Comments by "EebstertheGreat" (@EebstertheGreat) on "Absolute Infinity - Numberphile" video.

If your model of set theory has an inaccessible cardinal, you can define the universe of sets up to that cardinality (in the von Neumann hierarchy). That universe doesn't contain its own cardinality, and there is no set in the universe as large as the universe itself, so you do essentially "run out of sets." Or if you use the whole universe V, you can discuss in a philosophical sense the "size" of V, and that can't possibly be the size of a set (because that would have to be a universal set). Rather, it's the size of a proper class. It's consistent that all proper classes have the same size, but it's also consistent that they have different sizes. But even if they all have the same size, that size is not a cardinal, because you can't form an equivalence class of proper classes.
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@tomkerruish2982 This actually contains two errors at once. First, it contains the error of assuming the generalized continuum hypothesis. Second, it confuses functions with plane curves. A curve in a plane is the image of some continuous map from the unit interval to the plane. There are only continuum-many of those. Specifically, you can easily map it to a subset of C(R,R^2) by picking some parameterization for each curve. What Gamow meant was the set of all functions from the interval to the plane, which is (R^2)^R and has cardinality 2^2^(ℵₒ). But since most of these aren't continuous (or even connected), they aren't curves. This second error is very surprising to tell the truth, and it has been repeated in some otherwise quite reliable and serious books.
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@whataboutthis10 Space-filling curves are not bijections. What makes them special is that they are continuous, but they actually intersect themselves, so they are not injective. You can create a bijection between R and R^2, but it will be discontinuous almost everywhere.
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@TianYuanEX If k is a cardinal, then Vₖ doesn't contain k. It does contain every cardinal less than k. And yes, it contains all their powersets, and powersets of powersets, etc., but that never reaches k. Because k is inaccessible. There could even be an inaccessible cardinal j < k, and if so, then Vₖ does contain inaccessible cardinals, including j. But it still doesn't contain k.
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@frankharris3308 In general, for cardinal numbers a and b with b ≤ a, if a is infinite, then we have a + b = b + a = a. So for instance, ℵ₁ + 1 = ℵ₁ + ℵ₀ = ℵ₁. That also applies to multiplication, so for instance, 2 * ℵ₀ = ℵ₀. But you can build larger cardinals using exponents. a^b = a if b < a (with a infinite), but if a ≤ b, then a^b > a. So for instance, (ℵ₀)² = ℵ₀, but 2^(ℵ₀) > ℵ₀. And since |R| = 2^(ℵ₀), the real numbers are uncountable. The reason for these rules is that the arithmetic is defined based on set operations. If sets A and B have cardinalities |A| = a and |B| = b, then their disjoint union A⊔B has cardinality a + b and their Cartesian product A×B has cardinality ab. For exponentiation, the set of all functions from B to A is written A^B, and its cardinality is a^b. These are all correct for both finite and infinite cardinals. But for infinite cardinals, some of these don't change the size. If you consider the set of natural numbers N, adding a finite number of elements to it clearly won't change its size. If I want to map {1,2,3,...} to {x,y,z,1,2,3,...} bijectively, that's easy to do by mapping 1 to x, 2 to y, 3 to z, and each other number n to n + 3. Similarly, multiplying N by any finite number just gives a finite number of copies of each number. Like, 2×N = {(0,1),(1,1),(0,2),(1,2),(0,3),(1,3),...}. Again, it's very easy to form a bijection between this and N. And even for N², the set of ordered pairs of natural numbers, it's not that hard. But for 2^N, which is the set of binary sequences, there is no bijection. All the numbers between 0 and 1 have unique binary expansions, and there are uncountably many of those. So |2^N| > |N|, i.e. 2^ℵ₀ > ℵ₀.
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@TianYuanEX If κ is an inaccessible cardinal, then V_κ is a model of ZFC. It contains everything in the cumulative hierarchy before κ.
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@TianYuanEX But the post you responded to began "If your model of set theory has an inaccessible cardinal, you can define the universe of sets up to that cardinality (in the von Neumann hierarchy). That universe doesn't contain its own cardinality."
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@625tvroom Think of C as just R^2, where the first element is the real part and the second element is the imaginary part. We can create a bijection from [0,1) to [0,1)^2 in the following way: Given two numbers a and b in [0,1)^2, write their decimal expansions and then "interleave" them to get a number c in [0,1). For instance, the pair (e-2, pi-3) = (0.71828..., 0.14159...) maps to 0.7114812589.... This map is an injection, because if (a,b) and (x,y) map to the same number c, that means every digit of a matches every digit of x and every digit of b matches every digit of y, so a = x and b = y. The map is also a bijection, because every number in [0,1) has a decimal expansion that can be constructed in this way. So it's a bijection. Then you can tile this bijection to fill the plane, giving a bijection between R and R^2, and thus between R and C. If you understand this, you should also see that we can make a similar bijection between R and R^n for any n. In fact, any interval can be mapped to any space. So they all have the same cardinality 𝔠.
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