Cantor Confusion
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From: David Marcus on 23 Jan 2007 16:59 Andy Smith wrote: > David Marcus writes > > >However, I think we can understand the fact that (in some sense) there > >are more reals than naturals without mentioning sets. A list is a bunch > >of rows where each row is labeled with a natural number, e.g., > > > >0. adfna > >1. afdkj > >2. dfajhadf > >3. adfj;df > >... > > > >The question is can we construct a list such that every real number in > >[0,1] appears on the list? We argue as follows. Let's agree to write > >real numbers in [0,1] as infinite decimals. Some infinite decimals have > >two representations, one ending in 0's, the other ending in 9's. Let's > >agree to use the one ending in 0's, except let's use 0.999... for 1. > >Suppose we have a list of reals: > > > >0. 0.1234... > >1. 0.4893... > >2. 0.3839... > >... > > > >For convenience, let's number the digits (after the decimal point) > >starting with zero. We construct a real x as follows. Let the n-th digit > >of x be 4 if the n-th digit of the n-th number on the list is 5. > >Otherwise, let the n-th digit of x be 5. > > > >Is x on our list? Well, it can't be the 0th number because it has a > >different 0th digit. It can't be the 1th number because it has a > >different 1th digit. In fact, it can't be any of the numbers. So, it > >isn't on our list. (Note that x can't be equal to a number on our list > >without the digits matching because the only way it could do that would > >be if it ended in 9's.) > > > >Therefore, given any list of real numbers in [0,1], we see there is a > >real number in [0,1] that is not on the list. Hence, no such list can be > >complete. > > > Yes, but, as you have observed to me in the past (and is my natural > understanding) that arguments which work with finite quantities can turn > into paradoxes or nonsense when things become infinite. That's not a reason to doubt everything. It just means to think before you leap. > The nub of my > original question was "why is this valid reasoning". Why not? It seems perfectly reasonable to me. What part do you consider questionable? > >Notice that I haven't used the word "set" or said anything about the > >"sizes" (or "cardinalities") of "infinite sets". > > > >> If I was asked to sum it up, at present I would say that my > >> understanding is that you can't have an actually infinite integer, > > > >What in the world would an "actually infinite integer" be and where did > >you get the idea there could be such a thing? It seems so > >counterintuitive. > > Yes, but on a simplistic view of infinity (e.g. mine as of 2 weeks ago) > there is no reason not to think that you can have ...1111 as well as > .1111... How do you go from "have ...1111" to concluding that "...1111" denotes a natural number? We said above what a natural number is. Why do you think the string "...11111" identifies or names a natural number? As for "0.111...", this is simply notation for sum_{i=1}^oo 10^{-i}, assuming you meant to use decimal notation. And, we can prove this sum converges to a real number. This should be quite intuitive if you identify the real numbers with the number line. > - or that the Universe has to have had a beginning (everything > has to succeed to something previous, so there must be an origin, so you > can't have infinite negative time - but can conceive of an infinite > future time, Aristotle again. Huh? > Hence my (ignorant) initial perspective > that there ought to be negative integers, and that in some sense > everything ought to wrap-around) Of course, there are negative numbers. You need to think about what your words mean. What do "in some sense" and "ought to wrap around" mean? > did you read my comment on transcendentals with an infinite random bit > string (a "conceptual" number, as I suggested)? Any one of such numbers, > that make up essentially all reals, requires an actually infinite number > of bits to represent it - it cannot be represented with a finite set of > integers. Worrying about the information content of digit strings will only confuse you. It is completely irrelevant to calculus. > BTW, my copy of Spivak turned up - it is beautifully presented and > written, absolutely at my level, Feynmann would have approved of it. You're welcome. -- David Marcus
From: Andy Smith on 23 Jan 2007 18:34 David Marcus writes >> But, to cut to the chase, if you don't have enough integers to define >> even one real, what chance of counting them all? > >What in the world do you mean by "don't have enough integers to define >even one real"? Do you have some problem defining sqrt(2)? > I meant that the infinite non-repeating irrational binary expansion of sqrt(2) requires an actually infinite set of numbers to define its location on the line (you don't buy my idea of conceptual numbers, so just regard all reals as necessarily requiring an infinite number of bits to describe them). Since the integers are finite, you cannot represent a real requiring an actually infinite number of bits, is what I meant. Maybe that is too simplistic? -- Andy Smith
