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Background

Definition: A ring $R$ is said to satisfy the ascending chain condition (ACC) for left (right) ideals if for each sequence of left (right) ideals $A_1,A_2,\ldots$ of $R$ with $A_1\subseteq A_2\subseteq\cdots,$ there exists a positive integer $n$ (depending on the sequence) such that $A_n=A_{n+1}=\dots$.

Questions

If I want to translate the portion where it says: "if for each sequence of left (right) ideals $A_1,A_2,\ldots$ of $R$ with $A_1\subseteq A_2\subseteq\cdots,$ there exists a positive integer $n$ (depending on the sequence) such that $A_n=A_{n+1}=\dots$" in the above Definition, does it go as follow:

$$(\exists n\in \Bbb{N})(\forall i>n )(\forall \{A_i\}_{i\in \Bbb{N}}\subset R)(A_1\subseteq A_2\subseteq\cdots \Rightarrow A_n=A_{n+1}=\dots)$$

If so, I don't think the translation is complete, since I am not sure how to capture the $\ldots$ notation in symbolic logic notation.

Bill Dubuque
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Seth
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  • Though the question arose for ideals, it makes sense for any (po)set so I removed those tags. – Bill Dubuque Jun 14 '24 at 19:43
  • @BillDubuque thank you for the editing help. I was not sure at first what the appropiate tags should be. – Seth Jun 15 '24 at 00:51
  • @ryang I just voted for the answer. Sorry for the delay. I also have a minor question for my post on superset notation where you posted an answer. – Seth Jun 19 '24 at 18:14
  • @ryang thanks for pointing it out. I first thought in this post, I did not accepted any answers. I was just going through my old posts to double check – Seth Apr 24 '25 at 07:29

3 Answers3

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I don't think the translation $$(\exists n\in \Bbb{N})(\forall i>n )(\forall \{A_i\}_{i\in \Bbb{N}}\subset R)(A_1\subseteq A_2\subseteq\cdots \Rightarrow A_n=A_{n+1}=\dots)$$ is correct. It seems to say that you can find an $n\in\mathbb N$ such that any chain $A_1\subseteq A_2\subseteq \dots$ stabilizes with the same $n$. To capture $n$'s dependence on the chain of ideals, the quantifier $(\forall\{A_i\}_{i\in\mathbb N})$ should appear before the others, i.e. a better translation is $$\left(\forall \{A_i\}_{i\in\mathbb N}\subseteq R\right)\left[{\color{blue}{\left(\forall k\in\mathbb{N}\right)\left(A_k\subseteq A_{k+1}\right)}}\rightarrow{\color{green}{\left(\exists n\in\mathbb N\right)\left(\forall i\geq n, A_i=A_{i+1}\right)}}\right]$$ The blue part of the if-then checks if your collection of ideals is ascending, and the green part says that the chain stabilizes.

