Tangent Cone is a cone?

Question

I first give two definitions.

Def1: A set $S$ is a cone if $x \in S, \lambda \geq 0 \implies \lambda x \in S$.

Def2: Let $S$ be any set (we may assume $\mathbb{R}^n$ with the usual Euclidean norm) and $\bar{x} \in S$. The tangent cone to $S$ at $\bar{x}$ is defined as

$$T_S(\bar{x}) = \overline{\{ h : \bar{x} + \lambda h \in S, \text{ for some } \lambda >0\}}$$

Prove that the tangent cone is indeed a cone.

Answer 1

Proof: Let $(h^n) \in T_S$, where $h^n \to h$; then we have $\bar{x} + \lambda^n h^n \in S$ for some $(\lambda^n) \geq 0$. Now consider $\gamma \geq 0$, then notice that $\bar{x} + (\lambda^n/\gamma) \gamma h^n \in S$, implying that $(\gamma h^n )\in T_S$. But by definition, $T_S$ is closed, and hence $\gamma h \in T$ proving the statement.

So it was a silly move in the end, and that convexity was not necessary.