The complement of O(n) is path connected

Idea. The origin is outside $O(n)$, and almost every ray from the origin avoids $O(n)$ altogether. The few bad rays can be avoided by a small preliminary detour.

Step 0: when is a matrix orthogonal in terms of singular values? $B\in O(n)$ iff $B^TB=I$ iff all singular values of $B$ equal $1$.

Step 1: rays. For $t>0$ the singular values of $tA$ are $t\sigma_i(A)$. So $tA\in O(n)$ iff $\sigma_1(A)=\dots=\sigma_n(A)=1/t$. Consequently:
- if the singular values of $A$ are *not all equal*, then $tA\notin O(n)$ for all $t\in[0,1]$ (for $t=0$ we get $0\notin O(n)$);
- so $t\mapsto tA$ joins $A$ to $0$ inside $M_n(\mathbb{R})\setminus O(n)$.

Step 2: the bad case. Suppose all singular values of $A$ are equal to $c\ge0$. Then $A^TA=c^2I$. If $c=0$ then $A=0$. Otherwise $U=A/c\in O(n)$, so $A=cU$ with $c\ne1$ (since $A\notin O(n)$).
- If $0<c<1$: for $t\in[0,1]$, $tA=tcU$ has all singular values equal to $tc<1$, so $tA\notin O(n)$. The ray works again.
- If $c>1$: consider $A_s=U\operatorname{diag}(c+s,c,\dots,c)$, $s\in[0,1]$. Because $n\ge2$ and $U$ is orthogonal, the singular values of $A_s$ are $c+s,c,\dots,c$: for $s>0$ they are not all equal, and for $s=0$ they are all $c>1$. In both cases $A_s\notin O(n)$ (at least one singular value is $>1$... precisely, the *smallest* singular value is $c>1$ so none equals $1$). So $s\mapsto A_s$ is a path in the complement from $A$ to $A_1$, whose singular values are *not all equal*. Then apply Step 1 to $A_1$ to go to $0$.

Conclusion. Every point of the complement is joined to $0$ inside the complement, so it is path connected.

The case $n=1$. $O(1)=\{\pm1\}$ and $\mathbb{R}\setminus\{\pm1\}=(-\infty,-1)\cup(-1,1)\cup(1,\infty)$ has three components. For $n=1$ Step 2 fails because $\operatorname{diag}(c+s)$ again has all singular values equal.

Remark. The result expresses that $O(n)$ has codimension $\frac{n(n+1)}{2}\ge3$ in $M_n(\mathbb{R})$, which is enough to prevent it from separating space. The general principle: a closed submanifold of codimension $\ge2$ never disconnects a connected manifold.