Intereting Posts

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Let $V = \mathbb{R}^n$, $v_1, \dots, v_s \in V$ and let $\sigma = \text{Cone}(v_1, \dots, v_s) = \{r_1v_1 + \dots + r_sv_s \mid r_i \geq 0\}$ be the associated convex polyhedral cone in $V$.

Why is $\sigma$ closed in the euclidean topology?

My attempts:

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$1.$ Define $\varphi: \mathbb{R}^s \to V$ as the unique linear mapping satisfying $e_i \mapsto v_i$. Then we have $\sigma = \varphi(\sigma’)$, $\sigma’ = \text{Cone}(e_1, \dots, e_s)$. Unfortunately, linear maps do not send closed sets to closed sets. But what is with closed sets which are themselves convex polyhedric cones?

$2.$ Define $\psi: \text{span}(v_1, \dots, v_s) \to \mathbb{R}^s$ as the unique linear mapping satisfying $v_i \mapsto e_i$. Than $\sigma = \psi^{-1}(\sigma’)$. But $\psi$ isn’t well defined, so this naive try won’t work. A friend of mine suggested to define kind of a lexicographic order to make it well-defined, but so far I do not know if it works.

Hopefully there will be a clever solution.

Thanks!

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Here is another proof without induction on dimensions.

Let $A$ be the matrix with column vectors $v_1\dots v_s$.

**Claim:** The cone

$$

\sigma:=\{ x: \ Ay = x, \ y\ge 0\}

$$

is closed.

*Notation:*

For an index set $I\subset\{1\dots s\}$ define $\| y \|_I^2 := \sum_{i\in I} y_i^2$.

*Proof:*

Let $(x_k)$ in $\sigma$ be given, such that $x_k\to x$.

The set $\tau_k:=\{ y\ge 0: \ Ay = x_k\}$ is non-empty and closed (the mapping $f:y\mapsto Ay$ is continuous, hence $f^{-1}(\{x\})$ is closed). Hence for each $k$ there is $y_k$

satisfying

$$

\|y_k\| = \inf_{y\in \tau_k}\|y\|.

$$

If $(y_k)$ does contain a bounded subsequence, then we are done.

It remains to consider the case $\|y_k\|\to \infty$.

Denote $\tilde v_k:=\frac1{\|y_k\|}y_k$.

By compactness, it contains a converging subsequence. W.l.o.g.

let $(\tilde v_k)$ be converging to $v$ with $\|v\|=1$, $v\ge 0$.

Moreover, $Av =0$.

Denote $I_1:=\{j: v_j>0\}$, $I_2:=\{1\dots n\}\setminus I_1$.

Since $\|v\|=\|v\|_{I_1}$, it follows $\frac{\|y_k\|_{I_1}}{\|y_k\|} \to 1$.

Hence, the sequence $(v_k)$ defined by $v_k:=\frac1{\|y_k\|_{I_1}}y_k$ converges to $v$, too.

Then there is an index $M$ such that

$$

v_{k,i}\in \left[ \frac12 v_i, \frac54 v_i\right]

$$

for all $k>M$, $i\in I_1$.

In the following let $k>M$.

The above inclusion is equivalent to

$$

0\le y_{k,i} – \frac{\|y_k\|_{I_1}}2 v_i\le \frac34 \|y_k\|_{I_1} v_i \quad \forall i\in I_1.

$$

Define $z_k:= y_k – \frac{\|y_k\|_{I_1}}2 v$. Then it holds $z_k\ge 0$, $Az_k = x_k$, which implies $z_k\in \tau_k$.

Now, let us show that the norm of $z_k$ is strictly less than the norm of $y_k$, which would give a contradiction to the minimum norm property of $y_k$.

We have

$$

\begin{split}

\|z_k\|^2

& = \|z_k\|_{I_1}^2 + \|z_k\|_{I_2}^2 \le \left(\frac34\|y_k\|_{I_1}\cdot \|v\|_{I_1}\right)^2 + \|y_k\|_{I_2}^2\\

&= \frac9{16} \|y_k\|_{I_1}^2 +\|y_k\|_{I_2}^2 \\

&= \|y_k\|^2 – \frac 7{16}\|y_k\|_{I_1}^2 < \|y_k\|^2,

\end{split}

$$

which is a contradiction to the minimal norm property of $y_k$.

*Note:* The claim only holds for a finite set of vectors $v_1\dots v_s$. If $A$ is a linear mapping on an infinite-dimensional space, then

the proof fails: We cannot prove that $\frac1{\|y_k\|} y_k$ converges strongly,

and we cannot prove that $\frac1{\|y_k\|} y_k$ converges weakly to a non-zero element $v$.

I think this problem has some intrinsic intricacies that you cannot get away from, but a careful induction proof will keep the intricacies down as much as possible.

