The set of all invertible matrices is a dense open subset of the set of all $n imes n$ matrices. The set of all unitary matrices is a compact subset of all $n imes n$ matrices. These two sets are also groups under multiplication. They are called the general linear group $GL(n)$ and the unitary group $U(n)$, respectively.
Solution.
(1). $GL(n)$ is a dense subset of $M(n)$, the set of all $n imes n$ matrices. Indeed, by the Schur triangularisation, for each matrix $A$, there exists a unitary $U$ such that $$ex A=Usex{a{cccc} vLm_1&&*\ &vLm_1&\ &&ddots&\ &&&vLm_s ea},quad vLm_i=sex{a{ccc} lm_i&&*\ &ddots&\ &&lm_i ea},quad lm_1=0,quad lm_i eq 0, 2leq ileq s. eex$$ We may just replace the $lm_1=0$ by $ve>0$ to get an invertible matrix $B$ such that $sen{A-B}_2=ve^2$.
(2). $GL(n)$ is an open subset of $M(n)$. In fact, by continuity, $$ex det A_n=0,quad A_n o A a det A=0. eex$$
(3). $U(n)$ is a bounded, closed subset of $M(n)$.