Convex set with empty interior is nowhere dense?

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).

• 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? – bof Dec 30 '14 at 20:54

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 $C$.

• I see that the intuitive argument works, but I still have trouble operationalizing it rigorously. I will keep trying. Thank you for the answer! – triple_sec Dec 31 '14 at 16:03
• 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. – Dave L. Renfro Jan 8 '15 at 17:07
• That works too. – user203787 Jan 8 '15 at 18:00

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 $$C$$.