Timeline for Is a function needed here?
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20 events
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| Apr 20, 2021 at 23:15 | comment | added | Dmitri Pavlov | @IvanFeshchenko: Yes, U is a set by separation: it is constructed as {u∈R| ...}, and ... can be any first-order formula, such as the one you used. | |
| Apr 20, 2021 at 19:44 | comment | added | Ivan Feshchenko | @DmitriPavlov Dear Dmitri, I have one more question. To define $\sup\{\|P_n...P_1-P_0\|\,|\,c_F(H_1,...,H_n)\leqslant c\}$ we have to consider the set $U$ of all real numbers $u$ such that $\|P_n...P_1-P_0\|\leqslant u$ for arbitrary Hilbert space $H$ and arbitrary system of closed subspaces $H_1,...,H_n$ of $H$ such that for some indices $i$ and $j$ $H_i\neq H_j$ (the Friedrichs number is defined only for such systems of subspaces) and $c_F(H_1,...,H_n)\leqslant c$. But why is $U$ a set? My argument: $U$ is a set by the axiom (scheme) of separation. Am I right? | |
| Apr 17, 2021 at 18:21 | comment | added | Dmitri Pavlov | @IvanFeshchenko: Yes, absolutely. In fact, the axiom of replacement was not used, so the proof works in the Zermelo set theory also. | |
| Apr 17, 2021 at 18:02 | comment | added | Ivan Feshchenko | @DmitriPavlov Thus we proved that two definitions of $\sup\{\|P_n...P_1-P_0\|\,|\,c_F(H_1,...,H_n)\leqslant c\}$ are equivalent. My question: I think that all arguments above are correct in ZF and I did not use the Axiom of Choice (I used only a few existential instantiations). Am I right? | |
| Apr 17, 2021 at 17:49 | comment | added | Ivan Feshchenko | @DmitriPavlov Now suppose that $u$ is not an upper bound for $A_n(c)$. Then there exists $a\in A_n(c)$ such that $a>u$. For this $a$ there exist $H;H_1,...,H_n$ such that $c_F(H_1,...,H_n)\leqslant c$ and $\|P_n...P_1-P_0\|=a$. Thus $\|P_n...P_1-P_0\|>u$. Thus $u\notin U$. It follows a real number $u$ is an upper bound for $A_n(c)$ if and only if $u\in U$, i.e., the set of upper bounds for $A_n(c)$ equals $U$. Then the minimal elements of the sets are equal, i.e., $\sup A_n(c)=\min U$. | |
| Apr 17, 2021 at 17:15 | comment | added | Ivan Feshchenko | @DmitriPavlov Let $U$ be the set of all real numbers $u$ such that $\|P_n...P_1-P_0\|\leqslant u$ for all $H;H_1,...,H_n$ such that $c_F(H_1,...,H_n)\leqslant c$. We will show that the set of upper bounds for $A_n(c)$ equals $U$. Suppose that $u$ is an upper bound for $A_n(c)$. Consider arbitrary $H;H_1,...,H_n$ with $c_F(H_1,...,H_n)\leqslant c$. Then $\|P_n...P_1-P_0\|\in A_n(c)$ and thus $\|P_n...P_1-P_0\|\leqslant u$. Thus $u\in U$. (To be continued.) | |
| Apr 17, 2021 at 17:01 | comment | added | Ivan Feshchenko | @DmitriPavlov Dear Dmitri, many thanks for your very helpful comments and answers to my questions. I have one more question. Now I see that $\sup\{\|P_n...P_1-P_0\|\,|\,c_F(H_1,...,H_n)\leqslant c\}$ can be defined in two different ways. The first way: the supremum is the smallest real number $u$ such that $\|P_n...P_1-P_0\|\leqslant u$ for all $H;H_1,...,H_n$ such that $c_F(H_1,...,H_n)\leqslant c$. The second way: consider the set $A_n(c)$, then our supremum is $\sup A_n(c)$. These two definitions are equivalent. My arguments are as follows. (To be continued). | |
| Apr 14, 2021 at 22:43 | vote | accept | Ivan Feshchenko | ||
| Apr 12, 2021 at 0:31 | comment | added | Dmitri Pavlov | @IvanFeshchenko: It appears that your insistence on using the auxiliary set A_n(c) may originate in one of the answers (mathoverflow.net/questions/375759/…) that was given here on MathOverflow. I posted a new answer (mathoverflow.net/questions/375759/…) to the same question, which should make it clear that no such auxiliary sets are necessary. | |
