Basis on a vector space with a filtration Let $V$ be a vector space over some field $F$. 
Consider a filtration of $V$ by subspaces $V_k$ ($k \in \mathbb{Z}$) such that $V_k \subseteq V_l$ for all $k \leq l$. That is, we have a filtration
$$\cdots \subseteq V_{-1} \subseteq V_0 \subseteq V_1 \subseteq V_2 \subseteq \cdots$$
Furthermore, assume that $V = \bigcup_{k \in \mathbb{Z}} V_k$. 
Does there exist a basis $B$ of $V$ such that $B \cap V_k$ is a basis of $V_k$ for all $k \in \mathbb{Z}$?
I guess this should be true and if so, should probably be a straightforward application of Zorn's lemma. But I cannot quite figure out the right argument. (In the finite-dimensional case the problem is easy).
 A: When we have an ascending filtration
$$V_0 \subseteq V_1 \subseteq V_2 \subseteq V_3 \subseteq \cdots$$
we can construct the suitable basis by first taking a basis of $V_0$, extending this to a basis of $V_1$, extending this to a basis of $V_2$, and so on. This inductive construction gives a basis with the desired properties.
So to the original question, we should still consider descending filtrations
$$V_0 \supseteq V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots$$
and here in general we cannot find a basis $B$ such that $B \cap V_k$ is a basis for all $k \in \mathbb{Z}$. If we had such a basis, then for all $k$ we have $V_k = W_k \oplus V_{k+1}$, where $W_k$ is spanned by $B \cap (V_k \setminus V_{k+1})$. Suppose $\cap_{k \geq 0} V_k = 0$ for simplicity. Then we have $$B = \bigcup_{k \geq 0} B \cap (V_k \setminus V_{k+1})$$ so $V \cong \bigoplus_{k \geq 0} W_k$ as vector spaces. But this might not be true. For example, $V$ might have dimension that is much larger than $\bigoplus_{k \geq 0} W_k$. 
That is exactly what happens if $V$ has uncountable dimension, and each $V_k/V_{k+1}$ is finite-dimensional. For a concrete counterexample like this, take $V = F^{\mathbb{N}}$, the vector space of all infinite sequences $$(a_0, a_1, a_2, a_3, \ldots)$$ with $a_i \in F$ for all $i \geq 0$. Let $V_k$ be the subspace of sequences such that $a_i = 0$ for all $i < k$. Then $$V = V_0 \supseteq V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots$$ and $V_k/V_{k+1}$ has dimension $1$ for all $k \geq 1$. Furthermore, $\cap_{k \geq 0} V_k = 0$. Since $V$ has uncountable dimension, there does not exist a basis $B$ of $V$ such that $B \cap V_k$ is a basis of $V_k$ for all $k \geq 0$.