From: David R Tribble on 23 Jan 2007 18:42 mueckenh wrote (2007-01-18): > 1. The union of all finite trees is an infinite tree. > 2. Every finite tree contains only a finite set of paths. > 3. The countable union of all paths of the finite trees is therefore the > countable union of all finite paths. > 4. The countable union of all finite paths is in the union of all finite trees. > 5. The "complete" tree containing all paths is identical to the union of > al finite trees, with respect to nodes and edges. > 6. Identical trees cannot contain different sets of paths. > 7. Therefore, both trees contain the same set of paths. > 8. Therefore the "complete" set of all path is countable. > 9. Therefore the set of all real numbers is countable. > 10. Therefore ZFC is inconsistent. Just wondering if you're still using this argument. If so, perhaps you can explain how you manage the leap from (8) to (9). It looks like you're talking about the union of all finite-length trees (whatever that means) in (1) through (8), then at (9) it looks like you somehow conclude that the set of all finite trees is equivalent to the set of all reals. Assuming you're using binary trees to represent binary real fractions in [0,1], you are omitting all the repeating rationals (like 1/3) and all the irrationals (like sqrt(2)/2), which all require infinite fractions to represent. Since your union only includes the finite-length trees, it only represents the finite-length binary fractions, which are obviously countable, but excludes the vast uncountable majority of the rest of the reals. Or are you using a different argument now? [Apologies if the alignment/spacing of this post is screwy; Google just updated their newsgroup viewer/editor, and it no longer has a 'preview' button.]
From: Andy Smith on 23 Jan 2007 18:48 David Marcus writes >> >> Yes, but on a simplistic view of infinity (e.g. mine as of 2 weeks ago) >> there is no reason not to think that you can have ...1111 as well as >> .1111... > >How do you go from "have ...1111" to concluding that "...1111" denotes a >natural number? We said above what a natural number is. Why do you think >the string "...11111" identifies or names a natural number? > I just meant, that on a simplistic lay view of infinity, that there is no reason that you shouldn't have an infinite number of digits in a number. OK, that isn't correct, but on a simple view of infinity it seems plausible. Doesn't work with natural numbers. > >> - or that the Universe has to have had a beginning (everything >> has to succeed to something previous, so there must be an origin, so you >> can't have infinite negative time - but can conceive of an infinite >> future time, Aristotle again. > >Huh? People cannot conceive of an infinite past, but are content with an infinite future. There is something in common here with the natural numbers and the Peano axioms - there are no negative numbers, all numbers are the successor of some successor to 0. Perhaps you would object to the idea of an infinite past on the grounds that this would imply the existence of infinite integers (e.g. a clock started infinity ago , but still ticking and showing an infinite integer on its dial?) > -- Andy Smith
From: stephen on 23 Jan 2007 18:55
Andy Smith <Andy(a)phoenixsystems.co.uk> wrote: > David Marcus writes >>> But, to cut to the chase, if you don't have enough integers to define >>> even one real, what chance of counting them all? >> >>What in the world do you mean by "don't have enough integers to define >>even one real"? Do you have some problem defining sqrt(2)? >> > I meant that the infinite non-repeating irrational binary expansion of > sqrt(2) requires an actually infinite set of numbers to define its > location on the line (you don't buy my idea of conceptual numbers, so > just regard all reals as necessarily requiring an infinite number of > bits to describe them). > Since the integers are finite, you cannot represent a real requiring an > actually infinite number of bits, is what I meant. Why not? There are an "actually infinite" number of integers. There is exactly one integer for each of the "actually infinite" number of positions needed to represent the sqrt(2). There is no "actually infinite" integer, but that is a different thing. > Maybe that is too > simplistic? You seem to be confused about an infinite set containing all finite quantities. The properties of a set, and the properties of the elements of a set are different things. Stephen |