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Suppose that $\mathbb R$ denotes the set of all real numbers, $\mathbb Q$; the set of all rational numbers.

Let $V=\mathbb R(\mathbb Q)$ denote the vector space of real numbers over field $\mathbb Q$.

Suppose on the contrary, that $V$ is finite dimensional. It follows that $V$ has a finite basis $\mathbb B=\{b_i\in \mathbb R:1\le i\le n\},$ where $n$ is a fixed natural number.

Every $r\in \mathbb R$ can be written uniquely as linear combination of vectors in $\mathbb B$.

That is, for any $r\in \mathbb R$, there exist unique $q_{ir}\in \mathbb Q, 1\le i\le n$ such that

$r=q_{1r}b_1+q_{2r}b_2+\cdots+q_{nr}b_n$

Let $f: \mathbb R\to \mathbb Q^n$ be a function defined as

$f(r)=(q_{1r},q_{2r},\cdots,q_{nr})$

$f$ is one-one and therefore $\mathbb R$ has cardinality not exceeding cardinality of $\mathbb Q^n$, i.e. aleph-null. This is a contradiction as $\mathbb R$ is uncountable.

Hence by contradiction, $V$ must be infinite dimensional vector space.

Is my proof correct? Thanks.

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    $\begingroup$ Looks good to me. $\endgroup$ Sep 16, 2021 at 5:03
  • $\begingroup$ @JairTaylor: Thanks a lot for reviewing the proof. :) $\endgroup$
    – Koro
    Sep 16, 2021 at 5:07
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    $\begingroup$ It's an alright proof, but there's no need to mention cardinality: all you had to use is the fact that $\mathbb{R}$ is not countable while $\mathbb{Q}^n$ is (of course, I recognize I'm saying the exact same thing you did, but in a simpler way). $\endgroup$ Sep 16, 2021 at 5:09
  • $\begingroup$ @MatheusAndrade: Thanks for the comment. Yeah, I thought about that but that somehow involved cardinal arithmetic ($r=\sum q_i b_i$ so $|\mathbb R|\le |\mathbb Q|^n$ as $q_i$ can be chosen in $|\mathbb Q|$ ways) so I wanted to avoid that. $\endgroup$
    – Koro
    Sep 16, 2021 at 5:13
  • $\begingroup$ @Koro Sorry for the mistake, but $\mathbb{R}$ is absolutely not countable, I mistyped but corrected the mistake in a later edit (it's correct now). If there were an injection from $\mathbb{R}$ to a countable space then $\mathbb{R}$ would be countable, which is not the case. $\endgroup$ Sep 16, 2021 at 5:14

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