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Topic: Does a nonlinear additive function on R imply a Hamel basis of R?
Replies: 8   Last Post: Apr 21, 2013 7:59 PM

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lugita15@gmail.com

Posts: 101
Registered: 4/1/05
Does a nonlinear additive function on R imply a Hamel basis of R?
Posted: Apr 20, 2013 11:28 AM
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A function is additive if f(x+y)=f(x)+f(y). Intuitively, it might seem that an additive function from R to R must be linear, specifically of the form f(x)=kx. But assuming the axiom of choice, that is wrong, and the proof is rather simple: you just take a Hamel basis of R as a vector space over Q, and then you define your function f to be different in at least two distinct elements of the basis.

But my question is this: if there is no Hamel basis of R, then must f be linear? To put it another way, does ZF + the existence of a nonlinear additive function imply the existence of Hamel basis of R?

I checked the Consequences of the Axiom of Choice Project, a database of choice axioms and their relationships here, and it said that it didn't know.

Any help would be greatly appreciated.

Thank You in Advance.



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