Ep. 2 Least Upper Bound Property and Uncountability of Reals: Navigating Metric Space Topology

We grapple with the definition of the set of real numbers, eventually arriving at the statement of its least upper bound property. We also present the Cantor's diagonalization proof that the set of real numbers is uncountable.

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[mihirvora]

Great video as always! I just want to provide some more context regarding countability for those who are interested:

Georg Cantor (the founder of set theory) referred to the number of elements in any set (empty, nonempty, finite, infinite) as the set’s cardinality. From the same logic presented in the video, it can be proven that the set of natural numbers is also countable. Cantor called the cardinality of the natural numbers something called “aleph-null” and referred to any set with cardinality ≤ aleph-null as a countable set. Hence, the set of real numbers (as we saw in the video) must have a cardinality that exceeds aleph-null. An interesting consequence of this is that if you have a bijective function f : A —> B (where A and B are sets defining the domain and codomain of the function, respectively), then A and B must have the same cardinality. For this reason, there exists no bijective function f that associates an element of say the reals with an element of the natural numbers (because R is uncountable while N is countable, which by definition requires that they have different cardinalities, namely R has cardinality > aleph-null and N has cardinality = aleph-null).

Hope this info sheds some more light on the already great material presented in the video! Will definitely try out the practice problems! :-)

[MathematicalEndeavors]

Thank you very much for making this series! Here are some of my thoughts on the first four problems. (I wish the Youtube comment section had TeX compatability.)

1) R is the union of Q and (R - Q), so if (R - Q) is countable, then R is the union of two countable sets, hence countable, which is a contradiction. This can be shown using essentially the same argument you gave for Q. If sets A and B are countable, I can write them in sequences (a_n) and (b_n), possibly with repeats, and then alternate (a_1, b_1, a_2, b_2, ), so A U B is countable. A diagonal argument could also be used to construct a bijection directly between N x N and A x B.

2) If max A exists, it is an upper bound for A by definition; as sup A is the least upper bound, we must have max A <= sup A. Furthermore, by definition, max A is an element of A, and sup A >= x for every x in A, hence sup A >= max A. So sup A = max A.

3) On this one, I'm not 100%. It depends on whether we define countable as "finite and countably infinite" or just "countably infinite." In the latter case, the set A is infinite, so if the sequence has repeats, I can just construct a subsequence which drops the repeats. Conversely, a sequence without repeats satisfies the original definition. If finite sets are countable, this doesn't work because, if A were finite, I only have finitely many distinct terms -- unless finite sequences are allowed. Another way to view it is: the first definition says that A is countable if and only if there exists a surjection from N to A, sending n --> a_n. The second definition says that a set is countable if and only if there exists an injection from A into N. For a nonempty set, these can be shown to be equivalent.

4) a) Max A = 6 = sup A; inf A = 5. min A does not exist, because 5 is not in the set and, given any epsilon > 0, I can choose N sufficiently large so that 5 + 1/N < 5 + 1/epsilon.

b) I believe this is the same set.

c) The elements of the set are the partial sums of the infinite geometric series sum (1/2)^i, which converges to 2. The sum is increasing, so min A = 1/2 = inf A. As a sequence in n, this is bounded above and increasing, so it converges to sup A = lim (a_n) = 2, the infinite sum, but infinity is not a number, so 2 is never obtained and there is no max.

d) Min A = 1 = inf A; the set is not bounded above, so it has no max and sup A = infinity.

e) The set has no min or max, so inf A = - infinity and sup A = +infinity.

[zmaj]

I'll put problem 5 here. I assume part a in problem 5 has a typo, where S is meant to be A.

Part a: Since A is nonempty, so is A'. Since A is bounded below, a lower bound b exists. For all t in A', t = -s for some s in A, and b <= s. Therefore, -b >= -s = t, and so -b is an upper bound for A'. By the least upper bound property, sup A' exists. (Note that for any lower bound b of A, -b is an upper bound of A'. We use this later.)

For all s in A, -s is in A', and thus -s <= sup A' --> s >= -sup A'. -sup A' is then a lower bound on A. For the sake of contradiction, suppose there exists a real number c such that c is a lower bound of A, and c > -sup A'. Since c is a lower bound of A, -c is an upper bound on A'. We also have -c < sup A', but this contradicts the definition of supremum. Thus, no such c exists, and so -sup A' is the greatest lower bound of A.

Part b: Since A is bounded below, a lower bound exists, so L is nonempty. Since A is nonempty, consider s in A. s is an upper bound on L, because all b in L are lower bounds of A, so they satisfy b <= s. Thus, M = sup L exists.

For the sake of contradiction, suppose M is not a lower bound of A. Then, some t in A satisfies t < M, but t is an upper bound of L. This contradicts the fact that M = sup L, so M is a lower bound of A. Also, there does not exist m > M that is a lower bound of A, since all lower bounds of A are in L, and thus must be at most M. M is therefore the greatest lower bound of A, equal to inf A.

[Westinvasion]

Would it be valid to prove problem 1 as follow: we take the exact way of proving uncountability of R but we multiply those sets of decimals with some irrational value i_0. So we multiuply x_0, x_1, x_2, ..., x_n, ... with i_0 and we get another irrational number. And after that we make new irrational number with diagonal elements. And that irrational number is not included in the initial set.