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This is my (silly) proof to a claim on top of p. 54 of Rotman's "Homological algebra".

For $k$ an infinite field (the finite case is trivial) prove that $k^\mathbb{N}$, the $k$-space of functions from the positive integers $\mathbb{N}$ to $k$, has uncountable dimension.

Lemma. There is an uncountable family $(A_r)_{r \in \mathbb{R}}$ of almost disjoint infinite subsets of $\mathbb{N}$, i.e., $|A_r \cap A_s| < \infty$ for $r \neq s$.

The proof is standard, let $f : \mathbb{Q} \stackrel{\sim}{\to} \mathbb{N}$ and $A_r := \{f(r_1), f(r_2), \ldots\}$ for $(r_j)_{j \in \mathbb{N}}$ a sequence of distinct rationals whose limit is $r$. Of course, these must be chosen simultaneously for all $r$, but any choice will work.

Now the family $f_r : \mathbb{N} \to k$, $f_r(x) = 1$ for $x \in A_r$ and $0$ elsewhere is linearly independent, since $a_1f_{r_1} + \cdots + a_kf_{r_k} = 0$ yields $a_1 = 0$ if applied to $x \in A_{r_1} \backslash (A_{r_2} \cup \cdots \cup A_{r_k})$, etc.

Curiously, this shows that $\dim(k^\mathbb{N}) \ge |\mathbb{R}|$, which is "a bit more". I'd like to see the folklore trivial "one-line argument", since I don't remember to have learned about it.

Thanks in advance. Also, greetings to stackexchange (this being my first topic here).

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Don't know about one-liners. You could also argue that the subspaces $V_r$ defined for all real numbers $r$ as $$V_r:=\{s:\mathbb{Q}\rightarrow k \mid s(q)=0\,\forall q>r\}$$ form an uncountable chain of subspaces (without repetitions): $V_{r_1}\subset V_{r_2}$ whenever $r_1<r_2$. – Jyrki Lahtonen Jul 29 '12 at 14:02
For $\lvert k\rvert <\mathfrak c$ it is trivial just like in the finite case, by cardinality considerations. – tomasz Jul 29 '12 at 14:11
@JyrkiLahtonen: why wouldn't the same argument work for the countably infinite-dimensional space? – tomasz Jul 29 '12 at 14:15
@tomasz: If $V$ has a countable dimension then there is no strictly increasing chain of $2^{\aleph_0}$ many subspaces. – Asaf Karagila Jul 29 '12 at 14:23
@AsafKaragila: I know, but I'm asking about the particular argument JyrkiLahtonen used. It seems to me it should work just as well in that case, so that's what I'm asking about. I'm either missing something obvious, or there's some important part he's glossed over (no fault in that, given that it is a comment, of course). – tomasz Jul 29 '12 at 14:27

One liner argument which uses a much more difficult theorems (swatting gnats with cluster bombs kind of proof):

$k^\mathbb N$ is the algebraic dual of the polynomials in one variable, $k[x]$ which has a countable dimension. If $k^\mathbb N$ had a countable basis then $k[x]$ would be isomorphic to its dual, and since this cannot be we conclude that $k^\mathbb N$ has a basis of uncountable size.

The arguments given in Arturo's answer show that the above is indeed a proof (in particular Lemma 2 with $\kappa=\aleph_0$).

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It seems to me that the question you linked is actually a generalization of this one (sans the "one liner" part), so using it is more than just swatting gnats with cluster bombs. ;) – tomasz Jul 29 '12 at 14:41
@tomasz: Well, yes. The one-liner is simply the application of the general case to this particular case. It is a one-linear though... should I delete it? I did feel a bit as if it's cheating to use this argument... :-) – Asaf Karagila Jul 29 '12 at 14:43
well, I'd leave it to the asker, but I doubt that's the kind of answer he was looking for. ;-) – tomasz Jul 29 '12 at 14:46
I just read (same book) about $\dim(V^{\ast}) > \dim(V)$ in the infinite-dimensional case but didn't notice that $k^{\mathbb{N}} \simeq (k^{\oplus \mathbb{N}})^{\ast}$... it seems that I was wrong about the "folklore" statement though. – Chindea Filip Jul 29 '12 at 15:06
@ChindeaFilip: I'm not sure what you are trying to say... – Asaf Karagila Jul 29 '12 at 15:08

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