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From Vector Spaces to Periodic Functions (2019)
- andi999 6y agoDidnt Bona say:"The Axiom of Choice is obviously true, the Well–ordering theorem is obviously false; and who can tell about Zorn's Lemma?"
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- Koshkin 6y agoThat R as a vector space over Q has a finite basis is an amazing fact.
- fractionalhare 6y agoWhat do you mean? The real vector space defined over the rationals is infinite-dimensional; it can't have a finite basis. I think any basis of R over Q would have to be not only infinite, but uncountably so. Do you mean that any given vector of the space can be represented using a linear combination of a finite subset of the basis elements with rational scalar coefficients?
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- giraj 6y agoYou're right, any basis of R over Q has to be uncountably infinite. If a countable basis existed, we could for example write R as the union of the subspaces spanned by the n first basis elements, indexed by n. That would mean R is countable, since a countably-indexed union is countable.
- AnHonestComment 6y agoWhat...? Every set {1/n^k | k in N} for n in N is a basis for R over Q, in the sense that any r in R is a Sum[0, inf)[q_i * b_i]. It’s not a finite basis, but it’s definitely countable. It’s not generally the case that a countable union of countable sets is countable — that only holds in the finite case, as we see with power series representations of real numbers.
- a1369209993 6y ago> any basis of R over Q has to be uncountably infinite. It's actually trivial to give a (infinite but) countable set of (non-orthogonal) basis vectors: 2^i for integer i. 0 and 1 are both rationals (scalars), and every real number has a (possibly infinite) binary expansion, eg e = 10.1011011111100001... = 1·2^1 + 0·2^0 + 1·2^-1 + 0·2^-2 + ...
- giraj 6y agoThat's not a basis of R at all. If you want to create a basis of R you better choose irrational elements in your basis, otherwise the Q-span is contained in Q. What I think you want to say is that "any real number has a binary expansion". Which is true, but the binary sequences don't form a vector space over R, but a Z/2-module. And as a Z/2-module, your { 2^i for integer i } isn't even a basis because you need infinite expansions to express most real numbers. The span of a basis are only the finite linear combinations. (FWIW, I think you've given a description of the dyadic rationals.)
- a1369209993 6y ago> any given vector of the space can be represented using a linear combination of a finite subset of the basis elements with rational scalar coefficients? FWIW, this is definitely false: x = q₀·√2/1 + q₁·√3/3 + q₂·√5/9 + q₃·√7/27 + q₄·√11/81 + ... = Σᵢ(qᵢ·√πᵢ/3ⁱ) for rational qᵢ. Edit: primes approximate e^i, so 2^i isn't large enough.
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- fractionalhare 6y agoThis isn't false; it's a theorem. Using the axiom of choice, all vectors of any space can be represented as finite linear combinations. In the case of infinite-dimensional vector spaces, this means you would need at most n vectors from the infinite basis B, and n scalars from the underlying field F. I'm interested in what you're trying to demonstrate with your example though, or if we misunderstand each other, because this theorem is very well established.
- coverage 6y agoέςϧѯ ̑ ͜ ˀ ̑ ѯˀ ...//en.wikipedia.org/wiki/Tensor ...?
- denial 6y agoEach element is representable as a finite sum of basis elements; this is true of all vector spaces (they all have Hamel bases). However, this does not mean that the basis itself is finite.
- mathpadawan 6y ago> The axiom of choice is allowed. Why does this need to be stated explicitly? Why does the axiom of choice not enjoy the privilege of being a first-class axiom like other axioms? Why do many proofs need to state the assumption of this axiom explicitly before using it?
- generationP 6y agoBecause it gets you applications like this (or Banach-Tarski), which make you wonder how much it deserves its place. To some extent this is more due to Tertium Non Datur than to the axiom of choice, but my impression is ZF doesn't know enough about infinite sets to get to the really counter-intuitive stuff.
- hansvm 6y agoThere are counter-intuitive results in ZF without choice too: - Collection of non-empty sets with empty cartesian product. - Infinite set without a countable infinite subset. - There is a pair of sets such that neither is equinumerous with a subset of the other. More or less, our intuition about seemingly obvious ideas is completely thrown off without choice. Banach-Tarski at least has the property that it probably doesn't directly apply to the real world (if you can split an object into probably physically impossible sets and then rejoin them correctly then you can double the volume of the physical object) and so doesn't really violate our intuition -- the premise doesn't apply in the real world, so no conclusion really matters. It's like claiming that every element of the empty set is a leprechaun with a pot of gold -- it's true, but it doesn't matter in any meaningful sense.
- im3w1l 6y agoIiuc your examples are undecidable rather than true in ZF. That's why we can add Axiom of Choice without getting contradictions.
- hansvm 6y agoWell, sure. Without choice there exist models of ZF where those properties hold. Adding choice precludes those models -- i.e., it restricts our axioms in a way which prevents certain kinds of unexpected results.
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- im3w1l 6y agoI thought of this construction: Let f, the first function have period 1, g the second have period √2. Define the equivalence relationship that x~y if x = y + a + b√2 for some a and b. Pick for each class a unique representative using the Axiom of Choice. For each representative e, define f(a+b√2+e) = b√2, g(a+b√2+e) = a + e By construction, any number can be represented through a unique (a, b, e). And f and g manifestly have the right periods.
- a1369209993 6y agoDo you mean real x and y, and rational a and b? (Ie, "x~y if x = y + a + b√2 for some [rational] a and b".)
- im3w1l 6y agoMy bad, a and b are integers. That's why the periods work out.
- 6gvONxR4sf7o 6y agoI seem to be missing something. I’m reading this as looking for functions f and g from R to R, such that f(x)+g(x)=1 and f(x+p)=f(x). If I were reading that correctly, then for any periodic f, g(x)=1-f(x) would do, meaning any old periodic function would do. What an I missing? Am I also reading it right that the author is saying the axiom of choice is equivalent to the statement that every vector space has a finite basis? I don’t get how that allows infinite dimensional vector spaces. If not, and it’s just that every vector space has a basis that’s maybe infinite, then what’s the justification of the Hamel basis being finite? I feel like I’m missing a lot between the lines here.
- shakow 6y ago> What an I missing? Identity function means that f(x) = x, not f(x) = 1.
- 6gvONxR4sf7o 6y agoDoh. Thanks.
- leephillips 6y agoThis is where it lost me. Vector spaces can have infinite basis sets.
- kamilner 6y ago> Am I also reading it right that the author is saying the axiom of choice is equivalent to the statement that every vector space has a finite basis? Not quite. The axiom of choice is equivalent to saying every vector space has a Hamel basis, which is to say every element can be represented as a finite combination of elements of the Hamel basis. It doesn't imply that the Hamel basis is finite itself.