How to prove that a set is closed Consider the set $C$ of vectors in $\mathbb{R}^{N^2}$ which have symmetric rank-one matrix representation in $\mathbb{R}^{N \times N}$. In other words,
$$
C = \{ u \in \mathbb{R}^{N^2} | \exists v \in \mathbb{R}^{N}, u = vec(vv^T) \}
$$
where $vec(.)$ stacks the columns of a matrix into a vector.
Is the set $C$ closed? how can I prove that?
Thanks in advance!
 A: Excluding the zero vector makes $C$ not closed, as you can have a sequence of vectors $v_n\in \Bbb  R^N$ with $v_n\to \vec 0$, which would make a sequence $u_n = \operatorname{vec}(v_nv_n^T)$ in $C$ that converges to the zero vector.
Including the zero vector in $C$, on the other hand, changes the story.
You can express symmetry as a set of equations (a bunch of entries that must be equal to one another, or more usefully to us, a bunch of differences between pairs of entries that must be $0$). You can express rank-at-most-1 as a set of equations (all the $2\times 2$ minors must be $0$).
Each difference and each minor mentioned above is a continuous function $\Bbb R^{N^2}\to \Bbb R$. The set of elements in $\Bbb R^{N^2}$ where any one of them is equal to $0$ is closed. Thus their intersection is closed. But this intersection is $C$. So $C$ is closed.
A: First a remark: as $\mathbb R^{N^2}$ is isometric to the space of real matrices $M_N(\mathbb R)$, I will work in that later space.
Let $\{u_n\}$ be a sequence of $C$. By definition, it exists a sequence $\{v_n\}$ of vectors of $\mathbb R^N$ such that $u_n = v_n v_n^T$. If $\{u_n\}$ converges to $u \in M_N(\mathbb R)$, $\{u_n\}$ is bounded. As all norms are equivalent on a finite dimensional space, this implies that $\{v_n\}$ is also bounded. According to Bolzano–Weierstrass theorem, $\{v_n\}$ has a converging subsequence to let say $v \in \mathbb R^N$. As $v \mapsto vv^T$ is continuous, we get that $u = vv^T$ and finally that $C$ is indeed closed.
