Question: Given the real numbers as a set, does it require the (non-finite) Axiom of Choice to pick out an arbitrary single element? What about if we wanted to pick out an integer? What about if we wanted to pick out "0"?
Motivation: Those who have seen some of my previous questions here know that I am terrible at picking out when and where to apply the axiom of choice, isabel marant online and when it is being used in a proof. Perhaps isabel marant dicker boots slightly motivated by the Bertrand Russell quote
and perhaps slightly motivated by an argument I had over the AoC, I wanted to find out how much it is actually being used. In the reals, unless we distinguish an element, there doesn isabel marant seem to be a rule we could use to make our choice purely arbitrary. Same for a random integer. For 0, or mulberry sale for any number, there is clearly a choice function (pick that number). I unsure if I am stating these correctly, though. james Mar 27 '11 at 22:44
There a difference between the probability notion of "random integer" and "select an integer". In probability, to have "random integer" you would need some probability distribution over the integers that makes every integer "equally likely", but that is impossible (each integer would need to have probability zero of mulberry factory outlet being chosen, but selections being independent would make the probability of choosing any integer equal to $0$ as well by $\sigma$-additivity). But this is a probability obstacle, shop isabel marant not a set-theoretic one. Arturo Magidin Mar 27 '11 at 22:50
Given a finite number of nonempty sets $A_1,\ldots,A_k$, the Axiom of Choice is not needed to show that there are functions $f\colon\{1,\ldots,k\}\to\cup A_i$ such that $f(i)\in A_i$ for each $i$. That is, there is no need for the Axiom of Choice in order to select an element form finitely many nonempty sets.
In particular, you do not need the Axiom of Choice to show that you can celine luggage choose a real number (a single set).
Simply: 30675556 nike air max 09 since $A_1$ is nonempty, there exists $a_1\in A_1$. Since $A_2$ is nonempty, there exists $a_2\in A_2$. Likewise, we have $a_i\in A_i$, $i=3,\ldots, k$. And we can let $f=\{(i,a_i)\mid i=1,\ldots,k\}$ be the Choice function. We can write all of this down because there are finitely many $A_i$. Apparently there are subtle non-standard-set-theory issues here (thanks to Carl Mummert for the pointer); so instead, let's say that for a family of nonempty sets indexed by a celine sale natural number you do not need the Axiom of Choice to get a choice function, and this can be shown by induction on the index set.
The specter of the Axiom of Choice (so to speak) does not even begin to suggest itself until you have to make infinitely many choices. Even then, you may not need it.
Note, however, that using phrases like "arbitrary real number" may make it seem like you are talking about some kind of uniform probability distribution over all the real numbers that makes all "selections" equally likely. This is a completely different thing altogether, but not what you are talking about here. This is likewise the problem that arises when, in the context of probability, we talk about "selecting a random integer." The problem with "selecting a random integer" is that you cannot have a uniform isabel marant sale probability distribution on the integers: this would amount to a measure on the power set of the integers that is $\sigma$-additive, for which each integer has equal probability, and for which $\mu(\mathbb{Z})=1$; no such thing exists, because you would need 73338677 women nike air max to have $\mu(n)=0$ for each integer $n$, and $\sigma$-additivity would imply $\mu(\mathbb{Z}) = 0$ as well. This is what is behind the statement "you cannot 'choose a random integer'" true in that context; but this is an obstacle 12912883 where to buy nike shoes in the notion of "randomness", not a set-theoretic obstacle. Likewise, there celine handbags can be no uniform probability distribution over all the reals, because uniformity would imply that each interval $[n,n+1)$ with $n\in\mathbb{Z}$ would have the same measure, and $\sigma$-additivity together with the fact that the reals have finite total measure implies that each interval must have measure equal to $0$, and so the reals would also have total measure $0$. Again, this is a probability/measure-theoretic obstacle, 61176665 nike air max 2011 not an Axiom-of-Choice one.
There is a subtle point in the proof. ZFC does prove the principle in the first sentence of the answer, which means that the principle holds even in nonstandard models where the $k$ is not actually metafinite. In that case the argument of the third paragraph cannot work because you cannot actually write down the specified set. So to prove this in ZFC you need to go by induction on $k$ instead. But the answer is morally correct: the axiom of choice isn needed to "pick" an element of a nonempty set; that type of choice is just called "existential instantiation". Carl Mummert Mar 28 '11 at 2:48 |