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Tuesday, October 21, 2014

Basis of R=RN must be uncountable

Recall that a subset M of a vector space X is a basis if and only if every element in X can be written as a linear combinations of finitely many elements in M.

The title of this post has a simple proof: {aαR}α>0, where aα=(1α,2α,3α,), is an uncountable linearly independent subset, so every basis has to be uncountable.

I just want to point out knowledge in functional analysis quickly gives an intuitive answer.

Suppose R has M as a countable basis, then consider its vector subspace :={(x1,x2,)R:{xn} bounded}, a standard prototype on which we do functional analysis, must also have a countable basis. However, if we equip with a norm (xn)=supn1|xn|, then becomes a Banach space with countable basis.

This is a contradiction to the standard fact in functional analysis that every basis of an infinite dimensional Banach space must be uncountable, so every basis of R must be uncountable!

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