Let $K \subset \mathbb{R}^n$ be a closed convex set. Given a point $u \in K$, the tangent cone of $K$ at $u$ is defined as (or characterized by)
$$
T\_K(u) := \mathrm{cl}(\left\{ t (v - u) \mid v \in K, t \geq 0 \right\}).
$$


My question is about the cases where $K$ is a closed convex **cone**.
The equation right after Eq. (2.1) of [this paper](https://arxiv.org/abs/1510.08029) says that, if $K$ is a closed convex cone, we have
$$
T\_K(u) = \left\{ v - tu \mid v \in K, t \geq 0 \right\}.
$$
I've tried to prove this identity, but I'm having trouble with the last step of the proof. So can anyone help me figure out the right way?


### What I've tried so far:


The inclusion
$\left\{ v - tu \mid v \in K, t \geq 0 \right\} \subset \left\{ t (v - u) \mid v \in K, t \geq 0 \right\} \subset T\_K(u)$ is easily proven with some algebra, so I will prove the opposite inclusion.
It is straightforward to see that
$$
\left\{ t (v - u) \mid v \in K, t \geq 0 \right\}
\subset
\left\{ v - tu \mid v \in K, t \geq 0 \right\}
$$
so taking the closures of both sides, we obtain
$$
T\_K(u) \subset \mathrm{cl}(\left\{ v - tu \mid v \in K, t \geq 0 \right\}).
$$
So it suffices to show that the set $\left\{ v - tu \mid v \in K, t \geq 0 \right\}$ is closed, but I can't figure out how I can do it.
Given an arbitrary sequence $\{v\_n - t\_n u\}\_{n=1}^\infty$ ($v\_n \in K, t\_n \geq 0$) that converges to some point, my goal is to prove the limit can be written in the form of $v - tu$ with $v \in K, t \geq 0$.
If both $\{v\_n\}$ and $\{t\_n\}$ converge, this is easy because $K$ is closed. But we can easily construct examples in which neither $\{v\_n\}$ nor $\{t\_n\}$ doesn't converge, e.g., $v\_n = nu$ and $t\_n = n$, so we cannot assume the convergence of $\{v\_n\}$ or $\{t\_n\}$.



Tangent cone of a closed convex cone
The equality
$$T\_K(u)=\{v-tu\colon v\in K,\, t\ge0\}$$
is false in general, because $T\_K(u)$ is closed by definition, whereas
$$S\_K(u):=K-\mathbb R\_+ u=\{v-tu\colon v\in K,\, t\ge0\}$$
can be not closed.


A counterexample is provided by [this answer](https://math.stackexchange.com/a/894816/96609). Indeed, let
$$K=\big\{(x,y,z)\in\mathbb R^3\colon z\ge\sqrt{x^2+y^2}\big\}$$
and $u=(-1,0,1)$. Then $K$ is a closed convex cone in $\mathbb R^3$ and $u\in K$.


Also, $w:=(0,1,0)\notin S\_K(u)$ -- otherwise, we would have $w+tu=(-t,1,t)\in K$ for some real $t$. However, letting
$$w\_t:=v\_t-tu$$
for real $t>0$ and $v\_t:=(-t,1+1/t,\sqrt{t^2+(1+1/t)^2})$, we have $v\_t\in K$ and hence $w\_t\in K-\mathbb R\_+ u=S\_K(u)$, whereas $w\_t\to w$ as $t\to\infty$. $\quad\Box$