First, let $W \subset V$ be the unique linear subspace of maximal dimension such that $W \subset \sigma$. Let $q :\mathbb{R}^n \to \mathbb{R}^n / W$ be the quotient by $W$. One checks that $q(\sigma) = \text{Cone}(q(v_1),….,q(v_s))$ and that $\sigma = q^{-1}(q(\sigma))$. This allows one to reduce to the case that $\dim(W)=0$: if this is not true for $\sigma$ in $V$, it *is* true for $q(\sigma)$ in $V/W$, allowing you to conclude that $q(\sigma)$ is closed, and therefore $\sigma$ is closed by continuity of $q$.

We may therefore assume that $W$ is trivial. It follows that there exists a codimension 1 linear subspace $P$ and a half space $H$ bounded by $P$ such that $\sigma \subset H$ and such that $P \cap \sigma$ is a single point. Let $P’$ be a hyperplane parallel to $P$ that is contained in the interior of $H$. Let $A_i$ be the intersection of $P’$ with the ray $[0,+\infty) \cdot v_i$. Denote the convex hull of the points $A_i$ by

$$\text{Hull}(A_1,…,A_s) = \{t_1 A_1 + \cdots + t_s A_s \bigm| t_i \ge 0, t_1 + \cdots + t_s = 1\} \subset P’

$$

If we can show that $\text{Hull}(A_1,…,A_s)$ is closed then it is easy to see that $\sigma$ is closed, using that each point of $\sigma$ except for the origin is uniquely represented in the form $r A$ where $r \in (0,+\infty)$ and $A \in \text{Hull}(A_1,….,A_s)$.

In fact the set $\text{Hull}(A_1,…,A_s)$ is compact, so it is closed. Compactness is proved in two steps. First one does one single special case, namely the “standard $s-1$ simplex” which is

$$\Delta^{s-1} = \text{Hull}(e_1,…,e_s) = \{(t_1,…,t_s) \in \mathbb{R}^s \bigm| t_i \ge 0, t_1 + \cdots + t_s=1\}

$$

where $e_1,…,e_s$ are the standard unit vectors in $\mathbb{R}^s$. Then one uses that to prove compactness of $\text{Hull}(A_1,…,A_s)$ using continuity of the function $\Delta^{s-1} \mapsto \text{Hull}(A_1,…,A_s)$ defined by $(t_1,\ldots,t_s) \mapsto t_1 A_1 + \cdots + t_s A_s$.

Finally some words about compactness of $\Delta^{s-1}$. It is a bounded set, and it is closed because it is the intersection of $s$ closed half-spaces of the hyperplane $P’$.

**Added later**: Regarding the existence of the half-space $H$ bounded by the hyperplane $P$, here is a proof by induction on dimension. Start by picking an affine half-space that contains $\sigma$ (any proper convex set is contained in a half-space). Translate that half-space to minimize its distance to $\sigma$. We obtain a half-space $H_1$ bounded by the hyperplane $P_1$ such that $\sigma \subset H_1$ and such that $\sigma$ has distance zero from $P_1$. From this one can argue that the origin is contained in $\sigma \cap P_1$. If $\sigma \cap P_1$ is just the origin, we’re done. Otherwise, one can argue that $\sigma \cap P_1$ is a face of $\sigma$, meaning the cone of a proper subset of $v_1,…,v_s$, and therefore by induction on dimension $P_1$ contains a halfspace $H_2$ bounded by a hyperplane $P_2$ such that $H_2$ contains the face $\sigma \cap P_1$. Note that $H_2$ has codimension 2 in $V$. We can now rotate $H_1$ around $H_2$ by an arbitrarily small amount so that those of the points $v_1,…,v_s$ that were in the interior $H_1-P_1$ stay in the interior, and those which were in $P_1$ are now in the interior.

I think the simplest way is to invoke Caratheodory’s lemma, which says that every $v\in\sigma={\rm Cone}(v_1,\dots,v_s)$ lies in some ${\rm Cone}(v_{i_1},\dots,v_{i_t})$ with $v_{i_1},\dots,v_{i_t}$ linearly independent. Since a cone generated by linearly independent vectors can be transformed into $[0,\infty)^t\times\{0\}^{n-t}$ by a linear self-homeomorphism of $\mathbb{R}^n$, it is closed. Since there are only finitely many choices of $1\le i_1,\dots,i_t\le s$, $\sigma$ is the union of finitely many closed cones, hence closed.

The proof is similar to what appears in https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_theorem_(convex_hull)

Edit: This is not a desirable solution, as you can see in the comments.

I do have a solution now that uses the following lemma. I’m at the beginning of Fulton’s book about toric varieties, and there he states this fact:

If $\sigma$ is a convex polyhedric cone and $v \notin \sigma$, then there is a $u \in V^{*}$ with $u(\sigma) \subseteq \mathbb{R}_{\geq 0}$ and $(u, v) < 0$.

But as $u$ is continuous, there is a small open neighbourhood $W$ of $v$ with $u(W) \subseteq \mathbb{R}_{< 0}$. Then $W \subseteq V \setminus \sigma$, so $\sigma$ is closed.

Unfortunately, the lemma is not easy to prove, but can be found in the literature…

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