large.

\[lem:infty\] The natural inclusion $\iota_n:\operatorname{Gr}(k,n)\to \operatorname{Gr}(k,n+1)$ is isometric, i.e., $$\label{eq:iso}
d_{\operatorname{Gr}(k,n)}(\mathbf{A},\mathbf{B})=d_{\operatorname{Gr}(k,n+1)}\bigl(\iota_n(\mathbf{A}),\iota_n(\mathbf{B})\bigr).$$ Repeated applications of yields $$\label{eq:iso1}
d_{\operatorname{Gr}(k,n)}(\mathbf{A},\mathbf{B})=d_{\operatorname{Gr}(k,m)}\bigl(\iota_{nm}(\mathbf{A}),\iota_{nm}(\mathbf{B})\bigr)$$ for all $m > n$ and if we identify $\operatorname{Gr}(k,n)$ with $\iota_n\bigl(\operatorname{Gr}(k,n)\bigr)$, we may rewrite as $$d_{\operatorname{Gr}(k,n)}(\mathbf{A},\mathbf{B})=d_{\operatorname{Gr}(k,m)}(\mathbf{A},\mathbf{B})$$ for all $m > n$.

If $a\in \mathbb{R}^n$, we write $\hat{a} = \left[\begin{smallmatrix}
a \\
0
\end{smallmatrix}  \right]\in \mathbb{R}^{n+1}$. Let $A = [a_1,\dots, a_k]$ and $B = [b_1,\dots,b_k]$ be any orthonormal bases of $\mathbf{A}$ and $\mathbf{B}$ respectively. By the definition of $\iota_n$, $\iota_n(\mathbf{A})$ is the subspace in $\mathbb{R}^{n+1}$ spanned by an orthonormal basis that we will denote by $\iota_n(A) \coloneqq [\hat{a}_1,\dots, \hat{a}_k] \in \mathbb{R}^{(n+1) \times k}$. Hence we have $$\iota_n(A)^\mathsf{T} \iota_n(B)
=
\begin{bmatrix}
A^\mathsf{T} B \\
0
\end{bmatrix}.$$ By the expression for Grassmann distance in , we see that must hold.

Since the inclusion of $\operatorname{Gr}(k,n)$ in $\operatorname{Gr}(k,n+1)$ is isometric, a geodesic in $\operatorname{Gr}(k,n)$ remains a geodesic in $\operatorname{Gr}(k,n+1)$. Given $\mathbf{A},\mathbf{B} \in \operatorname{Gr}(k, \infty)$, there must exist some $n$ sufficiently large so that both $\mathbf{A},\mathbf{B} \in \operatorname{Gr}(k, n)$ and in which case we define the distance between $\mathbf{A}$ and $\mathbf{B}$ in $ \operatorname{Gr}(k, \infty)$ to be $$d_{\operatorname{Gr}(k, \infty)}(\mathbf{A},\mathbf{B}) \coloneqq d_{\operatorname{Gr}(k, n)}(\mathbf{A},\mathbf{B}).$$ By Lemma \[lem:infty\], this value is independent of our choice of $n$ and is 