A union of finitely many closed convex sets is not necessarily locally star-shaped?

Question

A subspace $X\subset\mathbb{R}^n$ is said to be star-shaped if there is a point $x_0\in X$ such that for each $x\in X$, the line segment from $x_0$ to $x$ lines in $X$. A subspace $X\subset\mathbb{R}^n$ is said to be locally star-shaped if every point of $X$ has a star-shaped neighbourhood in $X$. In this question, the original poster claims that

A union of finitely many closed convex sets is not guaranteed to be a locally star-shaped set.

Could anyone give an example of a union of finitely many closed convex sets which is not locally star-shaped?

Edit:

The definitions are taken from A. Hatcher's algebraic topology, page 38:

If we can prove that a union of finitely many closed convex sets is necessarily locally star-shaped, this question would not make much sense.

Answer 1

Let $F_i$ be the finitely many closed and convex sets. $\forall x\in \bigcup_i F_i$, if $x$ is not in some of the $F_{j_k}$, then for each $k$, since $F_{j_k}$ is closed, take an open ball $B_k$ centered on $x$, disjoint from $F_{j_k}$. Let $B$ be the intersection of these finitely many open balls. It's still an open ball centered on $x$, and disjoint from all $F_{j_k}$.

Now, $B$ is star-shaped around $x$, as well as each $F_i$ where $i\neq j_k$ (since $x\in F_i$, and $F_i$ is convex), and so $B\cap F_i$ is star-shaped around $x$. Then, $$B\cap(\bigcup_i F_i) = B\cap(\bigcup_{i\neq j_k} F_i) = \bigcup_{i\neq j_k} (B\cap F_i) $$ is a neighborhood of $x$ in $\bigcup_i F_i$, and it's a union of star-shaped sets around $x$, so it's star-shaped itself.

So $\bigcup_i F_i$ is locally star-shaped.

Thus the claim in your question is wrong.

Answer 2

The statement seems very much true to me. After a quick search in the pdf, Hatcher himself never seems to claim otherwise.

Note: Defining locally star-shaped as: "for all $x\in S$ there is a neighbourhood $U_x$ of $x$ in $\Bbb R^n$ such that $U_x$ is star-shaped and $U_x\subseteq S$" makes no sense, in my opinion. Since $\Bbb R^n$ is locally convex, it is apparent that said definition co-implies being a non-empty open subset of $\Bbb R^n$. So, I'll assume, as it looks, that the definition talks about relative neighbourhoods.

Let $C_1,\cdots, C_n$ be said closed convex subsets and $C=\bigcup_{i=1}^n C_i$. Suppose the ones that contain $x$ are $C_1,\cdots, C_m$ and the ones which do not contain it are $C_{m+1},\cdots, C_n$. If $S=\bigcup_{i=m+1}^nC_{i}\ne\emptyset$, by closedness $d(x,S)>d>0$. Thus, if we consider the relative neighbourhood $B(x,d/2)\cap C$, it holds $$B(x,d)\cap C\subseteq \bigcup_{i=1}^m C_i.$$

That set is star-shaped in $x$: consider $y\in B(x,d)\cap C$. Then, $y\in C_j$ for some $1\le j\le m$. Since $C_j$ is convex and $x,y\in C_j$ by hypothesis, the entire segment joining them is in $C_j$, hence in $C$.

Answer 3

A finite union of closed convex sets $X=X_1\cup...\cup X_n$ is always locally star-shaped. For $x_0\in X$ let $\alpha$ denote the set of $k=1,...,n$ such that $x_0 \notin X_k$ and pick for each such $k$ an open ball $B_k$ centered at $x_0$ and disjoint from $X_k$. Then $$X \cap \bigcap_{k\in \alpha} B_k$$ is a star-shaped neighborhood at $x_0$.

I think the following provides an example of the claim in the case of non-closed sets: It suffices to construct a (non-closed) convex set $C$ with a boundary point $x_0$ such that $C\cup \{x_0\}$ is not locally star-shaped at $x_0$.

You may do this in the following way: Let $C_0$ be the open cone:

$$ C_0 = \{ (x,y,z) : 0<z<1, x^2+y^2<z^2 \}$$ To this we now add some decorations (a collection of external line segments) on the cone: $$ C_1 = \{ (x,y,z) : z>0,y<x^2, x^2+y^2<1, x^2+y^2=z^2\}$$

You may check that $C=C_0\cup C_1$ is a convex set. But when adding the origin $x_0=(0,0,0)$ it is no longer star shaped at $x_0$. The reason being that there are line segments on the boundary arbitrarily close to $x_0$.