Clarification regarding the convergence of a sequence of vector subspaces In my studies of linear algebra, my instructor has provided the following example to illustrate convergence of vectors and the spaces they span.
We consider the two sequences of vectors $$u_n = (1,2^{-n},4^{-n})$$ $$v_n = (1,2 \cdot 2^{-n},3 \cdot 4^{-n}).$$ We know that in the limit $n\to \infty$ we have $u_n\to(1,0,0)$ as well as $v_n\to(1,0,0)$. This much is clear to me. However, my instructor says that since for every natural $n$, $u_n$ and $v_n$ are linearly independent, the space $\text{span} \{u_n,v_n\}$ is two-dimensional, and the limit of this vector space in the limit $n\to\infty$ is not one-dimensional despite both spanning vectors converging to the same vector. My instructor then said that $\text{span} \{u_n,v_n\}$ converges to the space $\text{span} \{(1,0,0),(0,1,0)\}$. I do not understand what my instructor said here, as he explained it sort of heuristically with neither rigor nor actually defining what it means for a sequence of vector spaces to converge. It is hard for me to see how a sequence of vector spaces spanned by two vectors converging to the same vector can converge to a two-dimensional space. Also, I really do not understand how the limit space was determined to be $\text{span} \{(1,0,0),(0,1,0)\}$. I would appreciate any explanations to derive and clarify what my instructor has said, and I thank all helpers.
 A: To state that a sequence of spaces converges to a space, you need a topology on the set of spaces. One convenient way to define a topology is via a metric (see here for more on that). Two spaces are close if their distance according to the metric is small. In the present case, on the space of all two-dimensional linear subspaces of $\mathbb R^3$, you can use the angle that two planes form as the distance between them. (You can check that this satisfies all the properties required for a metric, including the triangle inequality, but that’s not actually necessary for defining a topology.)
The angle between two planes is the angle between their normal vectors, so  the cosine of the angle between $\operatorname{span}\{u_n,v_n\}$ and $\operatorname{span}\{e_1,e_2\}$ (with $e_i$ denoting the canonical basis vectors) is
$$
\frac{(u_n\times v_n)\cdot(e_1\times e_2)}{|u_n\times v_n|\cdot|e_1\times e_2|}=\frac{(u_n\times v_n)\cdot e_3}{|u_n\times v_n|}=\frac{2\cdot2^{-n}-2^{-n}}{\sqrt{(2\cdot2^{-n}-2^{-n})^2+O(4^{-2n})}}\to1\quad\text{for}\quad n\to\infty\;.
$$
Thus, the angle that measures the distance between the two spaces converges to $0$, so in this topology $\operatorname{span}\{u_n,v_n\}$ converges to $\operatorname{span}\{e_1,e_2\}$ for $n\to\infty$.
The intuitive way to see this is to note that the $z$ components of the basis vectors decay faster than the $y$ components, so $\operatorname{span}\{u_n,v_n\}$ converges to $\operatorname{span}\{(1,2^{-n},0),(1,2\cdot2^{-n},0)\}=\operatorname{span}\{e_1,e_2\}$.
