Timeline for Can we choose an element from a class?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| May 4, 2021 at 21:34 | comment | added | Ivan Feshchenko | @AsafKaragila Unfortunately, I do not know how to do this formally; I want to have precise formal arguments to do this. Please help me. | |
| May 4, 2021 at 21:23 | comment | added | Ivan Feshchenko | @AsafKaragila Dear Asaf, I have one more question. Let $n$ be a natural number and $c,a$ be real numbers. Suppose that there exist a complex Hilbert space $H$ and closed subspaces $H_1,...,H_n$ of $H$ such that for some indices $i,j$ $H_i\neq H_j$ (the Friedrichs number is defined only for such systems of subspaces), $c_F(H_1,...,H_n)\leqslant c$ and $\|P_n...P_1-P_0\|=a$. I need to choose such system of subspaces $(H;H_1,...,H_n)$ and work with this system. I understand that to do this I should use existential instantiation. To be continued. | |
| Mar 28, 2021 at 21:43 | comment | added | Noah Schweber | If however there may be multiple such $b$s, we don't (yet) have a choice-free argument for the existence of a function $f:A\rightarrow B$ such that $\varphi(a,f(a))$ holds for all $a\in A$. (This is how your AC argument breaks.) Now there is a subtlety here: what exactly counts as a "description" of the above type? The answer is not quite simply "a first-order formula with finitely many parameters" (every model $M$ of $\mathsf{ZF}$ satisfies finite choice even if $M$ has nonstandard naturals!), but that's a good approximation initially. | |
| Mar 28, 2021 at 21:41 | comment | added | Noah Schweber | @IvanFeshchenko To add to Asaf's comment, a good rule of thumb for when you need AC (or rather, when you don't need it) is: can you describe how you're making your choices in such a way that somebody else following that method would be guaranteed to produce the same objects? For example, if for each $a\in A$ there is exactly one $b\in B$ such that $\varphi(a,b)$ holds, then we can form the function $A\rightarrow B$ sending each $a$ to the corresponding $b$ without choice. (cont'd) | |
| Mar 25, 2021 at 23:26 | comment | added | Asaf Karagila♦ | @Ivan: You need to use AC when you need to choose things uniformly. When you just want to claim that something is independent of the choice, you can just pick representatives when necessary. Don't feel bad, it's confusing for a lot of mathematicians. You need time and practice to get used to it. | |
| Mar 25, 2021 at 23:16 | comment | added | Ivan Feshchenko | @AsafKaragila I am lost in these arguments and the Axiom of Choice. I do not understand when we need a (choice) function and when we do not need a (choice) function. Can you suggest me a literature, please, so that I can understand the essence of these things? | |
| Mar 25, 2021 at 22:09 | comment | added | Ivan Feshchenko | @AsafKaragila Thus $A$ is a needed transversal. Where am I wrong? | |
| Mar 25, 2021 at 22:07 | comment | added | Asaf Karagila♦ | @Ivan: As I said, that would be misleading to the reader, apparently also for you. Just because you wrote $f(a)$ does not mean that it is actually define a function. It is a notation, it tells you that $f$ depends on $a$ and that it is fixed within a given context. That is all. | |
| Mar 25, 2021 at 22:04 | comment | added | Ivan Feshchenko | @AsafKaragila I have a feeling that I do not understand something important...Ok, assume that I can write $(H(a);H_1(a),...,H_n(a))$. But then by similar arguments I can prove the Axiom of Choice: let $A_i, i\in I$ be a set of mutually disjoint nonempty sets. We will construct a transversal for this set as follows. Consider arbitrary $i\in I$ and fix it. Since $A_i$ is nonempty, $\exists x(i)\in A_i$. After this we consider one-element sets $\{x(i)\}$ and the set $A=\bigcup_{i\in I}\{x(i)\}$. Then $A$ is a set and $A$ contains exactly one element in common with every $A_j$. To be continued | |
| Mar 25, 2021 at 21:36 | comment | added | Asaf Karagila♦ | @IvanFeshchenko: Can you? Sure. Will it be syntactically correct? Also yes. Will it be misleading for any reader who thinks that those are chosen in advance? Probably yes. | |
| Mar 25, 2021 at 20:42 | comment | added | Ivan Feshchenko | @AsafKaragila Ok, assume that I fix $n,c,a$. Can I write $(H;H_1,...,H_n)=(H(a);H_1(a),...,H_n(a))$? | |
| Mar 25, 2021 at 19:09 | comment | added | Asaf Karagila♦ | @IvanFeshchenko: No. Fix $c,n,a$, then you know there is some $(H;H_i)$ for which the thing work. As I write, you don't have to choose everything in advance. Deal with your problems when they come up. | |
| Mar 25, 2021 at 18:36 | comment | added | Ivan Feshchenko | @AsafKaragila Dear Asaf Karagila, thank you for your answer. Now I do not understand one point. Essentially, the core of my worries is the following. I do not understand if the function $A_n(c)\ni a\mapsto (H;H_1,...,H_n)$ such that $c_F(H_1,...,H_n)\leqslant c$ and $\|P_n...P_2 P_1-P_0\|=a$, is needed in my arguments above (in the question) or not. | |
| Mar 25, 2021 at 1:27 | comment | added | Asaf Karagila♦ | @LSpice: It puts the saying "we're cooking up a proof" in a whole light. | |
| Mar 24, 2021 at 23:50 | comment | added | Asaf Karagila♦ | @Mike: Obviously. :-) | |
| Mar 24, 2021 at 23:50 | comment | added | Mike Shulman | I presume you mean "no proper class is finite". (-: | |
| Mar 24, 2021 at 21:45 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
added 2 characters in body
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| Mar 24, 2021 at 21:10 | comment | added | LSpice | "But it's not, and it doesn't have to be" → "But they don't, and they don't have to be"? (I love that mise en place metaphor.) | |
| Mar 24, 2021 at 19:26 | history | answered | Asaf Karagila♦ | CC BY-SA 4.0 |