The cone of rank-one matrices

1. Description. If $A=xy^T$ then every column is a multiple of $x$, so $\operatorname{rank}A\le1$. Conversely if $\operatorname{rank}A=1$, pick a nonzero column space generator $x$; each column of $A$ is $y_jx$ for a scalar $y_j$, so $A=xy^T$ (and $A=0=0\cdot0^T$ if the rank is $0$).

Closedness. $\operatorname{rank}A\le1$ iff all $2\times2$ minors $a_{ij}a_{kl}-a_{il}a_{kj}$ vanish. These are polynomials, so $\mathcal{R}_1$ is the common zero set of continuous functions and is closed. It is stable under $A\mapsto tA$.

(Alternative: a limit of $x_jy_j^T$ need not obviously be of this form if $x_j,y_j$ blow up; the minor characterisation avoids this issue.)

2. Connectedness of rank exactly $1$. The map $\Phi:(\mathbb{R}^n\setminus\{0\})^2\to M_n(\mathbb{R})$, $(x,y)\mapsto xy^T$ is continuous, and its image is exactly the rank-one matrices (by part 1, a nonzero $xy^T$ requires both $x,y\ne0$). For $n\ge2$, $\mathbb{R}^n\setminus\{0\}$ is path connected (it is homotopy equivalent to $S^{n-1}$, which is connected), hence so is the product, hence so is its continuous image.

Remark. For $n=1$, "rank exactly $1$" is $\mathbb{R}\setminus\{0\}$, which is disconnected. In general $\{\operatorname{rank}=r\}$ is connected for $1\le r\le n-1$ and splits into two components ($\det>0$ and $\det<0$) for $r=n$.