Intersection of vector subspaces and extension of the ground field. Let $V$ be a vector space over a field $F$ and $\{W_i\}_{i\in I}$ a collection of subspaces of $V$. Let $K$ be an extension field of $F$. If $I$ is finite then it is easy to see that $$\big(\bigcap_{i\in I} W_i\big)\otimes_F K=\big(\bigcap_{i\in I} W_i\otimes_F K \big).$$ 
Is this true also when $I$ is infinite? 
 A: Yes, but I don't have as nice an argument as one might want, seemingly because "intersection" as categorical object has the natural maps going to it, rather than from, while the tensor product has maps going the opposite direction...?
But, nevertheless, the intersection has a basis (Axiom of Choice?), and extends to a basis of each $W_i$. It is not hard to prove that an "extension of scalars" of a free module produces a free module with "the same" generators, in the sense that they are images of the originals.
Thus, the intersection of the $W_i\otimes_F K$'s is no larger than the tensor product with the intersection.
A: We will use the followig fact about the tensor product of $F$-vector spaces.

Lemma:
  Let $V, U$ be $F$-vector spaces and let $(u_j)_{j \in J}$ be an $F$-basis of $U$.
  Then every $x \in V \otimes_F U$ can be written as $x = \sum_{j \in J} v_j \otimes u_j$ for unique elements $v_j \in V$ (with $v_j = 0$ for all but finitely many $j \in J$).
Proof:
  We have that $U = \bigoplus_{j \in J} \langle u_j \rangle_F$ and therefore
  $$
      V \otimes_F U
=     V \otimes_F \left( \bigoplus_{j \in J} \langle u_j \rangle_F \right)
=     \bigoplus_{j \in J} ( V \otimes_F  \langle u_j \rangle_F )
=     \bigoplus_{j \in J} (V \otimes u_j)
\cong \bigoplus_{j \in J} V,
$$
  where $\langle u \rangle_F = \{\lambda u \mid \lambda \in F\}$ denotes the $F$-span of $u \in U$.

We can now prove the statement itself by showing both inclusions:
For every $j \in I$ we have that $\bigcap_{i \in I} W_i \subseteq W_j$, therefore
$$
            \left( \bigcap_{i \in I} W_i \right) \otimes_F K
  \subseteq W_j \otimes_F K,
$$
and thus alltogether
$$
            \left( \bigcap_{i \in I} W_i \right) \otimes_F K
  \subseteq \bigcap_{j \in I} (W_j \otimes_F K).
$$
For the other inclusion let $x \in \bigcap_{i \in I} (W_i \otimes_F K)$, and let $(b_j)_{j \in J}$ be an $F$-basis of $K$.
By using that $x \in V \otimes_F K$ we may write
$$
    x
  = \sum_{j \in J} v_j \otimes b_j
$$
for unique elements $v_j \in V$.
For every $i \in I$ it similarly follows from $x \in W_i \otimes_F K$ that
$$
    x
  = \sum_{j \in J} w^i_j \otimes b_j
$$
for unique elements $w^i_j \in W_i$.
For every $j \in J$ it follows from the uniqueness of these decompositions that $v_j = w^i_j \in W_i$ for every $i \in I$, and therefore that $v_j \in \bigcap_{i \in I} W_i$.
It thus follows that
$$
      x
  =   \sum_{j \in J} v_j \otimes b_j
  \in \left( \bigcap_{i \in I} W_i \right) \otimes_F K \,.
$$
This shows the other inclusion.
