Let $K$ be a field, say infinite, and denote by $L$ the $K$vector space $K^{\mathbb{N}}$. What is the maximal cardinality of a $K$linearly independent subset $X$ of $L$ such that any two distinct elements of $X$ agree on at most finitely many elements of $\mathbb{N}$?

2$\begingroup$ Forgive me, but what is the dimension of $K^{\mathbb{N}}$ as a $K$vector space? (Of course at least continuum, and exactly continuum when $K$ has size at most continuum; but what is it when $K$ is larger?) $\endgroup$ – Joel David Hamkins Aug 17 '13 at 12:26

$\begingroup$ The ErdosKaplansky theorem states that for any field $K$ and any set $I$, the linear dimension of the vector space $K^I$ is equal to its cardinality: $dim_K (K^I)=K^I$. $\endgroup$ – user38700 Aug 17 '13 at 16:20

1$\begingroup$ That can't be correct the way you say it, since for example $I$ could be finite. $\endgroup$ – Joel David Hamkins Aug 17 '13 at 16:37

$\begingroup$ Excuse me, I meant it for an infinite set $I$. So again, ErdosKaplansky theorem states that for any field $K$ and any infinite set $I$, the linear dimension of the vector space $K^I$ is equal to its cardinality: $dim_K(K^I)=K^I$ $\endgroup$ – user38700 Aug 17 '13 at 16:48

$\begingroup$ Yes, I see. Perhaps one can hope to modify the proof of ErdosKaplansky to answer the general case of your question. $\endgroup$ – Joel David Hamkins Aug 17 '13 at 16:54
You are asking for the maximal size of an almostdisjoint family in $K^{\mathbb{N}}$ that is also linearly independent in this space.
I claim that for any infinite field $K$, there is such a linearly independent almostdisjoint family $X$ of size continuum $2^{\aleph_0}$. Notice that for any countable field, and indeed, for any field of size at most continuum, this is optimal — $X$ has the largest conceivable size — since in this case the whole space $K^{\mathbb{N}}$ has size continuum.
Proof: We shall build a finitely branching tree $T\subset K^{\lt\mathbb{N}}=\bigcup_n K^n$, consisting of finite sequences from $K$, with the following properties:
 the $n^{th}$ level of the tree (sequences in $T$ of length $n$) consists of a linearly independent collection in $K^n$.
 the tree is splitting, in the sense that every node in $T$ has incomparable extensions in $T$.
 distinct sequences in $T$ have no agreement beyond their common initial segment. In other words, once two sequences disagree, they never agree again. This is equivalent to insisting that the values of $K$ arising on a given level of the tree are distinct.
Given such a tree, consider the collection $X$ of all paths through the tree. This will have size continuum, since the tree is splitting. Distinct paths through $T$ will have only finite agreement, since they disagree beyond their common initial segment, and so $X$ is an almost disjoint family. And any finitely many branches from $X$ will be linearly independent, since they will be linearly independent even when restricted to any level of the tree where those branches become distinct.
So let's build the tree. Suppose that it has been specified up to level $n$, consisting of $m$ linearly independent sequences in $K^n$. Extend each of these sequences by appending a distinct nonzero element of $K$ to it, choosing one of the nodes to receive two such extensions. I claim (it is a little linear algebra exercise) that the resulting family is still linearly independent in $K^{n+1}$.
Now, the point is that since the tree will have only countably many nodes altogether, we can arrange to choose the splitting nodes in such a way that every node leads eventually to a splitting node. For example, we could arrange that at the $n^{th}$ step of our construction, we ensure that the $n^{th}$ sequence (in some canonical enumeration) that we added to the tree has a splitting node above it.
In this way we build a tree with all the desired properties, and so there is a linearly independent almost disjoint family of size continuum, as desired. QED
I'm less sure what happens with larger fields.

$\begingroup$ Many thanks Mr Hampkins for your generous answer. I still need to know if the answer is always the optimal one, namely $K^{\aleph_0}$, where $K$ denotes the cardinality of $K$. If anybody knows the answer, I would be grateful. $\endgroup$ – user38700 Aug 17 '13 at 11:17

1$\begingroup$ You're welcome (although my name is Hamkins). For $K$ larger than the continuum, it will often happen that $K^{\mathbb{N}}=K$ (but definitely not always), which may make things easier to think about. $\endgroup$ – Joel David Hamkins Aug 17 '13 at 11:31