A characterization of closure of a certain class of sets in $\mathbb{R}^n$

Question

Consider a set $K\subset \mathbb{R}^n$ that is symmetric ($B = -B$) and verifies $aK\subset bK$ if $|a|<|b|$. Can I conclude that $\overline{K} = \cap_{a>1}aK$? If not, does the result hold assuming extra properties, like $K$ being convex?

Thoughts: when drawing it in the plane it seems pretty clear. I know this is not the best argument, but I don't think this is something that depends on dimension, so I think working in $\mathbb{R}^n$ would suffice. It seems a good idea to me to use that $x\in \overline{K} \iff d(x,K) = 0$, but I am not sure of how to write the proof.

I also am interested if the property holds in general normed vector spaces.

Answer 1

First off, notice that if

$x\in K$

Then

$\forall_{-1<a<1}[ax\in K]$

(By letting $b=1$ in your condition)

So if the set $K$ contains a point $p$, it must contain everything in the span of $p$, which has a length less than $p$. So any set $K$ satisfying this consists of symmetric rays around the origin, with (possibly) finite length. Note however that the rays need not include their endpoints, i.e. a valid set $K$ could be

$K= \{x \in \mathbb{R}^n|\ \lVert x\rVert < 1 \}$.

So we have some union of rays, possibly finite in length, possibly containing their endpoints, but not necessarily either.

This allows us to construct a counterexample, the claim that

$$ \bigcap_{a>1}aS = \overline{S} $$

Implies that every point in $\overline{S}$ is equal to $a \vec{v}$ for some $\vec{v}$ in $S$. I.e. every point in the closure is in the subspace spanned by one of the ray's that make up $S$. But consider the set in $\mathbb{R}^2$

$$ S_1=\{\vec{v}\in \mathbb{R}^2 : \lVert\vec{v}\rVert \leq 1 \mbox{ and } \theta({\vec{v})} = 1/n \}$$

Where $\theta(\vec{v})$ is the angle the point makes with the $x$-axis. And then define

$$ S = S_1 \cup -S_1 $$

You can check that $S$ satifies the conditions. But the closure of $S$ contains the ray

$$r = \{a(1,0) \in \mathbb{R}^2 : |a| \leq 1 \}$$

Which is an entirely new ray, and not contained in the intersection, contradicting our assumption that every point in the closure is in the subspace spanned by some ray.

Answer 2

An example in the plane.

If you don't restrict to $K$ convex, it seems false to me.

Consider a points $P = (x, y)$ where $x$ and $y$ are both nonzero. We can build the pair of rays $$ R(P) = \{ tP | |t| > 1 \} $$ starting at $\pm P$ and extending out to infinity, but not including the points $\pm P$.

Each such ray-pair satisfies your inclusion assumption, so the union of any number of them does so as well. So let's consider the set of points $$ U_n = (1, 1 - \frac{1}{n}) $$ and let $$ K = \cup_{n=1}^\infty R(U_n) $$ Then the point $(1, 1)$ is in $\bar{K}$, but not in the intersection you describe. (Indeed, each of the points $U_n$ is also in $\bar{K}$, but not in your intersection, unless you change $\alpha > 1$ to $\alpha \ge 1$.)

If you include convexity...then you could take $K = \cup_{y\in [0, 1)} R((1, y))$ and the point $(1,1)$ would still be in the closure but not in your specified intersection. So that condition seems not strong enough either.