0
$\begingroup$

I am trying to understand the proof for the following theorem from Apostol:

enter image description here

Here is the proof:

enter image description here

I don't understand the parts underlined in red. For the first part, why is it trivial if each $Q$ is finite? For the second part, how does $Q$ contain all points but possibly a finite amount of points of $A$? Does it have something to do with the first condition $Q_{k+1}\subseteq Q_{k}$?

$\endgroup$
2
$\begingroup$
  1. Suppose that $Q_i$ only has $n$ elements. Since $Q_i\supset Q_{i+1}\supset Q_{i+2}\supset\cdots$, you have$$n=\#Q_i\geqslant\#Q_{i+1}\geqslant\#Q_{i+2}\geqslant\cdots$$So, you have a decreasing sequence of elements of $\{1,2,\ldots,n\}$. But then the sequence $\#Q_i,\#Q_{i+1},\#Q_{i+2},\ldots$ must becaom stable after some point. So, for some $N\geqslant i$, you have $Q_N=Q_{N+1}=Q_{N+2}=\cdots$ and therefore $\bigcap_{n\in\mathbb N}Q_n=Q_N\neq\emptyset$.
  2. Yes, it has to do with that. Note that $x_1$ belongs to every $Q_n$, that $x_2$ belongs to every $Q_n$ except perhaps for $Q_1$, then $x_3$ belongs to every $Q_n$ except perhaps for $Q_1$ and $Q_2$ and so on. So, for each $Q_n$, all but possibly a finite number of elements of $A$ belong to $Q_n$.
$\endgroup$
1
$\begingroup$

If at least one $Q$ is finite, say $|Q_n|=N$, then $|Q_k|$, $k\ge N$, is a non-decreasing sequence of positive integers. It follows that $|Q_k|$ is constant for $k$ large enough. But then $Q_k$ is also constant for these large $k$.


By the nesting property, $x_j\in Q_k$ for all $j\ge k$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy