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I'm reading Apostol's Calculus. Given the axioms

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I got a silly doubt. A point is a measurable set with zero area, according to the axioms. Suppose I have a square with $a(S)=16$ and a point with $a(P)=0$, then I can use $(3)$ from the axioms above and subtract one from another.

Wouldn't that imply that I could apply $(3)$ repeatedly, remove all the points in $S$ and still have $a(S)=16$ even when $S$ has no points and should have $a(S)=0$?

Red Banana
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4 Answers4

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No, because no matter how many times you apply (3) removing one point, you'll only have removed finitely many points.

Henrik supports the community
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  • But in the axioms, nothing forbids me of having an infinite removal, no? Perhaps in the axioms of measure theory (which I'm still too stupid to understand). – Red Banana Jan 03 '15 at 11:16
  • You can't just say "I'll do this infinitely many times", math doesn't work that way, the easiest way to think about it is that you don't have an infinite amount of time. – Henrik supports the community Jan 03 '15 at 11:21
  • You're telling me that I can't. But you're not telling me why, which is the important point. Given that axiom system, there's nothing that forbids me of doing that. There is no axiom that says: It's not possible to have infinite uses of $(3)$ or something similar. The problem with your argument is that, you're assuming things based on hand waving. "It's not possible because it does not work that way, it's not possible because Jesus told me that in a dream." and so on. – Red Banana Jan 03 '15 at 11:23
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    This set of axioms doesn't list ZFC either, you can use that! It's a general rule that you can't use any theorem infinitely many times. – Henrik supports the community Jan 03 '15 at 11:27
  • General rule from what? I know that historically, finitism made more sense and infinite processes were developed more of less when set theory was developed. – Red Banana Jan 03 '15 at 11:29
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If you "apply (3)(3) repeatedly", you can remove only a countable subset $C$ and

$$\text{measure}(C) = \sum_{x\in C}\text{measure}(\{x\}) = \sum_{x\in C}0 = 0.$$ because the countable additivity of the measure. (In the case of Apostol is even worse: only finite additivity is required)

If you remove a uncountable subset $U$ with measure$(U) > 0$, $$\text{measure}(U)\ne\sum_{x\in U}\text{measure}(\{x\}) = \sum_{x\in U}0 = 0.$$ (and what means an uncountable sum isn't obvious)

Community
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Martín-Blas Pérez Pinilla
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You can never remove so many points that way that it will make a difference, since every subtraction removes exactly 0 area. There is no way to remove all points by removing points one at a time.

Dasherman
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  • The problem is that you made up the part of "there is no way", you're coming with that from somewhere else. The axioms I used do not imply that I can't. That's my problem, I also can invent premises from somewhere else that enable me to make sense of it. – Red Banana Jan 03 '15 at 11:18
  • There exists no bijection between the naturals and $(0, 4)^2$. If you could remove all points that way, you could make a (infinite) list of all the coordinates in that square by listing them in the order of removal. This would imply that there is a bijection. Contradiction – Dasherman Jan 03 '15 at 11:22
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I'm completely out of my league here, so bear with me. But I recently viewed a video from vsauce that seems to suggest that theoretically what you describe might actually be possible and would end up creating The Banach–Tarski Paradox

Stijn de Witt
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