# Intersection of infinite sets that contain each other

• Apr 25th 2009, 04:13 AM
eddi
Intersection of infinite sets that contain each other
If each A_i is a set containing infinite elements, and A_1 contains A_2 contains A_3 contains ... on and on, then is the intersection of all these sets infinite?

I think the intersection of all the sets is infinite cause cause no matter how far you go down the A_i's that particular A_i will contain infinite elements and the intersection of all the sets before it, and including it, will be equal to A_i. So is my reasoning correct? Because maybe maybe as i approaches infinity the sets somehow get smaller?
• Apr 25th 2009, 04:40 AM
Plato
Quote:

Originally Posted by eddi
If each A_i is a set containing infinite elements, and A_1 contains A_2 contains A_3 contains ... on and on, then is the intersection of all these sets infinite?

Is this what the question says: $\left( {\forall n} \right)\left[ {\left| {A_n } \right| = \infty \,\& \,A_n \subseteq A_{n + 1} } \right]$?
That is, each set is infinite and each set is a subset of the ‘next’ set in the indexing.

If that is correct, the notice $A_1 = \bigcap\limits_n {A_n }$.
• Apr 25th 2009, 04:50 AM
eddi
Quote:

Originally Posted by Plato
Is this what the question says: $\left( {\forall n} \right)\left[ {\left| {A_n } \right| = \infty \,\& \,A_n \subseteq A_{n + 1} } \right]$?
That is, each set is infinite and each set is a subset of the ‘next’ set in the indexing.

If that is correct, the notice $A_1 = \bigcap\limits_n {A_n }$.

I think the subset symbol goes the other way:
$\left( {\forall n} \right)\left[ {\left| {A_n } \right| = \infty \,\& \,A_{n + 1} \subseteq A_n } \right]$

So $A_\infty = \bigcap\limits_n {A_n }$. If that makes any sense...
• Apr 25th 2009, 05:41 AM
Plato
In that case, the intersection may well be empty.
Consider: $\begin{gathered} A_n = \left\{ {n,n + 1,n + 2, \cdots } \right\} \hfill \\ ex.\quad A_{10} = \left\{ {10,11,12, \cdots } \right\} \hfill \\ \end{gathered}$

Then $\emptyset = \bigcap\limits_n {A_n }$.
• Apr 25th 2009, 06:04 AM
eddi
Quote:

Originally Posted by Plato
In that case, the intersection may well be empty.
Consider: $\begin{gathered} A_n = \left\{ {n,n + 1,n + 2, \cdots } \right\} \hfill \\ ex.\quad A_{10} = \left\{ {10,11,12, \cdots } \right\} \hfill \\ \end{gathered}$

Then $\emptyset = \bigcap\limits_n {A_n }$.

I haven't slept yet so maybe that's why I can't really see why $\emptyset = \bigcap\limits_n {A_n }$. haha.

I guess I forgot to say that n is part of the natural numbers...

As n -> $\infty$ does $|A_n|$ -> 0? Because in that example I can't see how $A_n$ gets "less infinite" as n goes on forever.
• Apr 25th 2009, 06:46 AM
Plato
Quote:

Originally Posted by eddi
As n -> $\infty$ does $|A_n|$ -> 0? Because in that example I can't see how $A_n$ gets "less infinite" as n goes on forever.

You have some real misconceptions about infinity.
The idea of “less infinite” is simply finite.
Now it is true that $N = \left\{ {0,1,2,3 \cdots } \right\},\;B = [0,\infty ),\,N \subset B~\& ~ \left| N \right| < \left| B \right|.$
That is about cardinality of infinite sets. There is a injection from N to B, but no injection from B to N.

In addition, $\lim _{n \to \infty } \left| {A_n } \right| \to 0$ implies that $A_K$ is finite for some $K$.
• Apr 25th 2009, 02:18 PM
eddi
Sorry, what I meant to say was that after considering your example $

\begin{gathered} A_n = \left\{ {n,n + 1,n + 2, \cdots } \right\} \end{gathered}
$
, I could not see how $\bigcap^{\infty}_{n=1} {A_n}=\emptyset$.

How I interpret your argument is that since $\bigcap^{i}_{n=1} {A_n }=A_{i}$ if $\lim _{n \to \infty } \left| {A_n } \right| \to 0$ then $\bigcap^{\infty}\limits_n {A_n }=\emptyset$.

I can't see how $\lim _{n \to \infty } \left| {A_n } \right| \to 0$ happens in your example.

You pointed out that the cardinality of the positive reals is greater than that of the natural numbers. Does this have to do with what I'm missing?
(I've had some exposure to countable/uncountable sets but not much.)
• Apr 25th 2009, 02:44 PM
Plato
Quote:

Originally Posted by eddi
How I interpret your argument is that since $\bigcap^{i}_{n=1} {A_n }=A_{i}$ if $\lim _{n \to \infty } \left| {A_n } \right| \to 0$ then $\bigcap^{\infty}\limits_n {A_n }=\emptyset$.

I can't see how ${\color{red} \lim _{n \to \infty } \left| {A_n } \right| \to 0}$ happens in your example.

The whole point is ${\color{red} \lim _{n \to \infty } \left| {A_n } \right| \to 0}$ never happens with infinite sets.
You are confusing several different ideas.

Here is the proof. Suppose that $k \in \bigcap\limits_n {A_n } \,\& \,k \in \mathbb{Z}^ +$.
But that is a clear contradiction because $\color{blue}k \notin A_{k + 1}$
Therefore, the intersection must be empty, even though each set in the intersection is infinite.
• Apr 25th 2009, 03:07 PM
eddi
Oh that makes sense. Thank you.