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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).

Asaf Karagila
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Filip Chindea
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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
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    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
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    @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
  • @tomasz: That is a very good question. The best answer I can give at this time is that the function $s_r$ that is the characteristic function of the set $S_r:=(-\infty,r]\cap\mathbb{Q}$ is in $V_r$, but it is not in the union $\bigcup_{r'<r}V_{r'}$, and hence these functions form an uncountable linearly independent set. The same functions won't work in the countably infinite-dimensional case, because the sets $S_r$ are infinite, and hence the characteristic function is not a finite linear combination of functions with singleton support. – Jyrki Lahtonen Jul 29 '12 at 17:44
  • @JyrkiLahtonen: I opened a question about this – tomasz Jul 29 '12 at 19:54
  • Thanks, tomasz. A closer examination reveals that the argument outlined in my first comment does not prove anything unless it is complemented by the argument of the second comment. I apologize for not thinking it thru carefully (I was rushing to go and watch the darts final). – Jyrki Lahtonen Jul 29 '12 at 20:34

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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 $\kappa=\aleph_0$).

Community
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Asaf Karagila
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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. – Filip Chindea 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
  • @AsafKaragila - Can't we just aruge that $\aleph_0^{\aleph_0}=\aleph$ and since $k$ is infinite then $\aleph_0\leq|k|$ and thas it ? – Belgi Aug 28 '12 at 15:07
  • @Belgi: I'm not sure what you're trying to suggest. Note that $|\mathbb R^\mathbb N|=|\mathbb R|$. Just a simple cardinality argument is not sufficient. – Asaf Karagila Aug 28 '12 at 17:00