13. Prove $\mathbb{R}^n≅\mathbb{R}$ as vector spaces over $\mathbb{Q}$.
14a. Let $\mathcal{A}$ be a basis for the infinite dimensional vector space $V$ over $F$. Prove $V ≅ \oplus_{a∈\mathcal{A}}F$.
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 $|\mathcal{B}| ≤ |V|$. Letting $V_i$ for $i∈\mathbb{N}$ be the set of elements of $V$ whose basis sum includes exactly $i$ nonzero vectors, we have $V$ is the countable union of the $V_i$, so it suffices to show $|V_i| = |\mathcal{B}|$. We can observe$$|\mathcal{B}| = |\mathcal{B} \times \mathcal{B}| \geq |F \times \mathcal{B}| = |(F \times \mathcal{B})_1 \times ... \times (F \times \mathcal{B})_i| \geq |V_i|$$so that$$|\mathcal{B}| = |\bigsqcup_{i∈\mathbb{N}} \mathcal{B}| \geq |\bigsqcup_{i∈\mathbb{N}} V_i| = |V|$$and now $|\mathcal{B}|=|V|$.
(13) First we prove $\mathbb{R}^2≅\mathbb{R}$, so that by induction the proposition easily follows. First, assume the basis $\mathcal{A}$ of $\mathbb{R}$ is finite. Then $\mathbb{R}$ is isomorphic to the direct sum of a finite number of copies of $\mathbb{Q}$ by the next exercise, which would imply $\mathbb{R}$ is countable, a contradiction. Now, there is a basis $\mathcal{B}$ for $\mathbb{R}^2$ given by the basis $\mathcal{A}$ in each component whose cardinality is equal to $|\mathcal{A} \sqcup \mathcal{A}|$ so that $|\mathcal{B}|=|\mathcal{A}|$. 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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