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According to the last step in proof of the unmeasurability of Vitali_set, it said that summing infinitely many copies of the constant $\lambda(V)$ yields either zero or infinity, according to whether the constant is zero or positive. It sounds pleasing to my ear, but I still have a bit doubt in the reason of the sum of infinitely many copies of a positive real constant would definitely yield infinite.

Actually, $$< \sum_{i=0}^n c>_{n=0}^{\infty}$$ is indeed an strict increasing series when $c>0$. However, this fact seems cannot guarantee the inevitability of $\sum_{i=0}^\infty c=\infty$.

Take an example in infinite product. $1,2,4,8,16...$ is actually a strict increasing series too, but $$\prod_{i=0}^{\infty}2$$ can yield $0$ in some cases.

Moreover $9,99,999,\ldots$ is also a strict increasing series, but in some theory $...999$ is not a infinite but $-1$.

So my question on what basis can the conclusion that $$\sum_{i=0}^\infty \lambda(V)$$ is necessarily not between $1$ and $3$ be concluded?

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Where is the limit of $9, 99, 999,\dots$ equal to $-1$? –  Michael Greinecker Nov 11 '12 at 12:42
    
@MichaelGreinecker $...999$ sometimes be dealt as $-1$ (in 10-adic), not the limit of $9,99,999...$ –  Popopo Nov 11 '12 at 12:47
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@Popopo: I think that the problem is that you are mixing contexts. The real numbers are not $10$-adic numbers, and they are not cardinal numbers, and they are not anything except the real numbers. If you agree that measure theory is done in the context of real numbers you cannot give "counterexamples" from a different context. This is like saying "Oh, religious Jews don't eat pork, but Christians do. So Judaism is inconsistent", within the real numbers the products and sum you mentioned are infinite because they are limits of strictly increasing sequences and therefore larger than any number –  Asaf Karagila Nov 11 '12 at 12:57
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@AsafKaragila: Just in case Popopo remembers the minor bits of your comments while ignoring (as (s)he seems to have done) the main points, perhaps clarify that not every limit of a strictly increasing sequence of reals is larger than any number, i.e., some increasing sequences converge. –  Andreas Blass Nov 11 '12 at 14:02
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@Popopo: Yes. Numbers are mythical creatures if you prefer to think about them that way. But this makes even more sense. Real numbers are pixies and cardinals are trolls. The fact that there are pixies and trolls which behave similarly to unicorns (natural numbers) does not mean that pixies are trolls, nor that unicorns are either. –  Asaf Karagila Nov 11 '12 at 14:13
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1 Answer

up vote 3 down vote accepted

What is $\sum_{n=0}^\infty x_n$? It is $\lim_{k\to\infty}\sum_{n=0}^k x_n$. This is a limit of real numbers.

Suppose that $x_n=1$ for all $n$, then what is this limit? The partial sum $\sum_{n=0}^k 1 = k+1$, and therefore this is the limit $\lim_{k\to\infty}(k+1)$.

Replacing $1$ by any other positive constant has the same effect.


It seems that the questions stems from mixing up contexts. One should never do that in mathematics. Real numbers are real numbers, they are not $10$-adic, they are not ordinals and they are not cardinals.

True, the natural numbers can be represented as cardinals, ordinals, real, $10$-adic numbers, and more. However each system carries out its own rules. In particular in the behavior of infinitary operations such as infinite sums and multiplications.

Even cardinals and ordinals, which are often thought as the same, behave differently with respect to infinitary multiplications. Let alone real numbers and cardinals, or real numbers and ordinals.

In measure theory we work with real numbers which means that the sums taken are sums of real numbers, and when taking infinitary sums of real numbers one apply the definitions for sums of real numbers.

For example, in the real numbers I am allowed to do this: $$\frac12\sum_{i=0}^\infty 1=\sum_{i=0}^\infty\frac12$$ Where as summation of ordinals or cardinals cannot be done because the object $\frac12$ is neither an ordinal nor a cardinal number.

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What about $\prod_{n=0}^{\infty}x_n$ when $x_n=2$ for all $n \in \omega$? $\lim_{k \to \infty}\prod_{n=0}^{k}x_n=\omega$ but $\prod_{n=0}^{\infty}x_n=0$ in some cases... –  Popopo Nov 11 '12 at 12:40
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@Popopo: No. The product of real numbers is not the product of cardinals. Furthermore $\infty$ is not $\omega$. Also the product of countably many copies of $2$ is never countable (if not empty). –  Asaf Karagila Nov 11 '12 at 12:42
    
But does $\lim_{k \to \omega}\prod_{n=0}^{k}x_n=\sup\{\prod_{n=0}^{k}x_n|k < \omega\}=\sup\{2,4,8,16,...\}=\omega$ hold? –  Popopo Nov 11 '12 at 12:59
    
@Popopo: No!! You are taking a product of cardinals, not a product of real numbers! There can never be any real number corresponding to $\omega$. Not even a hyperreal number can correspond to $\omega$!! Read my comment to your question. Mixing up contexts is bad. This is like saying that a regular graph and a regular cardinal have something in common because both are called regular. –  Asaf Karagila Nov 11 '12 at 13:01
    
@Popopo: And as cardinals go, the product is not continuous. This means that the limit of finite products is not the product over the infinite index. On the other hand, in ordinal arithmetic this is true, and there $2^\omega=\omega$. But those are ordinals and cardinals and neither are real numbers. –  Asaf Karagila Nov 11 '12 at 13:03
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