1
$\begingroup$

Suppose $C\subseteq\mathbb R^n$ is a convex set and $C^o=\varnothing$. Is it necessarily true that $(\overline C)^o=\varnothing$? In general, is this true if $\mathbb R^n$ is replaced by a topological vector space $X$? Or can a counterexample be found?

I know that if $C^o\neq\varnothing$, then $C^o=(\overline C)^o$, so my question is whether this result can be generalized to the case when $C^o=\varnothing$.


Update: It's not true in general topological vector spaces. Indeed, if $X$ is a topological vector space and $Y$ is a proper dense subspace (such examples can be constructed), then $Y$ is convex and has empty interior, but $(\overline Y)^o=X^o=X$ is clearly not empty. Yet, I'm still not sure if the result holds for finite-dimensional vector spaces (in which every proper subspace is closed and thus no proper subspace is dense, so the preceding counterexample doesn't work).

$\endgroup$
  • 1
    $\begingroup$ Seems obvious that a convex subset of $\mathbb R^n$ with empty interior is contained in a hyperplane, which is a nowhere dense closed set. What am I missing? $\endgroup$ – bof Dec 30 '14 at 20:54
1
$\begingroup$

This is true in $\mathbb{R}^n$. Say $n = 3$. If $\overline{C}$ has non empty interior that $C$ is dense in some ball hence it contains the vertices of a tetrahedron (just take a tetrahedron inside the ball and move its vertices slightly to land on a point in $C$). But now by convexity, $C$ must contain the whole solid tetrahedron which contradicts the fact that $C$ has empty interior. In higher dimensions, you can start with a generalized cube and move its vertices slightly to enter $C$.

$\endgroup$
  • $\begingroup$ I see that the intuitive argument works, but I still have trouble operationalizing it rigorously. I will keep trying. Thank you for the answer! $\endgroup$ – triple_sec Dec 31 '14 at 16:03
  • $\begingroup$ Maybe I'm missing something, but it seems to me that you don't need to resort to "move its vertices slightly". If $C$ is dense in some ball, then $C$ must contain $4$ non-coplanar points (because no subset of a plane can be dense in a ball). Use these points as the vertices of a tetrahedron. $\endgroup$ – Dave L. Renfro Jan 8 '15 at 17:07
  • $\begingroup$ That works too. $\endgroup$ – user203787 Jan 8 '15 at 18:00
1
$\begingroup$

Here is a rigorous—and, accordingly, a bit lengthy—operationalization of the intuition offered by the answer by @OohAah.

$\textbf{Proposition}\phantom{---}$ Let $C\subseteq \mathbb R^n$ ($n\in\mathbb N$) be a convex set. If $C$ has no interior, then nor does $\overline C$.

$\textit{Proof}\phantom{---}$ Suppose that $(\overline C)^o\neq\varnothing$. I will show that $C^o$ is not empty, either. Let $\mathbf x_0\in (\overline C)^o$. Then, there exists some $\varepsilon>0$ such that $\mathbf x_0\in B(2\varepsilon,\mathbf x_0)\subseteq\overline C$, where $B(2\varepsilon,\mathbf x_0)$ denotes the open ball of radius $2\varepsilon$ around $\mathbf x_0$ with respect to the Euclidean norm. Let $\{\mathbf e_i\}_{i=1}^n$ denote the standard basis of $\mathbb R^n$ and define $\mathbf x_i\equiv \mathbf x_0+\varepsilon \mathbf e_i$ for each $i\in\{1,\ldots,n\}$. Clearly, $$\{\mathbf x_i\}_{i=0}^n\subseteq B(2\varepsilon,\mathbf x_0)\subseteq \overline C.$$ That is, $\{\mathbf x_i\}_{i=0}^n$ is included in the closure of $C$, so that, for each $i\in\{0,1,\ldots,n\}$, there exists some $\mathbf y_i$ such that

  • $\mathbf y_i\in C$; and
  • $\mathbf y_i\in B((2\sqrt n)^{-1}\varepsilon,\mathbf x_i)$.

For each $i\in\{1,\ldots,n\}$ define $\mathbf b_i\equiv \mathbf y_i-\mathbf y_0$. I claim that the vectors $\{\mathbf b_i\}_{i=1}^n$ are linearly independent. Indeed, suppose that $\lambda_1,\ldots,\lambda_n\in\mathbb R$ satisfy $$\sum_{i=1}^n\lambda_i\mathbf b_i=0.$$ Suppose, for the sake of contradiction, that not all of $\{\lambda_i\}_{i=1}^n$ are zero. Then, $$\sum_{i=1}^n|\lambda_i|>0\tag{1}.$$ In addition, \begin{align*} 0=&\,\sum_{i=1}^n\lambda_i\mathbf b_i=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf y_0)=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_i-\mathbf x_0+\mathbf x_0-\mathbf y_0)\\=&\,\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\varepsilon\mathbf e_i+\mathbf x_0-\mathbf y_0), \end{align*} or \begin{align*} -\varepsilon\sum_{i=1}^n\lambda_i\mathbf e_i=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_0-\mathbf y_0). \end{align*} Clearly, the Euclidean norm of the left-hand side is $\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}$, so the following chain of inequalities holds true: \begin{align*} &\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}=\left\|\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_0-\mathbf y_0)\right\|\leq\sum_{i=1}^n|\lambda_i|\left(\|\mathbf y_i-\mathbf x_i\|+\|\mathbf x_0-\mathbf y_0\|\right)\\ \underset{\text{see (1)}}{<}&\,\sum_{i=1}^n|\lambda_i|\left(\frac{\varepsilon}{2\sqrt{n}}+\frac{\varepsilon}{2\sqrt{n}}\right)=\sum_{i=1}^n|\lambda_i|\frac{\varepsilon}{\sqrt{n}}\leq\sqrt{\sum_{i=1}^n|\lambda_i|^2}\sqrt{\sum_{i=1}^n\frac{\varepsilon^2}{n}}=\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}, \end{align*} where I used the Cauchy–Schwarz inequality. This is a contradiction, which implies that $\lambda_1=\ldots=\lambda_n=0$. Hence, the vectors $\{\mathbf b_i\}_{i=1}^n$ are linearly independent, and since $\dim\mathbb R^n=n$, it follows also that they constitute a basis.


