13. Prove Rn≅R as vector spaces over Q.
14a. Let A be a basis for the infinite dimensional vector space V over F. Prove V≅⊕a∈AF.
Proof: Throughout these exercises we shall assume the fact that countable unions and finite direct products of an infinite set S fix the cardinality.
(12) By the inclusion mapping we clearly see |B|≤|V|. Letting Vi for i∈N be the set of elements of V whose basis sum includes exactly i nonzero vectors, we have V is the countable union of the Vi, so it suffices to show |Vi|=|B|. We can observe|B|=|B×B|≥|F×B|=|(F×B)1×...×(F×B)i|≥|Vi|so that|B|=|⨆i∈NB|≥|⨆i∈NVi|=|V|and now |B|=|V|.
(13) First we prove R2≅R, so that by induction the proposition easily follows. First, assume the basis A of R is finite. Then R is isomorphic to the direct sum of a finite number of copies of Q by the next exercise, which would imply R is countable, a contradiction. Now, there is a basis B for R2 given by the basis A in each component whose cardinality is equal to |A⊔A| so that |B|=|A|. Linearly extend the homomorphism induced by this bijection to obtain an isomorphism.
(14a) By mapping the coefficients of a vector sum of the basis to the proper coordinates in the direct sum we obtain the evident isomorphism. Now, the direct product is clearly a vector space by closure with componentwise multiplication by scalars of F.
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