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I claim that for any infinite field $K$, there is such a linearly independent almost-disjoint family $X$ of size at least continuum. This is optimal when the field $K$ has size at most continuum, since in this case $K^{\mathbb{N}}$ has size continuum.

Proof: We shall build a finitely branching tree $T\subset K^{\lt\mathbb{N}}=\bigcup_n K^n$ with the following properties:

• the $n^{th}$ level of the tree 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 non-zero element of $K$ to it, choosing one of the nodes to receive two such extensions. I claim 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, we can arrange to choose the splitting nodes in such a way that every node leads eventually to a splitting node. (e.g. we could handle the finitely many nodes at level $n$ in the tree one after the other.)

So we've built the tree as desired, and so there is a linearly independent almost disjoint family of size continuum. QED

You are asking for the maximal size of an almost-disjoint 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 almost-disjoint 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 non-zero 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.

If the field is countable, then I claim that there is such a family of size continuum, which is the largest it could possibly be, since $K^{\mathbb{N}}$ has size continuum.

To see this, we shall build a tree $T\subset K^{\lt\mathbb{N}}$ in the space of finite sequences from $K$ so as to ensure the following properties:

• the tree is splitting, in the sense that every node has incomparable extensions.
• any two distinct paths through $T$ have at most finite agreement.
• any finitely many paths through $T$ are linearly independent.

If we can do this, then the collection $X$ of all paths through $T$ will be a size-continuum linearly independent family with any two elements having at most finite agreement.

We build the tree recursively, by specifying it up to increasingly high finite levels. We will ensure that distinct paths have at most finite agreement by ensuring that on any given level of the tree, distinct nodes use different elements of $K$ on that level. So once two paths through the tree depart, the elements of $K$ that appear on them will be totally different.

We can ensure that the tree is splitting by periodically inserting a splitting level into the tree, where we make sure that every node on that level has two successors in the tree.

Finally, the difficult part, we ensure the linear independence property. The key is to realize that there are only countably many possible linear dependence equations. So at a given stage of constructing the tree, we may consider just one equation, which has the form, say, $k_0f_0+\cdots+k_n f_n=0$, where each $k_i\in K$ is nonzero. The point now is that for any $n$ distinct branches through the finite tree as constructed so far, up to some height $N$, we may extend the tree a bit more so that the extension of those branches those branches no longer satisfy this equation in the finite dimensional vector space of sequences of that length. In finitely many steps, I can take care of all the finitely many ways of having chosen $n$ distinct branches in the tree up to the previously constructed level, and thereby handle them all. So at the $n^{th}$ stage of the construction, I kill off this equation as a possible dependence for the branches through my tree $T$. In the end, all such dependencies are killed, and any finitely many branches through the tree are linearly independent, completing the argument.

A set-theoretic way to think about the argument is this: if you force to add mutually generic sequences, then they will be linearly independent. The construction of the tree amounts to a density argument in forcing, and any sufficiently generic collection of branches will satisfy the independence property. The tree argument is reminiscent of the fact that in the forcing extension obtained by adding a single Cohen real $V[c]$, there is a family of continuum many finitely-mutually-generic Cohen reals, which I explained in my answer to Reals added after Cohen forcing.

When $K$ is uncountable, then I believe that set-theoretic issues will likely arise. For example, probably under Martin's Axiom you can still get large families for fields of size less than the continuum, using MA to choose the family of functions by using the fact that each dependency equation corresponds to a dense set in the forcing I have in effect described.

I claim that for any infinite field $K$, there is such a linearly independent almost-disjoint family $X$ of size at least continuum. This is optimal when the field $K$ has size at most continuum, since in this case $K^{\mathbb{N}}$ has size continuum.

Proof: We shall build a finitely branching tree $T\subset K^{\lt\mathbb{N}}=\bigcup_n K^n$ with the following properties:

• the $n^{th}$ level of the tree 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 non-zero element of $K$ to it, choosing one of the nodes to receive two such extensions. I claim 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, we can arrange to choose the splitting nodes in such a way that every node leads eventually to a splitting node. (e.g. we could handle the finitely many nodes at level $n$ in the tree one after the other.)

So we've built the tree as desired, and so there is a linearly independent almost disjoint family of size continuum. QED

If the field is countable, then I claim that there is such a family of size continuum, which is the largest it could possibly be, since $K^{\mathbb{N}}$ has size continuum.

