Let $V$ be the set of real numbers. Regard $V$ as a vector space over the field of rational numbers, with the usual operations. Prove that this vector space is not finite dimensional.
We can prove a stronger result:
Let $(V,F,+,\cdot)$ be a vector space over field $F$. If $V$ is uncountable and $F$ is countable, then $(V,F,+,\cdot)$ is infinite dimensional vector space.
My attempt: Assume towards contradiction, $V$ is finite-dimensional vector space, $\mathrm{dim}(V)=n\in \Bbb{N}$. Let $B=\{\alpha_1,…,\alpha_n\}$ is basis of $V$. Then $\exists f:V \to F^n$ such that $f$ is bijective, here is the proof. Since $f$ is bijective, $|V|=|F^n|$. Since $V$ is uncountable, $F^n$ is uncountable. But by theorem 2.13 of Baby Rudin, $F^n$ is countable. Thus we reach contradiction. Hence $(V, F,+,\cdot)$ is infinite dimensional vector space. Is my proof correct?
Best Answer
Alternative:
Suppose $\dim(V) =n$ and $\{1,\pi,\pi^2,\ldots ,\pi^n\}$ is a set of $n+1$ vectors, hence dependent.
Then $\sum_{k=0}^{n} r_k\pi^k=0$ where $r_k\in \Bbb{Q}$ and not all $r_k$ zero. $(r_1\neq 0) $
Hence $\pi$ is a root of the polynomial $f(x) \in \Bbb{Q}[x]$ , $f(x) =\sum_{k=0}^{n} r_k x^k$
implies $\pi$ is algebraic which is a contradiction.
Existence of transcendental number forces $\dim(V) $ infinite.