# Union of sets of sets

I am not sure how one should interpret the union of Sets of sets. Is the union a set with the elements being the elements of the sets inside the Sets, or is it a set with the elements being the Sets?

• the latter one. May 8, 2014 at 11:14
• It seems you do not agree?
– Emil
May 8, 2014 at 11:22
• Okay then, since the other comment disappeared, unless someone says elsewise I will think of the elements as being the Sets. But then 1 is not an element of {(0,1), (2,3)}U{(0,5)} I take it?
– Emil
May 8, 2014 at 11:48
• Yes, Emil, you're correct. $1$ is not an element of your union. May 8, 2014 at 12:12

Let $A, B, C, D, E$ be sets.

$A = \{0, 1, 2, \cdots 9\}$

$B = \{0, 2, 4, \cdots, 18\}$

$C = \{1, 3, 5, \cdots, 19\}$

$D = \{10, 11, 12, \cdots 19\}$

$E = \{0, 5, 10, 15, 20\}$

Let $P = \{A, B, E\}$ and $Q = \{C, D\}$.

Then $P\cup Q = \{A, B, C, D, E\}$.

Here, the elements of $P, Q$ are sets, so their union is a set of sets (a set whose elements are sets). So their union is not a set of all the set elements contained in $A, B, C, D, E$. (I.e., none of the numbers $0$ to $20$ are in $P, \;Q,\;$ OR $P\cup Q$.)

If you want to describe the union of all elements in $A, B, C, D, E$, you would write, e.g., $A \cup B \cup C\cup D\cup E = \{0, 1, 2, 3, \ldots, 18, 19, 20\}$.

Perhaps using plain language here can confuse, so

(1) Interpreting Sets of sets literally, Let $\mathscr X$ and $\mathscr Y$ be "sets of sets", with $\mathscr X$ = {$X_1, X_2, ..$} and $\mathscr Y$ = {$Y_1, Y_2, ..$} where the $X_i$ and $Y_i$ are sets then $\mathscr X \cup \mathscr Y$ = {$X_1, X_2, ..Y_1, Y_2, ..$} and $\mathscr X \cup \mathscr Y$ is a set of sets.

(2) If we are talking about sets of sets generally so that $\mathscr X$ is one such "set of sets" and $\mathscr X$ = {$X_1, X_2, ..$} where each $X_i$ is a set = {$x_{i,1 }, x_{i,2 }, ...$} then you can form the union of the set of sets in $\mathscr X$, notated as $\cup \mathscr X$ = {$x_{1,1 }, x_{1,2 }, ... x_{2,1 }, x_{2,2 }, ...$}

• Yeah I think I needed to understand the literal version. I'd probably just call the latter a union of a set of sets.
– Emil
May 8, 2014 at 12:06
• @emil. Both are valid constructions so you just need to be sure which one is indicated in the context you are reading. May 8, 2014 at 12:11

A set (say $X$) of sets (say $A$) is just a set where all of it's elements are sets themselves. Say we are living in $\mathbb{R}^3$, then if we let $X_1=\{A|m(A)=1\}$, then $X$ is the set of all sets with measure $A$ equal to 1.

If we take $\bigcup_{\alpha\in\mathbb{R}^+}X_\alpha$, then this union of sets of sets, contains every set $X_\alpha$, which itself contains all the sets of measure $\alpha$.

It would be wrong to say that the elements of the union are all the sets with some positive measure $\alpha$, since the $A$'s are all elements of $X_\alpha$'s themselves. The elements of the union are all the sets of sets $X_\alpha$.

Hopefully this example helps :)

• Oh man. I feel measure theory is going to be very challenging to master if I need to keep this many layers of abstraction in mind.
– Emil
May 8, 2014 at 12:02
• @Emil: It will only be hard if you will insist on holding the layers. Instead, let go of the unnecessary "hands on" layers of examples and uses. Work with the definitions, and when need be, apply them to concrete objects. It takes some time to develop this sort of way of thinking, but it ends up quite useful -- not just for doing mathematics, but in general for solving problems. May 8, 2014 at 20:23
• @AsafKaragila That's true, I don't want it to seem like this is overly complicated. I think measure theory has different types of difficulties :S May 8, 2014 at 22:24

Is the union of a set with the elements being the elements of the sets inside the Sets...

Yes. More precisely, let $F=\{s_1, s_2, s_3, ...\}$ where the $s_i$ are sets. Then $x\in\cup F$ (the union of $F$) if and only if $x$ is an element of at least one of the $s_i$.

Likewise $x\in\cap F$ (the intersection of $F$) if and only if $x$ is an element of every one of the $s_i$.