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.

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    $\begingroup$ How have you had the space defined? $\endgroup$ Sep 5, 2013 at 9:15
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    $\begingroup$ $\mathbb{F}^n = {(x_1,x_2,...,x_n):x_j\in\mathbb{F} for j = 1,...,n}$ $\endgroup$
    – Danny
    Sep 5, 2013 at 9:19
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    $\begingroup$ If you define $\Bbb F^{\infty}$ with '$\ldots$' like you did, I don't see why going for $(1,0, \ldots ), (0,1,0,\ldots), \ldots$ wouldn't be permissible. $\endgroup$
    – Git Gud
    Sep 5, 2013 at 9:27
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    $\begingroup$ Ok. Now, you argument looks fine (note that the important part is showing that those vectors are linearly independent). But it should probably be noted that the notation is a bit ambiguous, since there are also infinite dimensional spaces of strictly larger dimension that this one (so instead of using $\infty$ people will often prefer to write one of $\mathbb{F}^{\omega}$, $\mathbb{F}^{\aleph_0}$ or $\mathbb{F}^{\mathbb{N}}$). $\endgroup$ Sep 5, 2013 at 9:28
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    $\begingroup$ @TobiasKildetoft $K^\infty = \{f: \mathbb N \to K \mid f(n)$ is eventually $0\} \subsetneq \{f: \mathbb N \to K\} = K^{\mathbb N}$. In particular $e_1=(1,0,\dotsc)$, $e_2=(0,1,0,\dotsc)$ etc. isn't a basis for $K^{\mathbb N}$. For example $(1,1,1,\dotsc)$ cannot be written as a finite linear combination of the $e_i$. $\endgroup$
    – kahen
    Sep 5, 2013 at 9:48

3 Answers 3


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.


An indirect proof can be simpler than the one you sketch.

Suppose that $K^\infty$ (or $K^{\mathbb N}$ -- that doesn't matter here) has finite dimension $n$. We know already (I hope) that in a finite-dimensional vector space no linearly independent set can have more members than the dimension. But $\{\mathbf e_1,\mathbf e_2,\ldots,\mathbf e_n,\mathbf e_{n+1}\}$ is clearly a linearly independent set of size $n+1$, so $n$ can't have been the dimension after all.


In this post, I will prove a special case. $F^{\infty}=\{(x_n)_{n\in \Bbb{N}}| x_n\in F\}$. Let $K$ be a countable subfield of $F$. Proof of existence of such $K$: Characteristics of a field $F$ is either $0$ or prime. If characteristics of a field $F$ is $0$, then $F$ contains the field of rational numbers $\Bbb{Q}$. If characteristics of a field $F$ is prime say $p$, then $F$ contains the finite field $\Bbb{Z}_p$. Hence $\exists K\subseteq F$ such that $K$ is countable subfield of $F$. There is a “subtle” reason, why we use subfield instead of an arbitrary field. Usually scalar multiplication operation on $F^\infty$ is defined by $c\cdot (x_n)_{n\in \Bbb{N}}=(c\cdot x_n)_{n\in \Bbb{N}}$. If field is arbitrary, then $c\cdot x_n$ don’t make sense. It’s easy to check $(F^\infty, K,+,\cdot)$ is a vector space over field $K$.

Claim: $(F^\infty, K,+,\cdot)$ is infinite dimensional vector space. Proof: Assume towards contradiction, $F^\infty$ is finite-dimensional vector space, $\mathrm{dim}(F^\infty)=n\in \Bbb{N}$. Let $B=\{\alpha_1,…,\alpha_n\}$ is basis of $F^\infty$. Then $\exists f:F^\infty \to K^n$ such that $f$ is bijective, here is the proof. Since $F$ is field, $\exists 0_F,1_F\in F$ such that $0_F\neq 1_F$. Let $A=\{ (x_i)_{i\in \Bbb{N}}|x_i=0_F \text{ or } 1_F\}$. By theorem 2.14 of Baby Rudin, $A$ is uncountable. Since $A\subseteq F^\infty$, we have $F^\infty$ is uncountable. Since $f$ is bijective, $|F^\infty|=|K^n|$. So $K^n$ is uncountable. But by theorem 2.13 of Baby Rudin, $K^n$ is countable. Thus we reach contradiction. Hence $(F^\infty, K,+,\cdot)$ is infinite dimensional vector space.

We can generalize the claim in following way:

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.


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