Why does $\bigcup_{X \in \mathscr{P}(\mathbb{N})}X = \mathbb{N}$, instead of $\mathscr{P}(\mathbb{N})?$

This is a practice problem I just came across, $$\bigcup_{X \in \mathscr{P}(\mathbb{N})}X =$$ I came up with $\mathscr{P}(\mathbb{N})$, but the book lists $\mathbb{N}$ as the solution. I would've thought the solution set would look something like this, $$\{ \emptyset, \{1\},\{2\},\{309,12\},\{14,900,8,22\},\{6\},... \}$$ as opposed to this, $$\{ 1,2,3,...812,...973,... \}$$ We're "unionizing" every possible element of $\mathscr{P}(\mathbb{N})$, each of which is a subset of $\mathbb{N}$, not an element of $\mathbb{N}$. More generally, if for set $S$ you were to take the union of all it's elements, wouldn't your result be set $S$ itself? $$\bigcup_{x \in S}x=S$$ Not sure where I'm going wrong here. Any help most welcome :)

Update: I get it now. Thanks for you helpful explanations. My mistake was, as Brian put it,

When you take a union of sets, you’re collecting the elements of those sets into one bit set; you’re not collecting the sets themselves.

So my "general" example of taking the union of all the elements in a set is wrong because you can't take the union of elements, only the union of sets. I blame it all on the curly brackets :)

• You are right . Feb 24 '13 at 18:07
• I don't agree with the statement "my 'general' example of taking the union of all the elements in a set is wrong because you can't take the union of elements, only the union of sets." Even if $S$ is a set of sets (and in axiomatic set theory everything is a set) the union $\bigcup_{x\in S}x$ (which by the way can be written more simply as $\bigcup S$) is not generally the same thing as $S$ itself. Feb 24 '13 at 19:13
• You're right to blame it on the curly brackets though. To go from $S$ to $\bigcup S$ you essentially "remove some curly brackets" as in Brian's example. Feb 24 '13 at 19:15

What does it mean to say that $$x\in\bigcup_{X\in\wp(\Bbb N)}X\;?$$ By definition it means that there is some $X\in\wp(\Bbb N)$ such that $x\in X$. In other words, there is some $X\subseteq\Bbb N$ such that $x\in X$. But then $x\in X\subseteq\Bbb N$, so $x\in\Bbb N$. In other words, every member of $\bigcup_{X\in\wp(\Bbb N)}X$ is a natural number, and

$$\bigcup_{X\in\wp(\Bbb N)}X\subseteq\Bbb N\;.$$

Note too that if $x\in X\in\wp(\Bbb N)$, then by definition $x\in\bigcup_{X\in\wp(\Bbb N)}X$, so for every $X\in\wp(\Bbb N)$ we have $X\subseteq\bigcup_{X\in\wp(\Bbb N)}X$. In particular, for any $n\in\Bbb N$ we have $\{n\}\in\wp(\Bbb N)$, so $n\in\{n\}\subseteq\bigcup_{X\in\wp(\Bbb N)}X$, and therefore $$\Bbb N\subseteq\bigcup_{X\in\wp(\Bbb N)}X\;.$$ It follows immediately that $$\bigcup_{X\in\wp(\Bbb N)}X=\Bbb N\;.$$

It may help to look at a finite example. What is $$\bigcup_{X\in\wp(S)}X\;,$$ where $S=\{0,1,2\}$? $\wp(S)=\big\{\varnothing,\{0\},\{1\}.\{2\},\{0,1\},\{0,2\},\{1,2\},\{0,1,2\}\big\}$, so $$\bigcup_{X\in\wp(S)}X=\varnothing\cup\{0\}\cup\{1\}\cup\{2\}\cup\{0,1\}\cup\{0,2\}\cup\{1,2\}\cup\{0,1,2\}=\{0,1,2\}=S\;.$$

When you take a union of sets, you’re collecting the elements of those sets into one big set; you’re not collecting the sets themselves.

• @ivan: You’re welcome! Feb 24 '13 at 17:39

If $X\in \mathcal{P}(\mathbb N)$, then $X \subseteq \mathbb N$. So the union of two $X,Y\in\mathcal P (\mathbb N)$ is a subset of $\mathbb N$.

Your second remark is (depending on notation) wrong. $\bigcup\limits_{x\in S} x$ is not defined if the element of S are not sets themselves. One way to circumvent this is to identify $x$ with $\{x\}$ in such a union, so the union becomes $$\bigcup\limits_{x\in S} x = \bigcup\limits_{\{x\}\subseteq S} \{x\}$$

• @Stephan - Ah, I see your point. I was totally misinterpreting the union operation in this example.
– ivan
Feb 24 '13 at 17:29

The union of sets combines all elements of those sets, which results in one set containing every element that belongs to any of the sets we are taking the union.

So are taking the union of all $X_i \in \mathcal{P}$ which is the set $$\bigcup_{X \in \mathscr{P}(\mathbb{N})}X = \bigcup_{X_i \in \mathscr{P}(\mathbb{N})}X_i \; =\{x \in X_1 \lor x \in X_2, \lor ...X_n \lor x\in \ldots \mid X_i \in \mathcal{P}(\mathbb N)\}$$ $$= \{x \in X_i \mid X_i \in \mathcal{P}(\mathbb N), \;\;\text{for some}\;\;i\in I\}= \mathbb N$$

The meaning of the symbol is $$\bigcup_{X \in \mathscr{P}(\mathbb{N})}X = \{ x : \text{x \in X for some X \in \mathscr{P}(\mathbb{N})} \}.$$ So this union is a subset of $\Bbb{N}$ (which takes care of any worry we might have because I have written $\{ x : \dots \}$ without specifying an ambient set), because $x \in X \in \mathscr{P}(\mathbb{N})$ is just a convoluted way to say $x \in \Bbb{N}$. And clearly, given any $x \in \Bbb{N}$, then $x \in \{ x \} \in \mathscr{P}(\mathbb{N})$.

Every element $X \in \mathscr{P}(\mathbb{N})$ satisfies $X \subset \mathbb{N}$, so you must have $\cup_{X \in \mathscr{P}(\mathbb{N})} X \subset \mathbb{N}$.

Theorem. Suppose that $\{A_i\mid i\in I\}$ is a collection of sets, and $B$ is such that for all $i\in I$, $A_i\subseteq B$. Then $\bigcup_{i\in I}A_i\subseteq B$.

Proof. Let $x\in\bigcup_{i\in I}A_i$ then there exists some $i\in I$ such that $x\in A_i$. We assumed that $A_i\subseteq B$, and therefore $x\in B$. $\square$

Therefore $\bigcup_{X\in\mathcal P(A)}X\subseteq A$. The other direction is obvious because $A\in\mathcal P(A)$ so it is a part of this union. Now apply this for $A=\Bbb N$.

By definition, $\cup_{a\in \cal A}a$, where $\cal A$ is a collection of subsets of a set $X$, is the set consisting of all elements who are in some set of $\cal A$. More precisely, $\cup_{a\in \cal A}a=\{x\in X:\exists a\in \cal A,\, x\in a\}$. You can't do $\cup_{x\in S}x$ unless you can join (who is an operation of sets) the elements of $S$.