| Apr 11, 2021 at 23:27 | comment | added | Dmitri Pavlov | @IvanFeshchenko: No, the precise meaning of this formula is not what you wrote. The supremum in this case is by definition the smallest number A≥0 such that ∥P_n⋯P_2P_1−P_0∥ ≤ A for all H_1, …, H_n, H such that c_F(H1,…,Hn)≤c. Absolutely nothing in this definition requires the domain of quantifiers to be sets as opposed to proper classes, and I do not understand why you think they should be, or why you continue to insist on using A_n(c), which are totally unnecessary. The supremum always exists. In particular, there is no "unfolding" hidden here. | |
| Apr 11, 2021 at 23:12 | comment | added | Ivan Feshchenko | @DmitriPavlov Dear Dmitri, I think that the choice of $(H;H_1,\ldots,H_n)$ is implicitly present in your arguments. Indeed, we write $f_n(c)=\sup\{\|P_n\cdots P_2 P_1-P_0\|\,|\, c_F(H_1,\ldots,H_n)\leqslant c\}$, but the precise meaning of this formula is that $f_n(c)=\sup A_n(c)$, i.e., $f_n(c)$ is the supremum of a certain set. Now if we "unfold" your arguments, we will get essentially my arguments in the Question. So I think that the choice of $(H;H_1,\ldots,H_n)$ is necessary. | |
| Apr 11, 2021 at 22:14 | comment | added | Ivan Feshchenko | @NoahSchweber Dear Noah, thank you for your comment. I was desperate to find answers to my questions on the Axiom of Choice...So, if I understand you correctly, my mistake in my "proof" of the Axiom of Choice is that I need to consider all $x(i)$ simultaneously, and this cannot be done without the Axiom of Choice. Am I right? | |
| Apr 9, 2021 at 22:21 | comment | added | Dmitri Pavlov | @IvanFeshchenko: Your argument is not similar. In my answer, the statement is proved for all H_i and H. In your comment, you have to choose x(i). In particular, “existential instantiation” simply does not work in “families” as described; there is no such rule in first-order logic. No such choices are made in my answer. In particular, I do not choose H_i and H for each c, but rather prove the inequality for all H_i and H. | |
| Apr 9, 2021 at 22:12 | comment | added | Noah Schweber | @IvanFeshchenko You go wrong as soon as you define $A$: intuitively this requires you to "unfix" your chosen element $i$. You'll see this if you sit down and try to write out your argument as a formal proof (say, in sequent calculus). Basically you have a first existential instantiation picking an $i$ and following that a second existential instantiation picking an $x$, but there isn't a logical rule letting you somehow "uniformize" this. (In fact you can think of $\mathsf{AC}$ as an additional logical rule in a precise, if technical, sense.) | |
| Apr 9, 2021 at 21:37 | comment | added | Ivan Feshchenko | Now we use the existential instantiation and write $x(i)$ for a new symbol (just a symbol) such that $x(i)\in A_i$. Consider one-element sets $\{x(i)\}$ and let $A=\bigcup_{i\in I}\{x(i)\}$. Then $A$ is a set and $A$ contains exactly one element in common with each $A_j$. Where am I wrong? | |
| Apr 9, 2021 at 21:29 | comment | added | Ivan Feshchenko | Dear Dmitri, thank you for your answer. One thing is confusing for me. Assume that my arguments (proof of $f_n(c)\leqslant g_n(c)$) are correct. Then by "similar" arguments I can prove the Axiom of Choice. The "proof" is as follows. Let $A_i, i\in I$ be a set of mutually disjoint nonempty sets. We will construct a set that contains exactly one element in common with each of the sets as follows. Consider arbitrary $i\in I$ and fix it. Since the set $A_i$ is nonempty, $\exists x$ such that $x\in A_i$. To be continued. | |
| Apr 8, 2021 at 3:40 | history | edited | LSpice | CC BY-SA 4.0 |
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| S Apr 8, 2021 at 1:21 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| Apr 8, 2021 at 0:36 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |