I am solving the following : Vector space $\mathbb{R}$ over rational number $\mathbb{Q}$ is infinite dimensional.

I proved this by using that $\mathbb{R}$ is uncountable, but my professor suggested a following slightly different method : Suppose that this vector space is finite-dimensional and let {$r_1,...,r_n$} be a basis for it. Then find an element in $\mathbb{R}$ which doesn't belong to $\operatorname{Span}${$r_1,...,r_n$}. I've tried several methods, but it vain. How can I prove this?


marked as duplicate by Asaf Karagila Jun 17 at 21:43

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  • $\begingroup$ There are probably a few other duplicates... $\endgroup$ – Asaf Karagila Jun 17 at 21:44

Suppose $\mathbb{R}$ is a finite-dimensional $\mathbb{Q}$-vector space, say dimension $n$.

Pick a real transcendental number, such as $e$, and consider $1,e,e^2,e^3,e^4,\dots,e^n$. This is a set $n+1>\dim_\mathbb{Q}\mathbb{R}$ elements so must be linearly dependent, but that means $e$ satisfies a polynomial with $\mathbb{Q}$ coefficients, contradicting $e$ transcendental.


Your approach is fine. The span over $\Bbb Q$ of any finite set of vectors (be they real numbers, or from some other vector space) is necessarily countable, and as such any $\Bbb Q$-vector space with uncountably many vectors must be infinite dimensional.

As for your professor's approach, the best way I can think of would be to first enumerate all elements in the span, and then by Cantor's diagonal argument construct a real number which is not in the span. This feels like it's just the same logic in reverse, though.

If your professor (or anyone else) has a more concise way of constructing a real number not in the span, I would love to hear it.


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