Uncountable linearly independet family in KNK^\mathbb{N}

An example I like is this: Say two sets of natural numbers are almost disjoint iff they are infinite but their intersection is finite. Suppose $\mathcal F$ is an uncountable family of almost disjoint sets. Then the collection of characteristic functions of the sets in $\mathcal F$ is an example. (If $A$ is a subset of $\mathbb N$, then its characteristic function $\chi_A$ is the infinite sequence of $0$s and $1$s whose $n$-th entry is $1$ iff $n\in A$.)

Indeed, if $A_1,\dots,A_n,B$ are distinct sets in $\mathcal F$, then there are infinitely many $n\in B$ but not in $\bigcup_i A_i$, so any linear combination of the $\chi_{A_i}$ is zero at all these entries, but $\chi_B$ is not.

To see that $\mathcal F$ exists, a cute argument is to note that it suffices to find such a family but made up of subsets of $\mathbb Q$. Now, to each real $r$ assign an increasing sequence $A_r$ of rationals converging to $r$, and take $\mathcal F=\{A_r\mid r\in \mathbb R\}$.

Another, more sophisticated example, uses the concept of independent family. A collection $\mathcal I$ of subsets of $\mathbb N$ is called an independent family iff for any distinct $A_1,\dots,A_k,B_1,\dots,B_n$ in $\mathcal I$, we have that the set $A_1\cap\dots\cap A_k\cap(\mathbb N\setminus B_1)\cap\dots\cap(\mathbb N\setminus B_n)$ is infinite.

Again, it is easy to check that if $\mathcal I$ is an independent family, then the collection of its characteristic functions is linearly independent.

As with the almost disjoint family from above, we can show that there is an independent family of the same size as the reals. But the argument is more involved; a sketch follows: As before, if $X$ is countable, it suffices to find an independent family as required, but consisting of subsets of $X$ (so we replace each ${\mathbb N}\setminus B_i$ with $X\setminus B_i$, of course).

Take $X=\{(a,A)\mid a\subseteq{\mathbb N}$ is finite, and $A\subseteq{\mathcal P}(a)\}.$ Check that $X$ is countable.

Now, given $S\subseteq{\mathbb N},$ let $t_S=\{(a,A)\in X\mid a\cap S\in A\}.$ The claim is that ${\mathcal I}=\{t_S\mid S\subseteq{\mathbb N}\}$ is as wanted.

To see this, check first that the assignment $S\mapsto t_S$ is injective.

To show that ${\mathcal I}$ satisfies the stronger independence requirement, suppose $X_1,\dots,X_n,Y_1,\dots,Y_m$ are infinite, pairwise distinct, subsets of ${\mathbb N}$. Write $A=t_{X_1}\cap\dots\cap t_{X_n}\cap(X\setminus t_{Y_1})\cap\dots\cap(X\setminus t_{Y_m})$. For each $i,j$ with $1\le i\le n$ and $1\le j\le m$ let $\alpha_{ij}\in X_i\triangle Y_j$ (that is, $\alpha_{ij}$ is in either $X_i$ or $Y_j$, but not in both). Show that for any finite $F\supseteq\{\alpha_{ij}\mid 1\le i\le n,1\le j\le m\}$ there is an ${\mathcal F}$ such that $(F,{\mathcal F})\in A.$ This gives the result.

Both these examples, using almost disjoint families, and using independence, are classical applications of infinitary combinatorics to analysis (in this case, to infinite dimensional linear algebra, that is usually studied in the setting of functional analysis). I once asked for the details of the sketch above on a homework problem.