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In p.76 of this book, an example 4 says "Take a strictly decreasing sequence $F_1 \supset F_2 \supset \cdots $ of closed sets in $X$ ".

I don't understand if there exists such the sequence. I think the following statement may be true in general topology.

Let $(X, \mathcal{O})$ be an infinite Hausdorff topological space. Then, there exists a strictly decreasing sequence $F_1 \supset F_2 \supset \cdots $ of closed sets in $X$.

I may be able to get the proof of this statement. The summary of the proof is this.

$X$ is clearly an closed set. So we can get $X$ as $F_1$. And let $U$ be an arbitary "infinite" open set in $X$. Then, existing V such that V is an open set in $X$ and $\overline{V} \subsetneq \overline{U}$. Since $X$ is also an open set in $X$ and a closure is closed set, we get $F_1 \supsetneq F_2 \supsetneq \cdots $ as a strictly decreasing sequence of closed sets.

Is my proof correct? And I concerned if I use the axiom of choice in the proof.

Introduction to commutative algebra page 76

Riley
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  • How do you justify the existence of V ? (It is false if U is empty) – Maxime CAILLEUX Feb 21 '24 at 11:10
  • Thank you for your comment. I forget a condition of U. Let U be an infinite open set. – Riley Feb 21 '24 at 11:15
  • This doesn't answer your question, but perhaps you might find this interesting: spaces, which do not have such a sequence, are called noetherian. It is well-known (and not very difficult to show) that this is equivalent to each subspace being compact. And yes, T2 noetherian spaces are finite. – Ulli Feb 21 '24 at 15:39
  • @Ulli Thank you for telling me new information. – Riley Feb 21 '24 at 16:04

1 Answers1

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Why does there exist such $V$? You've not tried to justify this, which of course would be one of the most important steps, and don't seem to have used assumptions on $X$. You've also not linked that to actually obtaining, say, $F_{1000}$ from $F_{999}$. For help with justifying the claim $\mathfrak{a}_n\subsetneq\mathfrak{a}_{n+1}$ in the rest of the example, see here.

If we choose all $F_\bullet$ to be infinite then they'll satisfy the same hypotheses as $X$ so for the sake of induction, starting from $F_1:=X$, I need only find an infinite closed proper subset $F\subsetneq X$.

Take any two distinct points $x,y$ in $X$; as $X$ is Hausdorff we may find disjoint open neighbourhoods $U,V$ of $x,y$. If $V$ is infinite, then $\overline{V}$ is also infinite, closed and it does not contain $x$ because the neighbourhood $U$ of $x$ avoids $V$, so $F:=\overline{V}$ is an infinite, closed proper subset and therefore does the trick. If $V$ is finite, then $F:=X\setminus V$ is necessarily infinite and also proper and closed so this $F$ works.

This does not use the full axiom of choice, at best it uses countable choice (but it does use LEM ;)). But, a lot of algebra, topology and analysis does ("in secret") and I wouldn't worry about it unless you're really interested in the intersections of these disciplines with matters of logic. Many theorems you might find in a book on any of these will use the axiom of choice, somewhere, without explicitly saying so.

FShrike
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  • Thank you for answering to my question. I understand your proof that we can take such V. By the way, let me know what "LEM" is. – Riley Feb 21 '24 at 13:45
  • @Riley Law of excluded middle, uh, I don't know my logic but I think of it essentially as the law that everything is either strictly true or strictly false. It's what permits proof by contradiction, I'm told. $\neg\neg p\implies p$ or something. For example, I think I use that to conclude that $X\setminus V$ is infinite if $X$ is infinite and $V$ is finite, but maybe there's a way without it. Most "conventional" mathematics and mathematicians don't worry about this stuff, I only mentioned it because you seem a bit interested – FShrike Feb 21 '24 at 13:54
  • Thank you for your reply. I understood about LEW. – Riley Feb 21 '24 at 14:34
  • @FShrike: Perhaps I've missed something, but I think your proof uses (AC), as in each induction step you first choose $x, y$ and, moreover, disjoint neighborhoods of $x, y$, respectively. – Ulli Feb 21 '24 at 15:27
  • @Ulli this is a finite choice. In each induction step, I make finitely many choices, so I have proven the induction step. Then the usual induction theorem goes through without AC. However I really have no expertise at all in foundations so there’s a chance I’m wrong, but I don’t see how. Certainly it is true a finite choice, which is really a finite string of invocations of “$\exists$”, is ok in basic ZF. – FShrike Feb 21 '24 at 17:06
  • @FShrike: yes, of course, the step itself does not need (AC), but this gives you only the existence of an $F$, hence in the induction you have to choose infinitely many times. Of course, countable choice suffices. I'm also not an expert for this, so I cannot prove, that at least some weak version of (AC) is needed, but I would be surprised, if not. – Ulli Feb 21 '24 at 17:45
  • @Ulli I considered this but I “feel like” (and you know it’s a bad sign when you can only do mathematics with feelings!) the fact I can verify each induction step separately, without choice, means I can now follow the proof entirely in ZF. Like, ZF shows $P(0)$, and it shows for any given $n$ that $P(n)\implies P(n+1)$, still without choice, so ZF therefore shows $P$ is always true. Ah, but maybe to then say I can assemble all my constructed closed sets into one sequence needs countable choice? Confusing – FShrike Feb 21 '24 at 18:29
  • I think even for this much weaker statement, you need some choice: Let $X$ be an infinite set. Then there exists a strictly decreasing sequence $(F_n)_{n \in \mathbb N}$ of subsets of $X$. As you said, the point is, that you have to end up with a function $\mathbb N \rightarrow \mathcal P(X)$. – Ulli Feb 21 '24 at 18:54