Timeline for Compactness of the Hilbert cube without the Axiom of Choice
Current License: CC BY-SA 3.0
15 events
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| Jan 21, 2013 at 9:50 | comment | added | Péter Komjáth | Similarly to Andrej's argument above, one can give an AC-free proof of the following result: if $X$ is a countable graph, every finite subgraph of $X$ has a good coloring with $k$ colors ($k$ is finite) then so has $X$. | |
| Jan 20, 2013 at 19:43 | comment | added | Andrej Bauer | Dedekind-infinite is the wrong definition anyhow, as it does not work in general (only in the very limited environment of classical set theory), so I do not particularly care about it. | |
| Jan 20, 2013 at 16:27 | comment | added | Asaf Karagila♦ | Existence of a countably infinite subset is equivalent to Dedekind-infinite. Without choice not all infinite sets are necessarily Dedekind-infinite. Certainly if a set is dense in the real numbers we cannot have that it is finite. But we can have it Dedekind-finite, i.e. infinite without a countably infinite subset. Philosophically I have no problem calling something infinite if it cannot be bijected with a finite ordinal; but I do have a problem calling something "finite" just because it has no countably infinite subset. | |
| Jan 20, 2013 at 16:00 | comment | added | Andrej Bauer | I got rid of all talk of infinite sets (note that "an infinite path" is about defining a function on $\mathbb{N}$) and replaced those by unbounded height. Now are you happy? By the way, if I define that a tree is infinite when it has an injection from natural numbers, then I do not see where choice comes in. But the thing which is unecessarily complicated is that we have to show that a non-infinite thing is finite, i.e., bounded by some $n$. It is better to just prove directly that the relevant tree has bounded height | |
| Jan 20, 2013 at 15:57 | history | edited | Andrej Bauer | CC BY-SA 3.0 |
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| Jan 20, 2013 at 13:49 | comment | added | Asaf Karagila♦ | Of course. But that would require a tiny bit of the axiom of choice. :-) You can simply replace it with "Dedekind-infinite", which means exactly that and then the result is choice free. I know I'm nitpicky, but we are all mathematicians, aren't we? :-) [Well, in the broad sense of the word... to include Ph.D. students!] | |
| Jan 20, 2013 at 13:32 | comment | added | Andrej Bauer | Ok, so if I say that "infnite" means "has an injection from natural numbers", would that be ok? | |
| Jan 20, 2013 at 13:07 | comment | added | Asaf Karagila♦ | Andrej, first I thank you for removing that line. Second, the reason I am coming up with things so irrelevant is that your phrasing is wrong. I gave you an example of an infinite binary tree without an infinite path. Infinite does not mean countably infinite, or even Dedekind-infinite without the axiom of choice. It is important to make this distinction because while I know it, and I am sure that you and Goldstern are well aware of this distinction not all the readers are. The relevant changes are not big and they are important. Next thing people insist that cardinals are always ordinals.. | |
| Jan 20, 2013 at 8:48 | comment | added | Andrej Bauer | I fixed the first sentence so as to make it as clear that nobody had doubts about choice as you made it clear in the comments that you had no such doubts ;-) | |
| Jan 20, 2013 at 8:48 | history | edited | Andrej Bauer | CC BY-SA 3.0 |
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| Jan 20, 2013 at 8:47 | comment | added | Andrej Bauer | Why are you coming up with stuff that is irrelevant to this question? | |
| Jan 19, 2013 at 23:38 | comment | added | Asaf Karagila♦ | Furthermore, let $S=\bigcup_{n\in\omega} P_n$ be a disjoint union of pairs such that no infinite set of pairs has a choice function. Let $T$ be the tree of finite choice functions from an initial segment of $\omega$, that is $g\in T$ is a function from some $n\in\omega$ into $S$ such that $g(i)\in P_i$ for $i<n$. This is an infinite binary tree, but it has no branch. | |
| Jan 19, 2013 at 23:11 | comment | added | Asaf Karagila♦ | No one in the comments to Goldstern's answer has any doubts that the axiom of choice is unneeded. | |
| Jan 19, 2013 at 23:06 | history | edited | Andrej Bauer | CC BY-SA 3.0 |
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| Jan 19, 2013 at 23:01 | history | answered | Andrej Bauer | CC BY-SA 3.0 |