Paths Through a Common Point in Star-Shaped Sets

One Center, Many Paths

A straight segment between two points stays in a convex set. Star-shapedness asks for less: there is one fixed point from which a segment to every point of the set stays inside it. That single center is enough to construct a path between any two points, even when the segment joining them directly leaves the set.

We use $\mathbb{R}^n$ with its Euclidean topology and the subspace topology on each subset $A\subseteq\mathbb{R}^n$. The unit interval $I=[0,1]$ has its standard topology. A path in $A$ from $x$ to $y$ is a continuous map $\gamma:I\to A$ with $\gamma(0)=x$ and $\gamma(1)=y$. The set $A$ is path-connected if it is nonempty and every pair of its points can be joined by a path in $A$.

Definition: For $p,x\in\mathbb{R}^n$, the line segment from $p$ to $x$ is $[p,x]=\{(1-s)p+sx:0\le s\le1\}$. A star-shaped set is a nonempty set $A\subseteq\mathbb{R}^n$ for which there is a point $p\in A$ such that $[p,x]\subseteq A$ for every $x\in A$. Such a point $p$ is called a center of $A$.

The parameter $s=0$ gives $p$, and $s=1$ gives $x$. Thus the definition requires the entire segment, including both endpoints, to lie in $A$. It does not require segments between arbitrary pairs of points in $A$ to lie in $A$. A center need not be unique.

The Path Through the Center

Theorem: Every star-shaped set $A\subseteq\mathbb{R}^n$ is path-connected. More precisely, if $p$ is a center and $x,y\in A$, then the following map is a path in $A$ from $x$ to $y$: $$ \gamma(t)= \begin{cases} (1-2t)x+2tp, & 0\le t\le \frac12,\\ (2-2t)p+(2t-1)y, & \frac12\le t\le1. \end{cases} $$

Proof. Fix $x,y\in A$ and a center $p\in A$. We first check that the two formulas agree where their domains meet. At $t=\frac12$, the first formula is $(1-1)x+p=p$, and the second is $p+0y=p$. Thus $\gamma$ is well-defined. At the endpoints, $\gamma(0)=x$ and $\gamma(1)=y$.

For $0\le t\le\frac12$, put $s=2t$, so $0\le s\le1$. The first formula becomes $(1-s)x+sp$, a point on $[x,p]=[p,x]$. This segment lies in $A$ by the center property. For $\frac12\le t\le1$, put $s=2t-1$, so $0\le s\le1$. The second formula becomes $(1-s)p+sy$, a point on $[p,y]$, which also lies in $A$. Hence every value of $\gamma$ is in $A$.

Each formula is affine in $t$, so it is continuous on its respective closed half-interval. The formulas agree at $t=\frac12$, so the piecewise map is continuous on $I$: for any closed set $F\subseteq\mathbb{R}^n$, the preimage on each half-interval is closed in that half-interval, and therefore closed in $I$; the full preimage is the union of those two closed sets. This proves continuity as a map into $\mathbb{R}^n$. Since its values lie in $A$, it is also continuous as a map into $A$ with the subspace topology: every open subset of $A$ has the form $A\cap U$ for some open $U\subseteq\mathbb{R}^n$, and its preimage is $\gamma^{-1}(U)$. We have constructed the required path for arbitrary $x,y\in A$, so $A$ is path-connected. $\square$

The first half of the parameter interval travels from $x$ to $p$; the second travels from $p$ to $y$. The two speeds need not match at $p$. Continuity requires the pieces to meet at the same value, not to have the same rate of travel.

Worked Example: A Path in a Cross-Shaped Set

Let $A=\{(u,0):0\le u\le2\}\cup\{(0,v):0\le v\le1\}\subseteq\mathbb{R}^2$, with the subspace topology. Show that $A$ is star-shaped and construct a path from $x=(2,0)$ to $y=(0,1)$.

Take $p=(0,0)$, which belongs to both segments. For any point $(u,0)\in A$, its segment to $p$ consists of $((1-s)u,0)$, where $0\le s\le1$. Since $0\le(1-s)u\le2$, this segment stays in the horizontal arm. For any $(0,v)\in A$, the segment to $p$ consists of $(0,(1-s)v)$; the inequality $0\le(1-s)v\le1$ shows that it stays in the vertical arm. Thus $p$ is a center.

The theorem gives $$ \gamma(t)= \begin{cases} (2-4t,0), & 0\le t\le \frac12,\\ (0,2t-1), & \frac12\le t\le1. \end{cases} $$ For the first half, $0\le t\le\frac12$ implies $0\le2-4t\le2$, so $\gamma(t)$ lies in the horizontal arm. For the second half, $\frac12\le t\le1$ implies $0\le2t-1\le1$, so $\gamma(t)$ lies in the vertical arm. At $t=\frac12$, both formulas give $(0,0)$. Also, $\gamma(0)=(2,0)=x$ and $\gamma(1)=(0,1)=y$. The path is continuous by the theorem’s piecewise-continuity argument.

Check: at $t=\frac14$, the first formula gives $(2-4\cdot\frac14,0)=(1,0)$, in the horizontal arm. At $t=\frac34$, the second gives $(0,2\cdot\frac34-1)=(0,\frac12)$, in the vertical arm. These values confirm the direction of travel and the membership checks on both pieces.

