# Infinite dimensional vector space [duplicate]

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

## 2 Answers

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.