Alann Rosas
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  • thank you for the correction. I thought I can just simply pull out $\exists n\in \Bbb{N}$ to the beginning of the statement when translating. – Seth Jun 14 '24 at 06:01
  • @Seth you can't do this in general. For example, the statement $(\forall n\in\mathbb N)[(\exists k\in \mathbb N)(n=k)]$ does not say the same thing as $(\exists k\in \mathbb N)[(\forall n\in\mathbb N)(n=k)]$. The first says that for each $n\in\mathbb N$, we can find a $k\in\mathbb N$ that equals $n$ — a true statement — but the second says we can find a single $k\in\mathbb N$ such that every $n\in\mathbb N$ is equal to $k$. – Alann Rosas Jun 14 '24 at 06:08
  • when doing these translation, is it better to stick close to the english as far as when the quantifier sign posts appear. Also, sometimes when i see statement: "$A_1,A_2,\ldots$ of $R$ with $A_1\subseteq A_2\subseteq\cdots,$ there exists", do I translate the "with" as "and" and $A_2\subseteq\cdots,$ there exists$, the missing 'then", I have to read it a few times to see if the "then" is inferred or not. How did you get so good at these translation exercsies. – Seth Jun 14 '24 at 06:17
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    @Seth: You should beware that natural language sentences very often mangle the position of quantifier sign posts, so you need to think carefully when doing these translations. It is not always better to stick close to the english. For instance, consider this English sentence: "$x < \epsilon$ for all $\epsilon > 0$". When properly translated into prenex normal form it becomes "$\forall \epsilon > 0, x < \epsilon$". – Lee Mosher Jun 14 '24 at 13:43
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    When making the translations, you should look at each quantifier and carefully identify which portion of the statement serves as the scope of that quantifier. Then, when translating into prenex normal form, you should pull the quantifier to the front of its own scope and no more than its own scope. – Lee Mosher Jun 14 '24 at 13:47
  • @LeeMosher sorry for my late response. I just learn the rule of prenex normal form recently. It has to do with how quantifiers distribute over statements when they don't contain the variable under a quantifier scope. I looked up multiple intro to math proof texts, and it doesn't seem to appear in many of them. – Seth Jun 16 '24 at 10:13
  • No, that's not the point of prenex normal form. It is simply a grammatical rule for how to properly write quantifiers: at the beginning of their scope, and with the scope clearly delimited. – Lee Mosher Jun 16 '24 at 12:54
  • @LeeMosher Forgive me if I'm misunderstanding, but you seem to be saying that Q → ∃xP(x) and (∃xP(x)) → Q are in Prenex form? If you compare Alann's answer with the formula in my answer, the former is more human-friendly, whereas it is mine that is in Prenex form. ∃x(Q → P(x)) and ∀x(P(x) → Q) are the Prenex forms of the foregoing formulae. – ryang Jun 18 '24 at 13:58
  • @ryang I think what Lee Mosher is tell me is that prenex normal form is something that i need to be aware and make use of when translating mathematical english into symbolic logic notations. I don't think his comments refers to your answer or that of Alann Rosas. – Seth Jun 19 '24 at 18:16
  • @Seth I was just using my answer to illustrate that LM does not seem to actually be describing Prenex form at all, in which case he is not in fact at all advising you to make use of Prenex form (certainly, that'd be sound). – ryang Jun 19 '24 at 18:38
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For the first $\ldots$, you could write $\forall j \in \mathbb{N}, A_j \subseteq A_{j+1}$, and similarly for the second.

I think you have the order of things a bit mixed up in your translation though. The $A_1 \subseteq A_2 \subseteq \ldots$ should go on the outside, before the $\exists (n \in \mathbb{N})$, like in the original statement.

Ted
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  • I thought I can have $\exists (n\in \Bbb{N}$ pull out to the beginning. – Seth Jun 14 '24 at 06:00
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    No, if you put $\exists(n \in \mathbb{N})$ at the beginning, you are saying that there is a single value of $n$ that works for all choices of $A_i$. That is not true. – Ted Jun 14 '24 at 14:42
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To make Alann's answer—which I agree with—really succinct (though not necessarily advocating for this level of pithiness): $$\forall \{A_i\}_{i\in\mathbb N}{\subseteq} R \:\: \exists k{,}n{\in}\mathbb N \:\: \forall i \; \Big(A_k\subseteq A_{k+1} \;\text{and}\; i\geq n \implies A_i=A_{i+1}\Big).$$

for each sequence of left ideals $A_1,A_2,\ldots$ of $R$ with $A_1\subseteq A_2\subseteq\cdots,$ there exists a positive integer $n$ (depending on the sequence) such that $A_n=A_{n+1}=\dots$.

I thought I can just simply pull out $∃n∈\mathbb N$ to the beginning of the statement when translating

No: as the boldfaced parenthetical portion of the theorem helpfully reminds you, $n$ depends on $A_1,A_2,A_3,\ldots;$ this means that the quantification $\forall \{A_i\}_{i\in\mathbb N}{\subseteq} R$ must precede the quantification $\exists n{\in}\mathbb N.$ In fact, that reminder is technically redundant, since in general $$\forall y \exists xP(x,y)\kern.6em\not\kern-.6em\implies \exists x \forall y P(x,y).$$ On the other hand, $$\exists x \forall y P(x,y)\implies \forall y \exists xP(x,y).$$

ryang
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    Why does the $k$ have an existential quantifier? We need the collection to ascend at least eventually, i.e. $A_k\subseteq A_{k+1}$ for all sufficiently large $k$, but this seems to say that the collection stabilizes as soon as we find a single containment $A_k \subseteq A_{k+1}$. Maybe I'm misunderstanding something? – Alann Rosas Jun 14 '24 at 18:03
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    @AlannRosas, $\Big((\forall x,Px){\implies} Q\Big)\equiv\Big(\exists x;(Px{\implies} Q)\Big),$ as proved in the appendix of this post. The logic statement that I wrote is the Prenex form of yours, which, again, I'm not asking the OP to translate that ascending-chain-condition definition into. – ryang Jun 14 '24 at 19:09
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    I just read your proof; it makes much more sense now. Thank you! – Alann Rosas Jun 14 '24 at 19:13