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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_12012-07-29
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    For $\lvert k\rvert <\mathfrak c$ it is trivial just like in the finite case, by cardinality considerations.2012-07-29
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    @JyrkiLahtonen: why wouldn't the same argument work for the countably infinite-dimensional space?2012-07-29
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    @tomasz: If $V$ has a countable dimension then there is no strictly increasing chain of $2^{\aleph_0}$ many subspaces.2012-07-29
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    @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).2012-07-29
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    @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 $\bigcup_{r'$S_r$ are infinite, and hence the characteristic function is not a finite linear combination of functions with singleton support. – 2012-07-29
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    @JyrkiLahtonen: I opened [a question about this](http://math.stackexchange.com/questions/176585/why-a-countably-infinite-space-does-not-have-an-uncountable-chain-of-subspaces)2012-07-29
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    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).2012-07-29

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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 $\kappa=\aleph_0$).

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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. ;)2012-07-29
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    @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... :-)2012-07-29
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    well, I'd leave it to the asker, but I doubt that's the kind of answer he was looking for. ;-)2012-07-29
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    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.2012-07-29
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    @ChindeaFilip: I'm not sure what you are trying to say...2012-07-29
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    @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 ?2012-08-28
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    @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.2012-08-28