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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_1.2012-07-29
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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', 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.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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