If $\mathbf z\in\mathbb R^n$, there exists a unique $(\mu_i)_{i=1}^n\in\mathbb R^n$ such that $\mathbf z=\sum_{i=1}^n\mu_i\mathbf b_i$. Define $\|\mathbf z\|_b\equiv\sum_{i=1}^n|\mu_i|$. Then, $\|\cdot\|_b$ is a norm and since $\mathbb R^n$ is finite-dimensional, it must be equivalent to the Euclidean norm. Therefore, there exists some $\xi_b>0$ such that $\|\cdot\|_b\leq \xi_b\|\cdot\|$.


Let $$D\equiv\left\{\sum_{i=0}^n\alpha_i\mathbf y_i\,\Bigg|\,\alpha_i\geq0\text{ for all $i\in\{0,1,\ldots,n\}$ and }\sum_{i=0}^n\alpha_i=1\right\}.$$ Since $\{\mathbf y_i\}_{i=0}^n\subseteq C$ and $C$ is convex, it follows that $D\subseteq C$. Define $$\mathbf w\equiv\sum_{i=0}^n\frac{1}{n+1}\mathbf y_i.$$ Clearly, $\mathbf w\in D$ and $$\mathbf w-\mathbf y_0=\sum_{i=1}^{n}\frac{1}{n+1}(\mathbf y_i-\mathbf y_0)=\sum_{i=1}^{n}\frac{1}{n+1}\mathbf b_i.$$


Let $$\delta\equiv\frac{1}{n(n+1)\xi_b}>0.$$ I will show that $B(\delta,\mathbf w)\subseteq D$. To this end, pick any $\mathbf z\in B(\delta,\mathbf w)$, so that $\|\mathbf z-\mathbf w\|<\delta.$ Since $\{\mathbf b_i\}_{i=1}^n$ is a basis, there exists a unique $(\mu_i)_{i=1}^n\in\mathbb R^n$ such that $\mathbf z-\mathbf y_0=\sum_{i=1}^n\mu_i\mathbf b_i$. Then, $$\mathbf z-\mathbf w=(\mathbf z-\mathbf y_0)-(\mathbf w-\mathbf y_0)=\sum_{i=1}^n\left(\mu_i-\frac{1}{n+1}\right)\mathbf b_i.$$ Consequently, for any $i\in\{1,\ldots,n\}$, it follows that \begin{align*} \left|\mu_i-\frac{1}{n+1}\right|\leq\sum_{j=1}^n\left|\mu_j-\frac{1}{n+1}\right|=\|\mathbf z-\mathbf w\|_b\leq \xi_b\|\mathbf z-\mathbf w\|<\xi_b\delta=\frac{1}{n(n+1)}. \end{align*} Hence, $$0\leq\frac{(n-1)}{n(n+1)}=\frac{1}{n+1}-\frac{1}{n(n+1)}<\mu_i<\frac{1}{n+1}+\frac{1}{n(n+1)}=\frac{1}{n},$$ for each $i\in\{1,\ldots,n\}$ so that $\sum_{i=1}^n\mu_i<1$. It follows that \begin{align*} \mathbf z=\mathbf y_0+\sum_{i=1}^n\mu_i\mathbf b_i=\mathbf y_0+\sum_{i=1}^n\mu_i(\mathbf y_i-\mathbf y_0)=\left(1-\sum_{i=1}^n\mu_i\right)\mathbf y_0+\sum_{i=1}^n\mu_i\mathbf y_i, \end{align*} so $\mathbf z\in D$.


Conclusion:

  • $\mathbf w\in B(\delta,\mathbf w)\subseteq D\subseteq C$; and
  • $B(\delta,\mathbf w)$ is a non-empty open set; so that
  • $\mathbf w\in C^o$.

In particular, $C^o$ is not empty. $\blacksquare$


Intuitively, the points $\{\mathbf x_i\}_{i=0}^n$ in $\overline C$ span an $n$-dimensional simplex, whose vertices are perturbed slightly so that the new vertices do not lie in the same hyperplane and are all in $C$. The convex hull of these new vertices gives rise to a “distorted simplex” fully contained in $C$. Then, the centroid of this distorted simplex can be surrounded by a small ball still included in the distorted simplex, and hence in $C$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.