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If $V$ is a finite dimensional $K$-vector space, then every set of subsets of subspaces contains a maximal element, i.e. a subspace which no subspace of the set contains properly, equivalently we have no infinitely ascending chain of subspaces, this is an easy consequence of finite-dimensionality.

For a vector space $V$, not every subgroup of $(V, +)$ is also a subspace, for example $\mathbb R$ considered as a vector space over itself has only the trivial subspaces, but among the subgroups are $\mathbb Q$ or $\mathbb Z$.

But is it possible that a subspace contains an infinite ascending (or infinite descending) sequence of subgroups? I am asking for a finite dimensional vector space $V$ with subspace $U \le V$, such that we have an infinitely ascending chain $$ A_1 < A_2 < A_3 < \ldots $$ of subgroups $(A_i, +) \le (V, +)$ of the additive group of $V$, but $A_i < U$ for all $i$.

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The abelian group $\mathbb{Q}$ is neither artinian nor noetherian, so it contains both an infinite descending chain and an infinite ascending chain (it's not difficult to find them explicitly).

If you have a vector space over a field $K$ of characteristic $0$, every nonzero subspace contains a copy of $\mathbb{Q}$ as an additive subgroup.

An infinite field of characteristic $p>0$ is infinite dimensional over the field with $p$ elements, so it's neither artinian nor noetherian as an abelian group and the argument is the same.

Of course, if the field is finite, the situation cannot happen.

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