Star-Shaped Does Not Mean Convex

A set is convex if $[x,y]\subseteq A$ for every $x,y\in A$. Every nonempty convex set is star-shaped: choose any $p\in A$; for each $x\in A$, convexity gives $[p,x]\subseteq A$. The nonempty condition matters. The empty set satisfies the universal condition in the definition of convexity vacuously, but it has no point that can serve as a center under our definition of star-shapedness.

Worked Example: A Nonconvex Star-Shaped Set

In $\mathbb{R}^2$, let $$ A=\{(u,v):u^2+v^2\le1\}\ \cup\ \{(s,0):0\le s\le2\}, $$ with the subspace topology. Show that $A$ is star-shaped but not convex.

The point $p=(0,0)$ belongs to $A$. If $(u,v)$ is in the closed unit disk, then each point of its segment to $p$ has the form $((1-s)u,(1-s)v)$, with $0\le s\le1$. Its squared norm is $$ ((1-s)u)^2+((1-s)v)^2=(1-s)^2(u^2+v^2)\le u^2+v^2\le1. $$ So the segment stays in the disk. If $(s_0,0)$ is on the added horizontal segment, each point on its segment to $p$ is $((1-r)s_0,0)$, with $0\le r\le1$. Because $0\le s_0\le2$, we have $0\le(1-r)s_0\le2$; this segment stays in the added part of $A$. Thus $p$ is a center.

To test convexity, take $x=(2,0)\in A$ and $y=(0,1)\in A$. Their midpoint is $$ \frac{x+y}{2}=\left(\frac{2+0}{2},\frac{0+1}{2}\right)=(1,\tfrac12). $$ Its squared norm is $1^2+(\frac12)^2=\frac54>1$, so it is not in the disk. It is not on the added horizontal segment either, because its second coordinate is $\frac12\ne0$. Hence the midpoint is not in $A$, and $A$ is not convex.

Check: the path through the center from $x$ to $y$ is the first path formula in the theorem with $p=(0,0)$. Its first half lies on the added horizontal segment, and its second half lies on the vertical radius of the unit disk. In particular, the path stays in $A$, despite the midpoint of the direct segment lying outside $A$.

Testing a Proposed Center

To verify that a point $p$ is a center, it is not enough that $p\in A$, nor that a few selected segments stay in $A$. The defining condition quantifies over every $x\in A$: the whole segment $[p,x]$ must be contained in $A$. Conversely, to rule out star-shapedness, one must rule out every possible center, not just a convenient candidate.

Worked Example: The Unit Circle Has No Star Center

Let $S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}$, with the subspace topology. We show directly that no point of $S^1$ is a center.

Suppose $p=(a,b)\in S^1$. Then $q=(-a,-b)$ is also in $S^1$, since $(-a)^2+(-b)^2=a^2+b^2=1$. The midpoint of $p$ and $q$ is $$ \frac{p+q}{2}=\left(\frac{a-a}{2},\frac{b-b}{2}\right)=(0,0). $$ But $0^2+0^2=0\ne1$, so $(0,0)\notin S^1$. Therefore $[p,q]$ is not contained in $S^1$. This argument works for every proposed $p\in S^1$, so no point can be a star center.

Check: for the particular choice $p=(1,0)$, its opposite is $q=(-1,0)$; the segment contains $(0,0)$, which is not on the circle. The general midpoint calculation shows that this obstruction is not limited to that one choice.

A Common Mistake: Using the Direct Segment

Star-shapedness guarantees segments from the center, not segments between arbitrary points. In the nonconvex example, it would be incorrect to use the direct interpolation $\eta(t)=(1-t)x+ty$ and claim that it stays in $A$. For $x=(2,0)$ and $y=(0,1)$, this gives $\eta(t)=(2-2t,t)$. At $t=\frac12$, $$ \eta\left(\frac12\right)=\left(2-2\cdot\frac12,\frac12\right)=(1,\tfrac12). $$ As computed above, this point is neither in the unit disk, since its squared norm is $\frac54>1$, nor on the added horizontal segment, since its second coordinate is nonzero. Thus the direct segment leaves $A$. The correction is to travel in two stages: from $x$ to the center, then from the center to $y$.

The construction is useful whenever a set has a common radial point but is not convex. It avoids having to test every segment between pairs of points: the center condition alone supplies the two segments needed for each path. When checking the resulting formula, verify all three parts separately: the endpoint values, agreement at the joining parameter, and membership of every point on each piece.

Check Your Understanding

Use the definition of a center and the explicit path construction to answer the following.

For the cross-shaped set in the first worked example, verify the center condition for a point $(0,v)$ with $0\le v\le1$ by writing its segment to $(0,0)$ as a parametrized set.
If $x=p$, simplify the first half of the theorem’s path and check that the construction still has the required endpoints and lies in $A$.
Explain why the empty set is convex under the universal segment condition but is not star-shaped under the definition given here.
For the nonconvex set in the second worked example, compute the direct interpolation at $t=\frac14$. Is the resulting point in the set?
Give a reason that testing only one proposed point cannot prove that the unit circle has no star center.