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I am reading Enderton's "Elements of Set Theory". He defines the union operation as $$\cup A = \{ x \;|\; x \text{ belongs to some member of } A\} = \{x \;|\; (\exists b \in A) x \in b\}$$ Maybe I am missing a subtle point from earlier on in the text, but what if the set in question is not a "set of sets"? For example, if $A = \{1,2,3\}$, then is $\cup A = \emptyset$?

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    $\begingroup$ In systems of set theory such as ZF, every set is "a set of sets". $\endgroup$ – Lord Shark the Unknown Mar 30 at 19:11
  • $\begingroup$ So then is my $A$ really $\{\{1\},\{2\},\{3\}\}$? $\endgroup$ – theQman Mar 30 at 19:12
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    $\begingroup$ No, it's really $\{1,2,3\}$. $\endgroup$ – Lord Shark the Unknown Mar 30 at 19:13
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    $\begingroup$ See Enderton's chapter on "Natural Numbers" to see the usual way of representing positive integers as sets. $\endgroup$ – Lord Shark the Unknown Mar 30 at 19:20
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    $\begingroup$ Even without the "everything-is-a-set" context, we can still make sense of this by carefully writing down the definition: see this earlier question. $\endgroup$ – Noah Schweber Mar 30 at 19:32
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When you're working in a system built out of the set theory, everything is a set. I assume the book will cover this later on, but one way of "building natural numbers" out of sets is called Von Neumann ordinals and the construction is as follows:

Let $0 \equiv \emptyset$. Let the number $n \equiv n - 1 \cup \{ n - 1\}$. That is, under this system: \begin{align*} &1 \equiv 0 \cup \{ 0 \} = \emptyset \cup \{\emptyset\} = \{\emptyset\} \\ &2 \equiv 1 \cup \{1 \} = \{\emptyset\} \cup \{\{\emptyset\}\} = \{\emptyset, \{\emptyset\}\} \\ &3 \equiv 2 \cup \{2\} = \dots = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\} \end{align*}

The point of the representation is that 0 is a set with zero elements, 1 is a set with 1 element, 2 is a set with 2 elements and so on. So we can define the "size of a set" as "the natural number it is in bijection with".

This goes down a rabbit hole, and we can then ask "well, now that I have the natural numbers, how do I define addition? Multiplication? What about the integers? rationals? reals?"

We can construct all of these things, and a set theory book usually describes these encodings in one of the later chapters.

But the point is that at this level, all the "stuff of math" is sets, and you can always write expressions such as $1 \ \cup 2 \cup 3 = \{\emptyset\} \cup \{ \emptyset, \{\emptyset\}\} \cup \{ \emptyset, \{\emptyset\},\{ \emptyset, \{\emptyset\}\} \} = \{ \emptyset, \{\emptyset\},\{ \emptyset, \{\emptyset\}\}\} = 3 $!

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  • $\begingroup$ +1.... In the most-favored system of set theory, not only is everything (that can be said to exist) a set, but in the formal language there is no word or definition of "set". Things ( with various properties that $can$ be stated in the language ) are merely asserted to exist or not exist. $\endgroup$ – DanielWainfleet Mar 30 at 19:39
  • $\begingroup$ Indeed, thanks! much appreciated $\endgroup$ – Siddharth Bhat Mar 30 at 20:18
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As Lord Shark the Unknown indicates, in nearly all systems of set theory, every object under the sun is a set. In particular, the elements of any set are, in turn, sets themselves. For the specific example you give, i.e. $$ \cup \{ 1,2,3 \}, $$ we need to be able to describe the natural numbers as sets. On page 67 of Enderton's text, the construction is outlined. He defines \begin{align} 0 &= \varnothing \\ 1 &= \{0\} = \{ \varnothing \} \\ 2 &= \{0,1\} = \{ \varnothing, \{\varnothing\} \} \\ 3 &= \{0,1,2\} = \{ \varnothing, \{\varnothing\}, \{ \varnothing, \{\varnothing\} \}\}, \end{align} and so on. The "and so on" is explained in somewhat more detail in chapter 4 of the text (starting on page 66). Note that we could write $2 = \{\varnothing, \{\varnothing\} \}$ over and over again, but it is likely easier to see what is going on if we work one level of abstraction higher. That is, $2 = \{0,1\}$. Once we accept this abstraction, we have $$ \cup\{1,2,3\} = 1 \cup 2 \cup 3 = \{0\} \cup \{0,1\} \cup \{0,1,2\} = \{0,1,2\} = 3. $$

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  • $\begingroup$ I see. Maybe I am getting ahead of myself since I am only in chapter 2. But I guess the construction you provided is pretty natural since at this point the only set that we can use as a building block is $\emptyset$ (Empty Set Axiom)? Then we can use "pairing" to form the singleton $\{\emptyset\}$, which is defined to be $\{\emptyset, \emptyset\}$? $\endgroup$ – theQman Mar 31 at 2:09

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