Timeline for Why accept the axiom of infinity?
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13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 9, 2012 at 6:50 | comment | added | Kaveh | One can work with recursive reals which are countable. ps: one doesn't need to be a finitist to reject the classical reals à la Dedekind/Cauchy. | |
| Jun 8, 2012 at 17:52 | vote | accept | pichael | ||
| Jun 8, 2012 at 7:49 | comment | added | Brian M. Scott | @William: There is absolutely no need to assume a bijection. The diagonal argument shows directly, without any appeal to contradiction, that no function from $\omega$ to $\Bbb R$ is surjective. | |
| Jun 8, 2012 at 7:30 | comment | added | William | @BrianM.Scott Be nicer to finitist! For all we know, the axiom of infinity could lead to inconsistencies of set theory. | |
| Jun 8, 2012 at 7:25 | comment | added | William | @BrianM.Scott (sorry too long; continuing) This usually the spirt of the diagonalization argument. Note that the assumption that the set binary sequence is countable is essential. How does your argument go? | |
| Jun 8, 2012 at 7:24 | comment | added | William | @BrianM.Scott How do you prove the uncountability of $\R$? Usually for simplicity, we will prove binary sequence $\omega \rightarrow \{0,1\}$ are uncountable? Before even going on, a finitistic would be believe such function exists, but lets return to Cantor Argument, if you like you can think this as the Cantor Set insider $\mathbb{R}$. Usually you assume there is a bijection $f$ from $\omega$ into the set of all binary sequences. Then you go along diagonal and produce a binary sequence that is unequal to $f(n)$ for all $n$. Contradiction. | |
| Jun 8, 2012 at 7:13 | comment | added | Brian M. Scott | @William: I think that you missed my point. I’m not arguing about the beliefs of finitists; I frankly don’t care, since I’m not interested in doing mathematics with one hand tied behind my back. I’m merely pointing out that the diagonal argument does not require the assumption that there is a bijection between $\omega$ and $\Bbb R$. As for the machine, it’s as nice as the class of functions you allow. | |
| Jun 8, 2012 at 7:06 | comment | added | William | @BrianM.Scott Also there is a subtle distinction. Finistic do believe in the formal proof Cantor Argument of the uncountability of $\mathbb{R}$ (or any other theorem of ZFC) in the formal language of ZFC. This is because proofs are essentially finite objects. So finitist will agree that Cantor proof is correct as purely logical finite manipulation of symbols. But they probably won't believe that your manipulation or your axiom describe anything real and interesting. | |
| Jun 8, 2012 at 7:04 | comment | added | William | @BrianM.Scott What is a machine? Finistic certainly believe in Turing Machines. However, how will you represent a real number. Finistic believe in anything that can be encoded as a natural number. But if you encode a real number as decimal expansion, binary expansion, equivalence class of Cauchy Sequence, Dedekind cut, you will formally require the existence of at least $\omega$. Functions on $\omega$ are also infinite objects. | |
| Jun 8, 2012 at 6:59 | comment | added | Brian M. Scott | @William: It doesn’t assume the existence of such a bijection; it’s a machine that when given a function from $\omega$ to $\Bbb R$ constructs a real number not in the range of the function. (This may be just as objectionable to a finitist, of course.) | |
| Jun 8, 2012 at 6:39 | comment | added | William | I am not sure if a finitist would even believe in the Cantor Diagonalization argument since it require you to assume the existence of a bijection from $\omega \rightarrow \mathbb{R}$ and they do not believe that either exists as a formal object. | |
| Jun 8, 2012 at 6:36 | comment | added | William | Note that the problem with the Cauchy and Dedekind Cut construction in finitistic mathematics come well before Cantors uncountability argument. Finistist have $\omega$, as an abbreviation, but $\omega$ is not a real object. However in the Cauchy Construction, your objects are equivalent class of sequence. Sequences are function on $\omega$. So you need $\omega$ to exists for this construction. Similary, Dedekind Cut construction are pairs of infinite sets of rational numbers, which again do not exists in finite math. | |
| Jun 8, 2012 at 6:28 | history | answered | A.S | CC BY-SA 3.0 |