Re: Why can't such an example be given?

On Sat, 17 May 2008 01:32:35 -0700 (PDT), mike3 <mike4ty4@xxxxxxxxx>
wrote:

On May 15, 4:16 am, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Thu, 15 May 2008 01:33:20 -0700 (PDT), mike3 <mike4...@xxxxxxxxx>
wrote:
<snip>
But how does ZF _with_ AC still not give enough to _construct_
(although it gives enough to _prove_) a non-measurable
function?

Well, this depends on what you mean by "construct".
Usually when people prove something exists using
AC they simply don't say they've "constructed" it.
Because, for example, if you've constructed something
then you should be able to say exactly what it is.
If you use AC to prove that there exists a non-
measurable set E then you cannot tell me whether
or not 0 is an element of E; given that most people
wouldn't say you've "constructed" E.


But can it be proven impossible to construct E in ZFC?
For example is it provably impossible to construct the
sets described by Vitali's theorem, or the Banach-
Tarski ballcutting "paradox", etc.?

Also, does this mean whether or not we can _construct_ a
non-measurable function is independent of/undoable in ZF and
ZFC just like how the Continuum Hypothesis is?

Whether or not a non-measurable set exists is independent
of ZF. It's not clear whether that gives a yes or no to your
question, since it's not clear exactly what _you_ mean
by "construct". In most people's terminology it is
impossible to construct a non-measureable set in ZF.

By "construct" I mean just that, construct, as in be able
to write down a procedure or method or something that
describes/produces such a set. Or whatever "most people's"
definition of "construct" is. At least that's what I think
of when I think of "construct".

Well, if you don't use AC then the construction would
also work in ZF, which is impossible. And if you do use
AC then most people simply don't call it a "construction".

So why is it impossible to construct non-measurable sets
in ZF or even ZFC?

David C. Ullrich
.

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