I'm supposed to prove that $\mathbb{F}^\infty$ is infinite-dimensional. I was planning on doing a proof by induction to show that $(1,0,…),(0,1,0,…),…$ is a basis. Is this permissible? Also, I think I could do a proof by contradiction and suppose $\mathbb{F}^\infty$ to be finite-dimensional, and thus have a finite-length basis, and then show that there exists some $v\in\mathbb{F}^\infty$ such that $v$ is not in the span of the basis. I'm not quite sure how to show that rigorously though.
[Math] Proving that $\mathbb{F}^\infty$ is infinite-dimensional.
linear algebravector-spaces
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Best Answer
Here's a really simple proof. Let $f\colon \mathbb{F}^{\infty}\rightarrow \mathbb{F}^{\infty}$ be given by $f(x_0,x_1,x_2,\ldots)=(x_1,x_2,\ldots)$. It's easy to see that $f$ is linear. Note that $\mbox{Im}\:f =\mathbb{F}^{\infty}$ but $\ker f=\{(x_0,0,0,\ldots)\mid x_0\in\mathbb{F}\}\cong\mathbb{F}$ and so we have a surjective linear map from a vector space to itself which is not injective. This is not possible for finite dimensional vector spaces by the rank nullity theorem and so $\mathbb{F}^{\infty}$ is infinite dimensional.