# Tangent cone of a closed convex cone

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 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 $$\{t_n\}$$.

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. 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 $$t\to\infty$$. $$\quad\Box$$