If closure of $C$ is a subvectorspace then so is $C$

Question

Let $C$ be a convex subset of $\mathbb{R}^n$.

It is asked to prove that if the closure $\overline{C}$ is a subvectorspace of $\mathbb{R}^n$ then so is $C$.

Any ideas on where to start are welcome.

Thanks.

Answer 1

This is fairly typical of many results in convex analysis that are geometrically obvious but very tedious to prove.

Choose a basis $b_1,...,b_m$ of $ \overline{C}$ and any point $p \in \overline{C}$.

Note that $p = {1 \over m+1} ((p-(b_1+\cdots+b_m)) + (p+b_1)+ \cdots + (p+b_m))$.

It is straightforward to check that the matrix $\overline{B} = \begin{bmatrix}1 & 1 & \cdots & 1 \\ p-(b_1+\cdots+b_m) & p+b_1 & \cdots & p+ b_m\end{bmatrix}$ is injective and $\overline{B} \ \overline{\lambda} = \begin{bmatrix}1 \\ p \end{bmatrix}$, where $\bar{\lambda} = ({1 \over m+1} ,{1 \over m+1} , \cdots , {1 \over m+1})^T$. In particular, we have $\bar{\lambda} = ( \overline{B}^T \overline{B})^{-1} \overline{B}^T \begin{bmatrix}1 \\ p \end{bmatrix}$.

Note that $p$ is in the convex hull of $p-(b_1+\cdots+b_m) , p+b_1 , \cdots , p+ b_m$ (in fact, in the relative interior).

Now choose points $c_0(n),...,c_m(n) \in C$ such that the matrix $\overline{B}_n = \begin{bmatrix}1 & 1 & \cdots & 1 \\ c_0(n) & c_1(n) & \cdots & c_m(n) \end{bmatrix}$ converges to $\overline{B}$. Note that $\overline{\lambda}(n) = ( \overline{B}_n^T \overline{B}_n)^{-1} \overline{B_n}^T \begin{bmatrix}1 \\ p \end{bmatrix} \to \overline{\lambda}$, so for sufficiently large $n$ we see that $\overline{\lambda}(n) >0$ and so $p = \sum_k \overline{\lambda}_k(n) c_k(n)$ with $\sum_k \overline{\lambda}_k(n) = 1$ and so $p$ is in the convex hull of $c_0(n),...,c_m(n)$ and so is in $C$.

Hence $C = \overline{C}$.

Answer 2

Here's a hint for the special case where $n=2$ and $\overline C$ is the whole plane $\mathbb R^2$. My goal is to convince you that $C$ itself is the whole plane, so consider any point $p$ in the plane. Draw a triangle, with vertices $a,b,c$, having $p$ in its interior. Since $a,b,c\in\overline C$, there are points $a',b',c'\in C$ very close to $a,b,c$, respectively. If "very close" is close enough, then triangle $a'b'c'$ will be so close to triangle $abc$ that $p$ is in its interior. By convexity, since $C$ contains $a',b',c'$ it must contain $p$ as well.

I claim further that this proof generalizes from the plane to any dimension $d$ of $\overline C$. Instead of a triangle with its three vertices, you'll use a $d$-dimensional simplex with $d+1$ vertices. But the idea is unchanged.