Given vector space $V$ over $\mathbb R$ such that the elements of $V$ are infinite-tuples. How to show that any basis of it is uncountable?
Uncountability of basis of $\mathbb R^{\mathbb N}$
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linear-algebra
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0How do you define "infinite $n$-tuple?" What is $n$? Do you just mean elements which are infinite sequences of real numbers? – 2012-10-16
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0For example (a_1,a_2,...) would be an element of V,where each a_i is a real number. – 2012-10-16
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1Then you don't mean $n$-tuple, which means a sequence $n$ real numbers. – 2012-10-16
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0Based on literature which I have read it can be called n-tuple as well. – 2012-10-16
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0No. n-tuple means length n, e.g. 3-tuple (5,2,7). For your things you might say "infinite tuple", or "countably infinite tuple", although as already pointed out the term "sequence" is more in use. – 2012-10-16
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0Let $V$ be the vector space with just one element, namely the sequence $(0,0,0,0,\ldots)$. Then the elements of $V$ are certainly infinite tuples, but $V$ has a finite (in fact empty) basis. – 2012-10-16
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0There is a technical difference here @Pilot The sequences you are considering fall under $\mathbb{R}^\mathbb{N}$ whereas by $\infty$-tuple, most people refer to $\mathbb{R}^\infty$ which are sequences which are eventually zero. The first is uncountable while the second countable. – 2012-10-16
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1My bad. Let me clarify it then: V contains all possible infinite tuples. – 2012-10-16
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2@Pilot As several people have mentioned, it's probably best to not use the term "tuple". I would just stick with "infinite sequence of real numbers". – 2012-10-16
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1I found an older question, which is a generalization of this one: [Vector dimension of a set of functions](http://math.stackexchange.com/questions/95028/vector-dimension-of-a-set-of-functions) – 2012-10-16
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0Possibly a duplicate: [What is the standard proof that $\dim(k^{\mathbb N})$ is uncountable?](http://math.stackexchange.com/questions/176475/what-is-the-standard-proof-that-dimk-mathbb-n-is-uncountable/) – 2012-12-10