A cone generated by finitely many Euclidean vectors is closed

Take a representation $x={\sum }_{i=1}^N{\lambda }_i{g}_i$ with every ${\lambda }_i\ge 0$ and let $S=\{i:{\lambda }_i>0\}$. If the sublist indexed by $S$ is dependent, [F2] gives coefficients $c_i$, not all zero, with ${\sum }_{i\in S}{c}_i{g}_i=0$. Reverse all signs if needed so $J=\{i\in S:c_i>0\}$ is nonempty, and set $t={\min }_{i\in J}{\lambda }_i\mathrm{/}{c}_i>0$ using [F4]. Define ${\lambda }_i^{\mathrm{\prime }}={\lambda }_i-t{c}_i$ for $i\in S$ and ${\lambda }_i^{\mathrm{\prime }}=0$ for $i\mathrm{\notin }S$. For $i\in S$ with $c_i>0$, the minimum choice makes ${\lambda }_i^{\mathrm{\prime }}\ge 0$ and makes at least one such coefficient zero; when $c_i\le 0$, ${\lambda }_i^{\mathrm{\prime }}\ge {\lambda }_i>0$. Thus every ${\lambda }_i^{\mathrm{\prime }}\ge 0$, the positive support strictly shrinks, and ${\sum }_{i=1}^N{\lambda }_i^{\mathrm{\prime }}{g}_i=x$. Each reduction loses at least one index and adds none to the support, which starts with at most $N$ indices, so after at most $N$ reductions the support is empty or linearly independent; the empty support is independent as well.

Fix a nonempty independent sublist indexed by $S=\{i_1,\dots ,i_s\}$, let $A:{\mathbb{R}}^s\to {\mathbb{R}}^r$ have these columns, and let $G=A^{\mathsf{T}}A$. If $Gz=0$, then $0=z^{\mathsf{T}}Gz=\mathrm{\parallel }Az{\mathrm{\parallel }}_2^2$, so [F1] gives $Az=0$ and [F2] gives $z=0$. Thus $G$ is invertible by [F3]. Put $B=G^{-1}{A}^{\mathsf{T}}$ and $P=AB$, so $BA=I$. The cone $C_S$ generated by this sublist is exactly $\{x\in {\mathbb{R}}^r:Px=x,(Bx{)}_k\ge 0\text{ for }k<s\}$: if $x=Az$ with $z\ge 0$, then $Bx=z$ and $Px=x$; conversely those displayed conditions give $x=Px=A(Bx)$ with nonnegative coefficients. Each coordinate of $Bx$ and $Px-x$ is a finite linear form $\mathrm{\ell }(x)={\sum }_{j<r}{c}_j{x}_j$. For $u\in {\mathbb{R}}^r$ and $k<r$, [F7] gives $u_k^2\le {\sum }_{j<r}{u}_j^2$; by [F5] and [F8], $\mathrm{\mid }{u}_k{\mathrm{\mid }}^2=u_k^2\le \mathrm{\parallel }u{\mathrm{\parallel }}_2^2$, and both $\mathrm{\mid }{u}_k\mathrm{\mid }$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_2$ are nonnegative, so [F8] gives $\mathrm{\mid }{u}_k\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_2$. For any finite list $a_0,\dots ,a_{n-1}$, induction using the two-term triangle inequality in [F8] gives $\mathrm{\mid }{\sum }_{j<n}{a}_j\mathrm{\mid }\le {\sum }_{j<n}\mathrm{\mid }{a}_j\mathrm{\mid }$: the empty case is equality and the successor step is the two-term triangle inequality. Applying this, the absolute-value product law, multiplication by nonnegative coefficients, and finite-sum monotonicity in [F7,F8] gives $\mathrm{\mid }\mathrm{\ell }(x)-\mathrm{\ell }(y)\mathrm{\mid }\le {\sum }_{j<r}\mathrm{\mid }{c}_j\mathrm{\mid }\mathrm{\mid }{x}_j-y_j\mathrm{\mid }\le ({\sum }_{j<r}\mathrm{\mid }{c}_j\mathrm{\mid })\mathrm{\parallel }x-y{\mathrm{\parallel }}_2$. If an equation $\mathrm{\ell }(x)=0$ fails at $x_0$, then $C_{\mathrm{\ell }}:={\sum }_{j<r}\mathrm{\mid }{c}_j\mathrm{\mid }>0$: otherwise [F7,F8] force every $c_j=0$, so $\mathrm{\ell }$ is identically zero. For $\mathrm{\parallel }x-x_0{\mathrm{\parallel }}_2<\mathrm{\mid }\mathrm{\ell }(x_0)\mathrm{\mid }\mathrm{/}(2C_{\mathrm{\ell }})$, the estimate gives $\mathrm{\mid }\mathrm{\ell }(x)-\mathrm{\ell }(x_0)\mathrm{\mid }<\mathrm{\mid }\mathrm{\ell }(x_0)\mathrm{\mid }\mathrm{/}2$, hence $\mathrm{\ell }(x)\mathrm{\ne }0$. The same positivity of $C_{\mathrm{\ell }}$ holds when an inequality $\mathrm{\ell }(x)\ge 0$ fails; throughout the ball of radius $-\mathrm{\ell }(x_0)\mathrm{/}(2C_{\mathrm{\ell }})$, the estimate gives $\mathrm{\ell }(x)<\mathrm{\ell }(x_0)\mathrm{/}2<0$. Thus every point outside $C_S$ has a ball in its complement; by [F5,F6] that complement is open and $C_S$ is closed. For the empty sublist, $C_{\mathrm{\varnothing }}=\{0\}$: if $x\mathrm{\ne }0$, then $d_2(x,0)>0$ by [F5,F9], and every $y$ with $d_2(x,y)<d_2(x,0)\mathrm{/}2$ is nonzero, since otherwise the triangle inequality would give $d_2(x,0)<d_2(x,0)\mathrm{/}2$. Thus the complement of $\{0\}$ is open by [F6], so $C_{\mathrm{\varnothing }}$ is closed.

By step 1.1, every $x\in C$ belongs to $C_S$ for some independent subset $S$ of the $N$ generators; conversely every such $C_S$ is contained in $C$. There are only finitely many subsets $S$, and the empty subset is included. For any $x\mathrm{\notin }C$, each closed $C_S$ has an open complement containing $x$; choosing one positive ball radius for each of the finitely many such complements and taking their positive minimum gives a ball about $x$ disjoint from their union. Thus $C$ has open complement and is closed. For $x,y\in C$, choose coefficients ${\lambda }_i,{\mu }_i\ge 0$ with $x={\sum }_{i=1}^N{\lambda }_i{g}_i$ and $y={\sum }_{i=1}^N{\mu }_i{g}_i$. Then $\alpha x+\beta y={\sum }_{i=1}^N(\alpha {\lambda }_i+\beta {\mu }_i)g_i\in C$ for all $\alpha ,\beta \ge 0$, proving the stated closure under nonnegative linear combinations.