Confusion regarding the proof of "Every PID satisfies the Ascending Chain Condition".

Question

I refer to this proof of the fact that Principal Ideal Domains satisfy the Ascending Chain Condition.

It says

Let $\bigcup\limits_{i=1}^{\infty}I_i=(a)$. As $a$ is present in $\bigcup\limits_{i=1}^{\infty}I_i$, it should be present in some $I_m\subseteq \bigcup\limits_{i=1}^{\infty}I_i$

I don't understand why. For example, in $\Bbb{Z},(2^8)\subset (2^4)\subseteq(2^2)\subseteq(2)$. We know $6\in (2)$, and $6\notin (2^8),(2^4),(2^2)$. Also, $(2)=(2)\bigcup(2^2)\bigcup(2^4)\bigcup(2^8)$. Hence, it is possible that an element is present in $(2)\bigcup(2^2)\bigcup(2^4)\bigcup(2^8)$, but not in any of its sub-ideals ($(2^2),(2^4)$ or $(2^8)$).

You might say this is an example in which the number of ideals is finite, and here we have an infinite number of ideals. However, I don't see why it should be any diferent for an infintie number of ideals. Any clarification would be great!

Thanks in advance!

Answer 1

There is no trouble that $2$ is not in the other ideals. That is, in this case $m=1$, or $2\in (2)$. It must be in one at least, and that is enough. Recall that if $\mathscr C$ is a collection of sets then $x\in\bigcup \mathscr C$, if and only if $x\in C$ for at least one set $C$ in $\mathscr C$.

Answer 2

The claim is that the generator of the union of all the ideals lies in one of the "bottom" ideals - so the claim is that $2$ lies in one of the ideals, which it does: $2 \in (2)$. It does not imply that every element of the "top" ideal lies in one of the others.