Cantor Sets

1. Nov 14, 2005

benorin

So the problem, and my partial solution are in the attached PDF.

I would like feedback on my proof of the first statement, if it is technically correct and if it is good. Any ideas as to how I can use/generalize/extend the present proof to proof the second statement, namely that E (the Cantor set) has the same cardinality as $$\mathbb{R}$$? Please, not the ternary expansion correspondence to the reals in [0,1] .

Attached Files:

• Show that Cantor set is measure zero.pdf
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2. Nov 14, 2005

Tide

You might consider summarizing your arguments here since your attachment isn't visible until it gets approved.

From the words you use it sounds like you are using the standard argument that the Cantor set consists of all ternary numbers expanded as decimals that do not contain the digit "1" Obviously, that set has the same cardinality as the reals using Cantor's diagonalization.

3. Nov 14, 2005

benorin

If I post it as a JPEG or BMP or other graphics file, will I still have to wait for this pending approval stuff?

Wrong dimensions! Hate image crap, what with the "waiting for approval" stuff ? Manual content approval or what?

Last edited: Dec 21, 2005
4. Nov 14, 2005

benorin

OK, so its no longer pending approval (and I have gotten some sleep). Please respond soon, this is due in the morning.

Thanks,
-Ben

Last edited: Nov 14, 2005
5. Nov 15, 2005

benorin

FYI, Theorem 2.20 (e) states that to every linear transformation $$T:\mathbb{R}^{k}\rightarrow\mathbb{R}^{k}$$ there exists a real number $$\Delta \left( T\right)$$ such that $$m\left( T\left( A\right) \right) = \Delta \left( T\right) m\left( A\right)$$ for every Lebesgue measurable set A.

6. Nov 15, 2005

NateTG

How do you know that the Cantor set actually has a Lebesgue measure? Beyond that, the proof looks OK.