What is the standard proof that $\dim(k^{\mathbb N})$ is uncountable?

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).

Solution 1:

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 $\kappa=\aleph_0$).