Tag Info

Hot answers tagged

12

As Asaf says, Choice is not needed. Another way to see it. Note that if $\bigcap_n \Omega_n = \varnothing$, then $\{ \mathbb C \setminus \Omega_n : n \in \mathbb N \}$ is a family of open subsets of $\mathbb C$ which covers $\Omega_1$. However no finite subfamily of this covers $\Omega_1$: if $n_1 < \ldots < n_k$ then $$( \mathbb C \setminus ...


10

This author is notorious for his claims about set theory, and mainstream mathematics does not take them seriously. For a detailed criticism of his style of argument, see this review (in German) by Franz Lemmermeyer of a book he wrote. Regarding the specific text you link to, arguments like this: The rational numbers are countable whereas the ...


9

Fortunately, the axiom of choice can be circumvented here. $\Bbb C$ is a Polish space, namely it is separable and completely metrizable. Moreover we are talking about compact sets and not sequentially compact sets. The equivalence between the two notion does require the axiom of choice, but it's not used here. Now. Compact sets are closed in $\Bbb C$. And ...


7

No. Of course not. Your proof is perfectly valid. This is also mentioned in the errata for the book.


6

I was bored, so I read the paper in order to figure out what's wrong with it. The paper purports to create an injection from $\mathbb{X}_+$ (the positive irrationals) to $\mathbb{Q}_+$ (the positive rationals) by repeatedly removing one element from each set. The glaring flaw I found was this (emphasis preserved): If the set $\mathbb{Q}_+$ were exhausted ...


5

HINT: If there is a surjection from $A$ onto $B$, then there is an injection from $B$ into $\mathcal P(A)$. This leaves the case where $n=0$, but that's easy.


4

The Skolem-Lowenheim theorem is equivalent to $\sf DC$, which is a strong countable choice principle. More generally, the following is true: Every structure has an elementary equivalent substructure of size $\leq\kappa$, if and only if $\sf DC+AC_\kappa$. You can find the proof here. In a nutshell, the idea is that given a family of sets, an ...


3

Yes, you need choice in order to pick a particular sequence for each of the $E_n$s -- remember that all you know about them for the purpose of this proof is that at least one sequence exists; this leaves open the possibility that there will be many sequences and no principled way to select one particular among them. You're given as an assumption that there ...


2

Showing $\mathbb Q$ is countable does not require AC. Showing that a countable union of countable sets is countable does require some form of AC. Yes, AC is used right at the start where he says let every element of the sequence etc. Saying $E_1$ is countable says that the elements of $E_1$ can be arranged in a sequence. There are many ways to do this; we ...


2

From wikipedia: $\:$ "In particular Krivine (1969) showed there was a model of ZFC in which every ordinal-definable set of reals is measurable, ..." Therefore ZFC does not prove that there is a definable nonmeasurable set. Using an explicit definable well-order on $\mathbb{R}\hspace{-0.04 in}\cap$OD to run a Vitali construction gives an explicit set ...


2

If you can prove that your definition defines something, then will in particular define something in Solovay's model, and that something will in particular not be a nonmeasurable set. So the best you can hope for is an explicit definition of a subset of $\mathbb R$, such that the thing defined will in some models of ZF be a nonmeasurable set. And this is ...


2

Every compact subset in $\mathbb {R}$ has a smallest element, so there is no trouble choosing points in compact subsets of $\mathbb {R}.$ Can't we do something similar in $\mathbb {C}?$ For example, if $K$ is a compact subset of $\mathbb {C},$ then we can let $a=\inf \{x: (x,y)\in K\}.$ Then set $b = \inf \{y: (a,y) \in K\}.$ Then $(a,b)\in K.$ We can think ...


2

No, there is no such thing. Of course in $L$, or even in other relatively tame models, you could prove there is such uniquely defined object. But in general, this is not the case. Force with $\operatorname{Col}(\omega,\omega_1)$ to make the old $\aleph_1$ countable. This forcing is homogeneous, therefore its weakest element decides the truth value of any ...


2

Well... how do you know that $X$ is non-empty to begin with? If $X$ is non-empty, then the definition of a measurable rectangle does not require the axiom of choice, since only finitely many arbitrary choices are made. And the definition of the smallest $\sigma$-algebra does not use the axiom of choice either. But it could be larger than you'd expect. And ...


1

This theorem implies the axiom of countable choice (and is probably equivalent to it, since I doubt its proof uses any more than countable choice). To show this, suppose $(Y_j)$ is a countable sequence of nonempty sets, and suppose the product $\prod Y_j$ is empty. Let $x$ be some point that is not in any $Y_j$ and let $X_j=Y_j\cup \{x\}$. Let ...


1

Suppose that $R$ is well-founded, and let $A$ be any nonempty subset of its field. We wish to show that $A$ contains an $R^t$-minimal element. Let $B$ consist of all elements of $A$, plus every element of every finite $R$-sequence that connects two elements of $A$. Because $R$ is well-founded, $B$ contains an $R$-minimal element $b$. The elements that were ...



Only top voted, non community-wiki answers of a minimum length are eligible