Arbitrary Intersection between $\mathbb{R, N, Z}$

Question

I would like to understand the functionality and intuition of performing an Arbitrary Union/Intersection on sets which contain sets, i.e their elements are sets. I would like some clarity on: A set(s) vs set(s) whose elements are sets.

$ G_1= \mathbb \{R\}, \ G_2= \mathbb \{Z\} \ G_3= \mathbb \{N\}\ $

Will: $ \bigcap_{i \in I}G_i = \varnothing$

I believe it to be so, because $ \bigcup_{i \in I} G_i = \{\mathbb R, \mathbb Z, \mathbb N\} $

Glad to be clarified on this concept.

Answer 1

If you write $$ G_1=\mathbb{N},\quad G_2=\mathbb{Z},\quad G_3=\mathbb{R} $$ then $G_1\cap G_2\cap G_3=\mathbb{N}$ and $G_1\cup G_2\cup G_3=\mathbb{R}$ simply because $\mathbb{N}\subset\mathbb{Z}\subset\mathbb{R}$.

On the other hand if you write $$ G_1=\{\mathbb{N}\},\quad G_2=\{\mathbb{Z}\},\quad G_3=\{\mathbb{R}\} $$ then $G_1\cap G_2\cap G_3=\emptyset$ and $G_1\cup G_2\cup G_3=\{\mathbb{N},\mathbb{Z},\mathbb{R}\}$.

The difference is in the idea that a set $A$ and a set whose only element is the set $A$ are two totally different things.

Answer 2

We have $x\in\bigcap_{i\in I}G_i$ if and only if $x\in G_i$ for every $i\in I$.

In your case e.g. $1\in G_i$ for $i=1,2,3$ so that $1\in\bigcap_{i\in I}G_i=G_1\cap G_2\cap G_3=\mathbb R\cap\mathbb Z\cap\mathbb N$.

Actually for $I=\{1,2,3\}$ we have: $$\bigcap_{i\in I}G_i=G_1\cap G_2\cap G_3=\mathbb R\cap\mathbb Z\cap\mathbb N=\mathbb N$$

Likewise we have $x\in\bigcup_{i\in I}G_i$ if and only if $x\in G_i$ for some $i\in I$.

In your case: $$\bigcup_{i\in I}G_i=G_1\cup G_2\cup G_3=\mathbb R\cup\mathbb Z\cup\mathbb N=\mathbb R$$

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edit (after edit of question).

If $A$ is a set then formally $\cup A$ is union of all elements of $A$.

That means that $x\in \cup A$ if and only if $x\in a$ for some $a\in A$. Actually we have: $$\bigcup_{i\in I}G_i:=\cup\{G_i\mid i\in I\}$$ so that $x\in\bigcup_{i\in I}G_i$ iff $x\in G_i$ for some $i\in I$.

In special case where $G_i=\{A_i\}$ the expression $x\in G_i$ is the same as $x=A_i$ so in that case the following statements are equivalent:

• $x\in\bigcup_{i\in I}G_i=\bigcup_{i\in I}\{A_i\}$
• $x=A_i$ for some $i\in I$
• $x\in\{A_i\mid i\in I\}$

Showing that: $$\bigcup_{i\in I}\{A_i\}=\{A_i\mid i\in I\}$$

We also have:$$\bigcup_{i\in I}\{A_i\}=\cup\{\{A_i\}\mid i\in I\}$$

Answer 3

I assume you mean $I=\{1,2,3\}. $

$G_1=\{N\}$ is an abbreviation for $\forall x\;(x\in G_1\iff x=N).$ Similarly for $G_2$ and $G_3.$

Let $S=\cap_{i\in I}G_i$. What this means is $\forall x \;(x\in S\iff \forall i\in I\;(x\in G_i)).$

In this case $I$ is finite and so is each $G_i.$ so we have $\forall x\;(x\in S\iff (x\in G_1\land x\in G_2\land x\in G_3)),$ which is equivalent to $\forall x\;(x\in S \iff (x=N\land x =Z\land x=R)). $ Since there is no $x$ such that $x=N=Z=R,$ there is no $x\in S,$ so $$\cap_{i\in I}G_i=S=\emptyset.$$

When I wrote "Let $S=\cap_{i\in I}G_i$" I was merely defining $S$ as an abbreviation for the RHS expression. This is often done to make the rest easier to read and comprehend.

There is also the notation $\cap X,$ often written with "\bigcap" as $\bigcap X$ In this notation, $X$ is a collection of sets, and $y\in \cap X$ means that $y$ belongs to every member of that collection. Similarly there is $\cup X$, or $\bigcup X$, where $z\in \cup X$ means $z$ belongs to at least one member of $X.$