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I 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 co
by 6gvONxR4sf7o 6y ago
I 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.