Comments by "MC116" (@angelmendez-rivera351) on "The Empty Set is a Subset of Every Set Proof" video.
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You can't take the negation of "for all" if there is no "all", i.e, no members of φ, to begin with; by definition of logical negation and empty set.
No, this is nonsensical. You should really sit down one of these days and read an introductory-level book in formal logic. You absolutely can negate quantifiers over an empty domain. Logical negation of the universal quantifier is actually equal to to the existential quantifier of the negation, and this remains true in the empty domain.
But also, this objection is genuinely stupid: he never used universal quantifiers in his proof.
The contradiction is that φ (the empty set) is not a subset of A, φ is empty to begin with.
0. φ does not denote the empty set. Stop calling it "phi". We do have notation for the empty set, namely ø and {}. φ is not one such valid notation.
1. {} is empty, which is why it is a subset of A. There is no contradiction here. Since the empty set has no elements, every element this empty has (a.k.a none) is an element of A, and this much is true.
If you assume {} has members to begin with, the contradiction doesn't change that.
It absolutely does. If {} actually has elements, then at least one of those elements is not an element of A. So {} is not a subset of A.
The contradiction applies to subset, not membership.
No, it applies to membership, because the contradiction occurs on the existential quantifier. Also, this is a silly objection: the subset relation is defined exclusively in terms of the membership relation.
Otherwise, your argument runs: Assume {} has members...
No, that is definitely not at all how his argument would run.
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@southali If the room has no vacuums, how can the vacuum be a subset of anything since it doesn't exist?
As I pointed out already, it does exist. But also, this argument is fundamentally fallacious, since this is inadequate analogy for sets. The problem here is that objects in space do not merely form sets. Sets are not compatible with the idea that elements appearing a different number of times changes the set, yet in this room, changing the number of particles of any given molecule type definitionally changes the mereological sum of the particles. Also, because the existence of any given configuration is also dependent on the amount of space, and not just the number of elements, there is no well-defined notion of a subset. For example, the configuration where we only consider the oxygen atoms is also not a valid subset, because it is not true, at least according to you, that the room only has oxygen atoms. This is the issue. So the vacuum is not the issue here: the scenario itself is incapable of working with a coherent notion of subset. Again, this is because the scenario you are describing is not at all analogous to a set. What you are describing is a differentiable manifold with topological deformities of a given type. This has nothing to do with sets, really.
Also, I should point out that it is impossible to refute a valid formal proof using analogies from intuition, regardless of how much sense you think those analogies make. Analogies, by their very nature, are necessarily flawed. This is why we use logic, rather than analogies, for evaluating the validity of proofs.
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π is defined as the ratio between the circumference of a circle and the diameter of said circle. One can prove said ratio is constant by first, noting that this ratio does not change if a circle is centered at the origin. A circle centered at the origin with radius r has equation x^2 + y^2 = r^2. The ratio between the circumference and the diameter is equal to the ratio between the arclength of the upper semicircle and the radius of the corresponding circle. The upper semicircle is given by y = sqrt(r^2 – x^2), and the arclength is given by the integral on (–r, r) of sqrt[1 + (y')^2]. y' = –x/sqrt(r^2 – x^2), hence (y')^2 = x^2/(r^2 – x^2), implying that sqrt[1 + (y')^2] = r/sqrt(r^2 – x^2) = r/sqrt(r^2·[1 – (x/r)^2]) = r/(r·sqrt[1 – (x/r)^2]) = 1/sqrt[1 – (x/r)^2]. Let t = x/r, hence x = r·t, hence dx/dt = r, and (–r, r) |—> (–1, 1), so the above integral is equal to r multiplied by the integral of 1/sqrt(1 – t^2) on (–1, 1). The integral on (–1, 1) of 1/sqrt(1 – t^2) is independent of r, so this is a constant ratio, and so the arclength is proportional to r. Therefore, the arclength divided by the radius r is simply this constant of proportionality: the integral on (–1, 1) of 1/sqrt(1 – t^2). This integral is the definition of π.
To get a better definition that we can use to prove that π is a real number that is not rational, we can first notice that if we define g(x) as being the integral on (–x, x) of 1/sqrt(1 – t^2), then π := g(1). One can then obtain the Maclaurin series expansion of g, which converges everywhere for g, and use the Lagrange inversion theorem to prove that [g^(–1)](π) = 0, and furthermore that [g^(–1)](z) = Im[exp(i·z)]. Hence g^(–1) can be analytically continued to the entire complex plane, and it can be shown that [g^(–1)](0) = 0, and that in general, exp(2·m·π·i) = 1. This gives us a new, more useful definition of π: it is the unique real number such that it is half of the imaginary period of exp. This can be used to prove all sorts of properties of π.
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