To see this, we shall build a tree $T\subset K^{\lt\mathbb{N}}$ in the space of finite sequences from $K$ so as to ensure the following properties:

• the tree is splitting, in the sense that every node has incomparable extensions.
• any two distinct paths through $T$ have at most finite agreement.
• any finitely many paths through $T$ are linearly independent.

If we can do this, then the collection $X$ of all paths through $T$ will be a size-continuum linearly independent family with any two elements having at most finite agreement.

We build the tree recursively, by specifying it up to increasingly high finite levels. We will ensure that distinct paths have at most finite agreement by ensuring that on any given level of the tree, distinct nodes use different elements of $K$ on that level. So once two paths through the tree depart, the elements of $K$ that appear on them will be totally different.

We can ensure that the tree is splitting by periodically inserting a splitting level into the tree, where we make sure that every node on that level has two successors in the tree.

Finally, the difficult part, we ensure the linear independence property. The key is to realize that there are only countably many possible linear dependence equations. So at a given stage of constructing the tree, we may consider just one equation, which has the form, say, $k_0f_0+\cdots+k_n f_n=0$, where each $k_i\in K$ is nonzero. The point now is that for any $n$ distinct branches through the finite tree as constructed so far, up to some height $N$, we may extend the tree a bit more so that the extension of those branches those branches no longer satisfy this equation in the finite dimensional vector space of sequences of that length. In finitely many steps, I can take care of all the finitely many ways of having chosen $n$ distinct branches in the tree up to the previously constructed level, and thereby handle them all. So at the $n^{th}$ stage of the construction, I kill off this equation as a possible dependence for the branches through my tree $T$. In the end, all such dependencies are killed, and any finitely many branches through the tree are linearly independent, completing the argument.

A set-theoretic way to think about the argument is this: if you force to add mutually generic sequences, then they will be linearly independent. The construction of the tree amounts to a density argument in forcing, and any sufficiently generic collection of branches will satisfy the independence property.

If the field is countable, then I claim that there is such a family of size continuum, which is the largest it could possibly be, since $K^{\mathbb{N}}$ has size continuum.

To see this, we shall build a tree $T\subset K^{\lt\mathbb{N}}$ in the space of finite sequences from $K$ so as to ensure the following properties:

• the tree is splitting, in the sense that every node has incomparable extensions.
• any two distinct paths through $T$ have at most finite agreement.
• any finitely many paths through $T$ are linearly independent.

If we can do this, then the collection $X$ of all paths through $T$ will be a size-continuum linearly independent family with any two elements having at most finite agreement.

We build the tree recursively, by specifying it up to increasingly high finite levels. We will ensure that distinct paths have at most finite agreement by ensuring that on any given level of the tree, distinct nodes use different elements of $K$ on that level. So once two paths through the tree depart, the elements of $K$ that appear on them will be totally different.

We can ensure that the tree is splitting by periodically inserting a splitting level into the tree, where we make sure that every node on that level has two successors in the tree.

Finally, the difficult part, we ensure the linear independence property. The key is to realize that there are only countably many possible linear dependence equations. So at a given stage of constructing the tree, we may consider just one equation, which has the form, say, $k_0f_0+\cdots+k_n f_n=0$, where each $k_i\in K$ is nonzero. The point now is that for any $n$ distinct branches through the finite tree as constructed so far, up to some height $N$, we may extend the tree a bit more so that the extension of those branches those branches no longer satisfy this equation in the finite dimensional vector space of sequences of that length. In finitely many steps, I can take care of all the finitely many ways of having chosen $n$ distinct branches in the tree up to the previously constructed level, and thereby handle them all. So at the $n^{th}$ stage of the construction, I kill off this equation as a possible dependence for the branches through my tree $T$. In the end, all such dependencies are killed, and any finitely many branches through the tree are linearly independent, completing the argument.

A set-theoretic way to think about the argument is this: if you force to add mutually generic sequences, then they will be linearly independent. The construction of the tree amounts to a density argument in forcing, and any sufficiently generic collection of branches will satisfy the independence property. The tree argument is reminiscent of the fact that in the forcing extension obtained by adding a single Cohen real $V[c]$, there is a family of continuum many finitely-mutually-generic Cohen reals, which I explained in my answer to Reals added after Cohen forcing.

When $K$ is uncountable, then I believe that set-theoretic issues will likely arise. For example, probably under Martin's Axiom you can still get large families for fields of size less than the continuum, using MA to choose the family of functions by using the fact that each dependency equation corresponds to a dense set in the forcing I have in effect described.