# Why is my answer incorrect? (Indexed Sets)

This is a problem from Section 1.8 of Richard Hammack's How to Prove It. I highly recommend this book for anyone viewing.

$$\bigcup_{X \in \mathcal{P}(\Bbb N)} X =$$

I wrote my answer as $$\mathcal{P}(\Bbb N)$$. The correct answer, however, is $$\mathbb{N}$$. I do not understand this because $$\emptyset \in \mathcal{P}(\Bbb N)$$ because if $$A$$ is a set, then $$\forall A, \emptyset \subseteq A$$. However, $$\emptyset \not \in \mathbb{N}$$. Any help would be greatly appreciated!

• The union of all subsets of $A$ is $A$. Try it with $A = \{1, 2\}$: we have $\bigcup_{X \in \mathcal{P}(A)} X = \{\} \cup \{1\} \cup \{2\} \cup \{1, 2\} = \{1, 2\}$. Commented Jul 23 at 13:43
• @NaïmFavier: I see, thank you very much. I have a greater understanding now! Commented Jul 23 at 13:47
• You shouldn't use $N$ and $\Bbb N$ to denote the same set. Commented Jul 23 at 13:47
• But $\emptyset \not \in \cup X$. If $A, B, C$ are sets then $A\not\in A\cup B\cup C$. The union is set consisting of the elements of the sets; Not the sets themselve. Basically if $x \in \cup_{X\in \mathscr P(\mathbb N)}X$ then $x \in X\subset \mathbb N$ for some subset $X$. That means $x$ is a natural number. Commented Jul 23 at 16:17
• In short you are confusing $\{A,B,C\}$ with $A\cup B\cup C$. This are very different concepts. $\{A,B,C\}$ is a listing of the sets. But $A\cup B\cup C$ is a set of all the elements "dumped" into a single set. The listing of of all the sets in the power set of $\mathbb N$ is $\{\emptyset, \{1\}, \{1,2\}, ... etc\}$ and this set is the power set. But that set is not the UNION of the powerset. The union is the set of all the elements of the sets in there. And that pretty clearly is just all the natural numbers. Commented Jul 23 at 16:24

The notation $$\bigcup_{X \in \mathcal{P}(\mathbb{N})} X$$ does not mean $$\{X \mid X \in \mathcal{P}(\mathbb{N})\}$$. Rather, it means the union of all $$X$$ in $$\mathcal{P}(\mathbb{N})$$. For example, $$\bigcup_{i\in \{1,2\}} A_i$$ equals $$A_1\cup A_2$$, not $$\{A_1, A_2\}$$.

In your case, taking the union of all subsets of $$\mathbb{N}$$ gives $$\mathbb{N}$$. This is actually true for any set.

• Perhaps it would be worth explicitly writing what $\bigcup_{X\in\mathcal P(\mathbb N)}X$ does mean: it is $$\{x\mid\text{there is an X\in\mathcal P(\mathbb N) such that x\in X}\} \, .$$
– Joe
Commented Jul 23 at 15:42
• Also, I think this answer blurs the distinction between the union of a set, and the union of an indexed family. If $\{A_i\}_{i\in I}$ is an indexed family of sets (officially, this is just a function $i\mapsto A_i$ with a domain of $I$), then $$\bigcup_{i\in I}A_i=\{x\mid\text{there is an i\in I such that x\in A_i}\} \, .$$
– Joe
Commented Jul 23 at 15:50
• On the other hand, if $\mathcal C$ is a set, then the union of the sets in $\mathcal C$ (written $\bigcup_{X\in\mathcal C}X$ or just $\bigcup \mathcal C$) is defined by $$\bigcup_{X\in\mathcal C}X=\{x\mid \text{there is an X\in\mathcal C such that x\in X}\} \, .$$ These are two slightly different notions, although they are of course closely related.
– Joe
Commented Jul 23 at 15:50

Recall that $$X\in\mathcal{P}(\mathbb{N})\iff X\subseteq \mathbb{N}$$. So the stuff becomes $$\bigcup_{X\subseteq\mathbb{N}}X$$ which is the union of all subsets of $$\mathbb{N}$$. This clearly gives $$\mathbb{N}$$.