Matrices with rational trace: dense with empty interior

Idea. The trace sends a neighbourhood of any matrix onto an interval; an interval contains both rationals and irrationals.

Let $T=\{A:\operatorname{tr}A\in\mathbb{Q}\}$.

Step 1: images of balls. Fix $A$ and $r>0$. For $\|H\|_F<r$ and $s\in(-r/\sqrt n,r/\sqrt n)$ take $H_s=\frac s{\sqrt n}I$: $\|H_s\|_F=|s|<r$ and $\operatorname{tr}(A+H_s)=\operatorname{tr}A+s\sqrt n$. Therefore
$$\operatorname{tr}\big(B(A,r)\big)\supset\big(\operatorname{tr}A-r\sqrt n,\ \operatorname{tr}A+r\sqrt n\big),$$
a non-empty open interval.

Step 2: dense. The interval above contains a rational number $q$. Choose $s$ with $\operatorname{tr}A+s\sqrt n=q$; then $A+H_s\in T$ and $\|A+H_s-A\|_F=|s|<r$. So every ball meets $T$.

Step 3: empty interior. The same interval contains an irrational number $x$ (the irrationals are dense), and choosing $s$ with $\operatorname{tr}A+s\sqrt n=x$ gives $A+H_s\in B(A,r)\setminus T$. So no ball is contained in $T$.

Remark.
- $T=\bigcup_{q\in\mathbb{Q}}\{\operatorname{tr}=q\}$ is a *countable union of closed hyperplanes*, each nowhere dense. This is an example of a dense *meagre* set: the Baire theorem (next problem) forbids a countable union of nowhere dense closed sets from being the whole space, consistent with $T\neq M_n$.
- The set $T$ is neither open nor closed; its complement (irrational trace) is a dense $G_\delta$.