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$Bbb{R}^n$ and $Bbb{R}$ are isomorphic as vector spaces over $Bbb{Q}$.

Mathematics Asked by user598858 on December 25, 2021

Show that $Bbb{R}^n$ and $Bbb{R}$ are isomorphic as vector spaces over $Bbb{Q}$.

My attempt:

Let $c=$ cardinality of $Bbb{R}$. Then we know that $Bbb{R}$ over $Bbb{Q}$ has basis of cardinality $c$. And for a basis $mathcal{B}$ of $Bbb{R}^n$ over $Bbb{Q}$, we have has $aleph_0<|mathcal{B}|leq |Bbb{R}^n|=c$. So from Cantor’s hypothesis $|mathcal{B}|=c$. Therefore bases for $Bbb{R}^n$ and $Bbb{R}$ have same cardinality when it is considered as a vector space over $Bbb{Q}$. Thus they are isomorphic.

Is this proof logically flawless? Please correct me if required. Thank you.

Edit I see use of Cantor’s hypothesis is not much justified. So how can we prove that the bases of both vector space have same cardinality without using this hypothesis?

2 Answers

Your proof is correct if you accept the axiom of choice which assures the existence of a basis $mathcal B_k$ of $mathbb R^k$ as a vector space over $mathbb Q$. Then $mathcal B_n$ and $mathcal B_1$ have the same cardinality. Now you invoke the following lemma:

Given vector spaces (over any field) $V_i$ with bases $mathcal B_i$ having the same cardinality. Then $V_1$ and $V_2$ are isomorphic.

Now cearly each function $f : mathcal B_1 to V_2$ extends to a unique linear map $f' : V_1 to V_2$. Now take a bijection $phi : mathcal B_1 to mathcal B_2$ and define $f(x) = phi(x)$. Then it is easy to verify that $f'$ is a linear isomorphism. Its inverse is induced by $phi^{-1}$.

Edited:

Let $B$ a basis for $V$. Then it is easy to show that the set of all pairs having the form $(b,0)$ and $(0,b')$ with $b,b' in B$ form a basis for $V times V$.

It is a well-known consequence of the axiom of choice that if $B$ is infinite, then there exists a bijection $B to Btimes{0} cup {0} times B$. Note that if $B$ is countable, then induction is sufficient to show this.

This shows that if $mathcal B$ is infinite, then $V$ and $V times V$ are isomorphic. Hence also $V$ and $V^n = V times ldots times V$ are isomorphic for any $n$.

Answered by Paul Frost on December 25, 2021

Note that $c$ depends on the underlying field ($mathbb{Q}$). The existence of a basis requires the axiom of choice. I do not know if the statement is correct. You are using the same cardinality to establish isomorphism property. This does only hold for finite-dimensional vectir spaces!

Answered by user7427029 on December 25, 2021

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