}^+\}$$ where $n\mathbb{Z}=\{\dots,-2n,-n,0,n,2n,\dots\}$.

Answer 7

Let $$S_k= \{j\in \Bbb N: j\ge k\}=\{k, k+1,k+2,\cdots\}$$

Then consider

$$S= \{S_k: k\in \Bbb N\}=\{S_1, S_2,\cdots\}$$

S is infinite and each $S_k$ are infinite too.

Answer 8

Here's a very simple example with no intersections between each subset:

$$ S = \left\{S_1,S_2,... \right\} $$

where

$$ S_n = \left\{p_n^1, p_n^2,...\right\} $$

and $p_n$ is the $n$-th prime.

Bonus: it's countable, just like the rationals.

Answer 9

How about a set of all lines on the plane, which are themselves sets of points?

Or a set of sets of natural numbers, greater than some natural number: $$\big\{\{1,2,3\dots\},\ \{2,3,4\dots\},\ \{3,4,5\dots\},\ \dots\big\}$$

Answer 10

In the Dedekind construction of the reals, the real line is such a set. Basically, each real number can be considered to split the set of rational numbers into a set of rational numbers greater than the real number, and rational numbers less than the real number. We can use this to define every real number in terms of rational numbers. So if you take the set of real numbers, and replace each of them with the set of rational numbers less than them, you have an uncountable set of countable sets.

Answer 11

How about several copies of the natural numbers where one number is missing from each copy:

$$\{\mathbb{N}-\{n\}\; |\; n\in \mathbb{N}\}$$