# Infinite intersection of indexed sets

1. Jul 16, 2011

### math771

Every element of a set $A$ can be written $a=w.a_1a_2a_3\ldots{a_n}\ldots$ with $w, a_n\in\mathbb{Z}$ and $0\leq a_n\leq9$ for every $n\in\mathbb{N}.$ If $A$ is bounded, there exists a greatest whole part $\overline{w}$ of the elements of $A,$ and because any set $S$ of elements $a_n$ is bounded, for every $n,$ there exists a greatest element $\overline{a_n}$ of $S.$

Let $A_0\subset{A}$ be the set of elements $a=\overline{w}.a_1a_2a_3\ldots{a_n}\ldots$ and $A_1\subset{A_0}$ the set of elements $a=\overline{w}.\overline{a_1}a_2a_3\ldots{a_n}\ldots,$ and define $A_{n+1}\subset{A_n}$ as the set of elements $a=\overline{w}.\overline{a_1}\overline{a_2}\overline{a_3}\ldots\overline{a_n}\overline{a_{n+1}}a_{n+2}\ldots.$

It would appear as though $\alpha\in\bigcap_{n=1}^\infty{A_n}$ contains a single element--namely, $\sup(A)$. However, this seems to lead to the absurd conclusion that for every set $A,\sup(A)\in{A}$. Perhaps we would say simply that $\bigcap_{n=1}^\infty{A_n}=\emptyset$ in case $\sup(A)\not\in{A}$?

Last edited: Jul 16, 2011
2. Jul 16, 2011

### micromass

Hi math771!

Here you went a bit too fast. It isn't necessary that $a=\overline{w}.a_1a_2a_3\ldots{a_n}...$ or $a=\overline{w}.\overline{a_1}a_2a_3\ldots{a_n}...,$ are in A. So it isn't necessary that $A_1\subseteq A$.

For example, take $A=\{222.000000..., 0.3333.....\}$, then it isn't true that $222.3333.....$ is in A, but that is what you are claiming!! Furthermore, 222.3333.... isn't even the supremum of A! So you'll need a different method to actually obtain the supremum.

3. Jul 16, 2011

### math771

I understand what you're trying to say. However, I meant to imply that A_0, for example, would be defined as the set of precisely those elements of A whose whole part is equal to $\overline{w}$ by stating that A_0 is a subset of A.

4. Jul 16, 2011

### micromass

Ah yes, I see now. Well, then I see no real reason why

$$\bigcap_{n}{A_n}$$

should be nonempty. If you use completeness, then this intersection is nonempty if the An are closed and nonempty. But the An aren't necessarily closed here!

5. Jul 16, 2011

### disregardthat

This is how I understand you:

A is a bounded subset of the reals. Let w be the largest integer such that w.a1a2a3.... for some sequence a_i is in A, and let A_0 be the subset of A such that w.a1a2... is in A for some sequence a_i. Let b0 be the largest integer in A_0 such that w.b0a1a2... is in A_0 for some sequence a_i. Recursively define A_n as such: let b_n be the largest integer such that w.b0b1b2...b_(n-1)a1a2a3... is in A_(n-1) for some sequence a_i, and let A_n be the subset of A_(n-1) of the elements on the form w.b0b1...bna1a2... for some sequence a_i.

Now what you seem to want to conclude is that w.b0b1b2... is in the intersection of the A_n's. But this is not necessarily so. For every n, there is a sequence a_i such that w.b0b1...bna1a2... is in A_n for some sequence a_i, but you don't know whether w.b0b1b2.... is in A_n for any n.