Proving cone(S) = lin(S) ⇔ ∀ x ∈ S : − x ∈ cone(S \ {x})

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Let $S\subset \mathbb R^n$. Show: that $\text{cone}(S) = \text{lin}(S) \iff \forall x \in S : − x \in \text{cone}(S \setminus {x})$.

Proof. For $S=\emptyset$ holds: $cone(\emptyset)=lin(\emptyset)=\{0\} \iff \forall x\in \emptyset : -x\in \{0\}$

For $S\neq \emptyset$ consider:

"$\implies$" Let $x\in lin(S)$ be arbitrary, then also holds $-x\in lin(S)$ Therefore because of $lin(S)=cone(S)$, $x_1-x\in cone(S)$ must apply. Next applies $S\subset lin(S)=span(S)$

So $\forall x\in S$ : -$x\in $cone(S{x}).

"$\Leftarrow$" $\forall x\in S$ : -$x\in $cone(S{x})

Let $x\in S$ be arbitrary. Then exists $x_1,...,x_k\in S${x} and $\lambda = (\lambda_1,...,\lambda_k)\in R^k_{\geq 0}$, such that $-x\sum^k_{i=1} \lambda_ix_i \iff x=-\sum^k_{i=1} \lambda_ix_i$

If $x=x_i$ for i=1,..,k, then switch.

If no $x=x_i \implies $ complete.

Is this proof correct? Thanks to anyone for feedback!