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: {aα∈R∞}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 ℓ∞:={(x1,x2,…)∈R∞:{xn} bounded}, a standard prototype on which we do functional analysis, must also have a countable basis. However, if we equip ℓ∞ with a norm ‖(xn)‖=supn≥1|xn|, then ℓ∞ 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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