Alfred001 9 Posted March 7, 2020 Share Posted March 7, 2020 If I understand the argument correctly, it's that if you compare a set of all whole numbers to a set of all whole + half numbers, when you look at each set up to number 2, set 1 would be 1 and 2 (and 0?) and set 2 would be 1/2, 1, 1 1/2 and 2 - so set 2 has more numbers, but that's only if you look up to 2. If you look at the whole sets the size of each is infinite, so neither is bigger than the other. What am I missing here? Link to post Share on other sites

studiot 2291 Posted March 7, 2020 Share Posted March 7, 2020 4 minutes ago, Alfred001 said: If I understand the argument correctly, it's that if you compare a set of all whole numbers to a set of all whole + half numbers, when you look at each set up to number 2, set 1 would be 1 and 2 (and 0?) and set 2 would be 1/2, 1, 1 1/2 and 2 - so set 2 has more numbers, but that's only if you look up to 2. If you look at the whole sets the size of each is infinite, so neither is bigger than the other. What am I missing here? Which argument would that be ? What do you understand by 'compare' ? I think you are talking about putting the elements of one set into one to one correspondence with the elements of another set. The wording and structure of set theory was very carefully crafted by Cantor (and later improved by others) when he discovered that there can be more members in one infinite set than there are in a different infinite set. This discovery came as a suprise to everyone. So it would help (you) to to try to be as careful as he was. Link to post Share on other sites

Strange 4273 Posted March 7, 2020 Share Posted March 7, 2020 33 minutes ago, Alfred001 said: If I understand the argument correctly, it's that if you compare a set of all whole numbers to a set of all whole + half numbers, when you look at each set up to number 2, set 1 would be 1 and 2 (and 0?) and set 2 would be 1/2, 1, 1 1/2 and 2 - so set 2 has more numbers, but that's only if you look up to 2. If you look at the whole sets the size of each is infinite, so neither is bigger than the other. What am I missing here? You can show that the set of integers, for example, is the the same "size" as the set of fractions between 0 and 1 because you can work out a way of mapping from integers to fractions (or vice versa). In other words there is a one-to-one mapping between them (there is a technical mathematical term for this, but I am not a mathematician; "bijection", maybe?) What Cantor showed is that it is impossible to form an equivalent mapping between the integers and the real numbers. However you try and do this, you can always invent another real number that fits between the two that you have mapped. (It is a bit like Euclid's proof that there are infinite primes, in that respect.) The conclusion is that there are infinitely many more real numbers than there are integers. Even though there are an infinite number of integers. He went on to show that that are multiple levels of infinity which are not equal to one another. In fact, an infinite number of them. Some explanations below, hopefully you will find one that makes sense to you http://jlmartin.faculty.ku.edu/~jlmartin/courses/math410-S09/cantor.pdf https://www.cs.virginia.edu/luther/blog/posts/124.html https://en.wikipedia.org/wiki/Cantor's_diagonal_argument Link to post Share on other sites

taeto 93 Posted March 7, 2020 Share Posted March 7, 2020 5 hours ago, Alfred001 said: If I understand the argument correctly, it's that if you compare a set of all whole numbers to a set of all whole + half numbers, when you look at each set up to number 2, set 1 would be 1 and 2 (and 0?) and set 2 would be 1/2, 1, 1 1/2 and 2 - so set 2 has more numbers, but that's only if you look up to 2. If you look at the whole sets the size of each is infinite, so neither is bigger than the other. What am I missing here? Maybe if we compare the set of all whole numbers from \(0\) to \(10\) with the set of all half numbers \(0, 1/2, 1, 3/2,\ldots, 19/2,10,\) then we see that the second set contains more numbers than the first. And yet each set has finite size, so in that sense, none is bigger than the other? It seems that you are missing that to say that two sets both have finite size, or both infinite size, does not imply that they have the exact same size. Link to post Share on other sites

MigL 1631 Posted March 7, 2020 Share Posted March 7, 2020 (edited) I always use whole numbers and even ( or odd, depending on my mood ) numbers to show the one-to-one correspondence. Most people would think that there are only half as many even numbers ( 2,4,6,8,10,...) as there are whole numbers ( 1,2,3,4,5,6,...). But you can easily see that each whole number can be doubled to give each even number, so in the case of infinite sets of both there is a one-to-one correspondence between each n, and 2n ( for whole and odd, n and 2n-1 ). That means the two infinities are equal. If you can't do a one-to-one correspondence then the two infinite sets cannot be equal and one must be greater than the other. Edited March 7, 2020 by MigL 1 Link to post Share on other sites

studiot 2291 Posted March 7, 2020 Share Posted March 7, 2020 4 hours ago, MigL said: I always use whole numbers and even ( or odd, depending on my mood ) numbers to show the one-to-one correspondence. Most people would think that there are only half as many even numbers ( 2,4,6,8,10,...) as there are whole numbers ( 1,2,3,4,5,6,...). But you can easily see that each whole number can be doubled to give each even number, so in the case of infinite sets of both there is a one-to-one correspondence between each n, and 2n ( for whole and odd, n and 2n-1 ). That means the two infinities are equal. If you can't do a one-to-one correspondence then the two infinite sets cannot be equal and one must be greater than the other. A good example +1 One point is of course defining an 'infinite set' This can be achieved using the above. A set can be considered infinite when some subset can be put into one to one correspondence with the whole set. For the Reals this can be strengthened from some subset to all subsets. Link to post Share on other sites

taeto 93 Posted March 8, 2020 Share Posted March 8, 2020 (edited) 11 hours ago, studiot said: For the Reals this can be strengthened from some subset to all subsets. Which should probably have said "all uncountable subsets", and even that is assuming the Continuum Hypothesis. Or simpler, all open intervals. Edited March 8, 2020 by taeto Link to post Share on other sites

studiot 2291 Posted March 8, 2020 Share Posted March 8, 2020 3 hours ago, taeto said: Which should probably have said "all uncountable subsets", and even that is assuming the Continuum Hypothesis. Or simpler, all open intervals. Yes I agree intervals is a better (more rigorous) use. But why only open intervals ? Are not both [0, 1] and ]0, 1[ isomorphic to R ? Or do you think ]0, 1[ has more elemens than R or less? Link to post Share on other sites

taeto 93 Posted March 8, 2020 Share Posted March 8, 2020 (edited) 6 minutes ago, studiot said: Yes I agree intervals is a better (more rigorous) use. But why only open intervals ? Are not both [0, 1] and ]0, 1[ isomorphic to R ? Or do you think ]0, 1[ has more elemens than R or less? With closed or half-open intervals, it is more awkward as to which real numbers to assign to the endpoints if you explicitly demand a 1-1 function. Mainly by specifying open intervals I would avoid the closed intervals of the form \([0,0]\) ... Edited March 8, 2020 by taeto Link to post Share on other sites

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