The rank metric

Idea. Rank takes only integer values, so a distance defined by the rank has a gap: distinct points are at distance at least $1$.

Step 1: $d$ is a metric.
- $d(A,B)=\operatorname{rank}(A-B)=0$ iff $A-B=0$ iff $A=B$.
- Symmetry: $\operatorname{rank}(-X)=\operatorname{rank}X$.
- Triangle inequality: $A-C=(A-B)+(B-C)$ and the column space of a sum is contained in the sum of the column spaces, so $\operatorname{rank}(A-C)\le\operatorname{rank}(A-B)+\operatorname{rank}(B-C)$.

Step 2: discrete. If $A\ne B$ then $d(A,B)\ge1$. Hence the open ball of radius $\frac12$ around $A$ is $\{A\}$: every singleton is open, which is the discrete topology.

Step 3: identity to the norm topology. From a discrete space every map is continuous. So $\mathrm{id}:(M_n,d)\to(M_n,\|\cdot\|)$ is continuous.

Step 4: the inverse. The sequence $A_k=\frac1kE_{11}$ tends to $0$ in the operator norm ($\|A_k\|=\frac1k$). But $d(A_k,0)=\operatorname{rank}(A_k)=1$ does not tend to $0$, so $A_k\not\to0$ in $(M_n,d)$. Thus $\mathrm{id}:(M_n,\|\cdot\|)\to(M_n,d)$ is not continuous.

Remark. The rank is only *lower* semi-continuous for the norm topology: the set $\{\operatorname{rank}\ge r\}$ is open. This is why rank "can only jump down" in the limit, as in $A_k\to0$. The discrete topology is strictly finer than the norm topology, which is why only one of the two identity maps is continuous.