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I'm trying to get a grasp of what a set represents from set-builder notation and have been working through two interesting problems. I've also attached my reasoning, althought I'm not sure if it's the correct way to approach the problem. I was hoping someone could clarify.

$\{x \in 2^{\mathbb{Z}} : 5 \in x \}$ This set represents the set of numbers $x$ which are a power of $2$ to some integer, where 5 is in $x$. $x$ is an integer and not a set, therefore $5$ cannot be in $x$, therefore this set represents the empty set, with cardinality 0.

$\{x \in 2^{\mathbb{Z}} : x \subseteq \{ 1, 2, 3\} \}$ This set represents the set of numbers $x$ which are a power of $2$ to some integer, where $x$ is also a subset of the set $\{1,2,3\}$. $x$ is an integer and not a set, therefore it cannot be a subset. Cardinality: $0$.

Is my assumption that this condition failing results in the empty set, or is this a vacuous truth and both sets are just $2^{\mathbb{Z}}$?

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    $\begingroup$ You're right that the set is empty if the condition is always false. But the expression $2^\Bbb Z$ often denotes a power set. $\endgroup$ Commented Oct 6, 2023 at 17:44
  • $\begingroup$ @Karl That's a great point, I didn't even think of that, thank you so much. How would I distinguish between the notation for the set of 2 to the power of the integers and the power set of Z? $\endgroup$ Commented Oct 6, 2023 at 20:15
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    $\begingroup$ Notation is often slightly ambiguous and you have to rely on context to interpret it correctly. The author should probably say what $2^\Bbb Z$ is, but it looks like they mean the power set since they're treating its elements as sets (but see Qiaochu's note). Also, operations defined on individual objects don't automatically apply to sets of objects, so it's not safe to assume $2^\Bbb Z$ can mean $\{2^x:x\in\Bbb Z\}$ unless that has been stated. $\endgroup$ Commented Oct 6, 2023 at 23:08
  • $\begingroup$ @Matthew: if this question is from a textbook then $2^{\mathbb{Z}}$ likely refers to the powerset of $\mathbb{Z}$, which would make this question make a lot more sense. Can you say more about where they come from? $\endgroup$ Commented Oct 7, 2023 at 2:06
  • $\begingroup$ @QiaochuYuan it does come from a text book-- Scheinerman. I definitely think it is referring to the power set, it was just an oversight on my end. $\endgroup$ Commented Oct 7, 2023 at 3:24

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$x$ is an integer and not a set

So you need to be careful about this for sort of awful reasons. The problem is that in the standard formalization of mathematics, namely ZF(C) set theory, everything is a set, including integers (I don't mean the set of integers, I mean individual integers). So the answer to the question of whether $5 \in x$ actually depends on the exact choice of sets used to build the integers, which is awful (since this shouldn't matter, really).

For example if you build the positive integers as the von Neumann ordinals then $0 = \emptyset, 1 = \{ \emptyset \}, 2 = \{ \emptyset, \{ \emptyset \} \}$ and in general $n+1 = n \cup \{ n \}$. This definition means that if $a, b$ are two positive integers (or more generally two ordinals) then $a \in b$ (as sets) iff $a < b$, which means that for example $5 \in 8$!

But we still need to be careful here since the sets $5, 8 \in \mathbb{N}$ are, strictly speaking, different from the sets $5, 8 \in \mathbb{Z}$, or the sets $5, 8 \in \mathbb{Q}$ since e.g. we need to build $\mathbb{Z}$ from $\mathbb{N}$ using ordered pairs and so forth.

So it's just terrible stuff. None of this should matter because the properties of numbers should not depend on the details of how they're constructed in set theory. So it's better to just not even ask questions that could potentially be sensitive to such details. This is a great example of the failure of set theory as a foundation to address a key aspect of mathematical practice, which is the informal type system in play (which you implicitly refer to when you say that an integer is not a set).

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  • $\begingroup$ Of course if $2^{\mathbb{Z}}$ actually refers to the powerset of $\mathbb{Z}$ then there are no longer any issues. $\endgroup$ Commented Oct 7, 2023 at 2:07

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