In article <0e0f648e-4ed3-4435-92ca-72143e174731@3g2000yqk.googlegroups.com> WM <mueckenh@rz.fh-augsburg.de> writes: > On 30 Jun., 16:36, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > In article <5c70e8d7-67e1-48e7-8dee-ec3d853bd...@r34g2000vba.googlegroups= > .com> WM <mueck...@rz.fh-augsburg.de> writes: > > > On 22 Jun., 15:11, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > ... > > > > > Every meaning of every word is defined by a language. > > > > > Every language is a finite definition. > > > > > The number of finite definitions is countable. > > > > > > > > For each word the meaning can indeed be countable. But that does > > > > not mean that the set of meanings for all the words in a language > > > > is countable. > > > > > > No? How can that be accomplished? > > > > I think by some negation of the axiom of choice. > > I did not ask how to "prove" that. I asked how to *do* it.
Apparently you do not know how mathematical proof works, especially if the proof includes something that is in itself not provable.
A: given a line and a point there is through that point exactly one line that does not meet the given line. B: eh? I do not think so, there may be more than one line. A: no? how can that be accomplished? B: I think by some negation of the parallel axiom. A: I did not ask how to *prove* that. I asked how to *do* it. B: by some negation of the parallel axiom.
My answer to your question above is similar. The question whether a countable union of countably many sets is countable is implied by the axiom of countable choice, which is not provable from ZF. So negating the axiom of countable choice (which implies negating the axiom of choice) we can have situations where the conclusion is false.
So it is up to *you* to prove that something is the case "the countable union of countable sets is countable", when I only state that I do not know that. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
