Explicit Contractions of Convex and Star-Shaped Sets

Linear Motion Toward a Point

To show that a space is contractible, it is useful to give the homotopy explicitly. In a subset of Euclidean space, the natural candidate is to move each point along the straight segment from that point to a chosen point. The main questions are whether every intermediate point stays in the subset and whether the resulting map is continuous.

Throughout, \(\mathbb{R}^n\) has its standard topology and Euclidean vector operations, \(I=[0,1]\) has its subspace topology from \(\mathbb{R}\), and every subset \(A\subseteq\mathbb{R}^n\) has the subspace topology. A homotopy from \(f:A\to A\) to \(g:A\to A\) is a continuous map \(H:A\times I\to A\) such that \(H(x,0)=f(x)\) and \(H(x,1)=g(x)\) for every \(x\in A\). A contraction of \(A\) to \(p\in A\) is a homotopy from the identity map \(\operatorname{id}_A\) to the constant map with value \(p\).

Definition: A subset \(A\subseteq\mathbb{R}^n\) is convex if \((1-t)x+ty\in A\) for every \(x,y\in A\) and every \(t\in I\). It is star-shaped with center \(p\in A\) if \((1-t)x+tp\in A\) for every \(x\in A\) and every \(t\in I\). Equivalently, in the star-shaped condition the entire line segment from each \(x\in A\) to \(p\) lies in \(A\).

Convexity controls the segment between every pair of points. Star-shapedness asks for this only when one endpoint is the specified center. Thus every nonempty convex set is star-shaped about each of its points, but a star-shaped set need not be convex.

The Explicit Formula and Its Continuity

Given \(p\in A\), the linear interpolation from \(x\) to \(p\) is

H(x,t)=(1-t)x+tp,\qquad x\in A,\quad t\in I.

At the start and end of the motion, direct substitution gives \(H(x,0)=(1-0)x+0p=x\) and \(H(x,1)=(1-1)x+1p=p\). In addition, this particular formula fixes the center at every time: \(H(p,t)=(1-t)p+tp=p\). This last property belongs to the displayed formula; it is not part of the definition of an arbitrary contraction.

For continuity, define \(F:\mathbb{R}^n\times I\to\mathbb{R}^n\) by \(F(x,t)=(1-t)x+tp\). Each coordinate of \(F\) is formed from coordinate projections and the continuous real operations of multiplication and addition, so \(F\) is continuous. To check that the restricted map is continuous as a map into \(A\), let \(A\cap V\) be an open subset of \(A\), where \(V\) is open in \(\mathbb{R}^n\). The preimage under the restriction is

(F|_{A\times I})^{-1}(A\cap V) =F^{-1}(V)\cap(A\times I).

Here \(F^{-1}(V)\) is open in \(\mathbb{R}^n\times I\), so its intersection with \(A\times I\) is open in the subspace \(A\times I\). This proves continuity of the restricted map whenever its values lie in \(A\). The remaining task is to verify that containment.

Convex Sets Have an Explicit Contraction

Theorem: Every nonempty convex subset \(A\subseteq\mathbb{R}^n\) is contractible. In fact, for any chosen \(p\in A\), the formula \(H(x,t)=(1-t)x+tp\) is a contraction of \(A\) to \(p\).

Proof. Assume \(A\) is nonempty and convex, and choose \(p\in A\). For any \(x\in A\) and \(t\in I\), both \(x\) and \(p\) belong to \(A\); convexity therefore gives \((1-t)x+tp\in A\). Thus the formula defines a map \(H:A\times I\to A\). The continuity argument above proves that \(H\) is continuous. Substitution gives \(H(x,0)=x\) and \(H(x,1)=p\) for every \(x\in A\), so \(H\) is a contraction. \(\square\)

Worked Example: Contracting an Open Disk

Let \(A=\{(u,v)\in\mathbb{R}^2:u^2+v^2<4\}\), with its subspace topology, and choose \(p=(1/2,0)\). The set is convex: if \(x,y\in A\) and \(t\in I\), the triangle inequality gives \(\|(1-t)x+ty\|\le(1-t)\|x\|+t\|y\|<2\), since both \(\|x\|\) and \(\|y\|\) are strictly less than \(2\). Thus the theorem applies, with the explicit contraction \(H((u,v),t)=((1-t)u+t/2,(1-t)v)\).

For a numerical check, take \(x=(1,1)\) and \(t=1/4\). Then \(H(x,t)=((3/4)(1)+(1/4)(1/2),(3/4)(1))=(7/8,3/4)\). Its squared norm is \((7/8)^2+(3/4)^2=49/64+36/64=85/64\), and \(85/64<4\) because \(85<256\). So this intermediate point is in the open disk. At \(t=0\), the formula returns \((1,1)\); at \(t=1\), it returns \((1/2,0)\), which is in the disk because its squared norm is \(1/4<4\).

Check: the strict inequality in the definition matters here: the boundary circle, where \(u^2+v^2=4\), is not part of \(A\). The general containment argument used the strict bound \(\|x\|,\|y\|<2\), so it establishes containment in the open disk, not just in its closure.

Star-Shaped Sets Need Only One Family of Segments

Theorem: If \(A\subseteq\mathbb{R}^n\) is star-shaped with center \(p\), then \(H(x,t)=(1-t)x+tp\) is a contraction of \(A\) to \(p\).

Proof. Assume \(A\) is star-shaped with center \(p\in A\). For every \(x\in A\) and \(t\in I\), the definition of star-shapedness gives \((1-t)x+tp\in A\), so the formula maps \(A\times I\) into \(A\). The restriction is continuous by the preimage argument above. The endpoint calculations are \(H(x,0)=x\) and \(H(x,1)=p\). These are exactly the conditions for a contraction to \(p\). \(\square\)

Worked Example: A Nonconvex Star-Shaped Cross

In \(\mathbb{R}^2\), let \(A=([-2,2]\times[-1/2,1/2])\cup([-1/2,1/2]\times[-2,2])\), with the subspace topology, and take \(p=(0,0)\). If \(x=(u,v)\) lies in the first rectangle, then \((1-t)x=((1-t)u,(1-t)v)\) still lies there: \(0\le1-t\le1\), so each coordinate’s absolute value can only decrease. If \(x\) lies in the second rectangle, the same coordinate bounds show that \((1-t)x\) stays in that rectangle. Hence every segment from a point of \(A\) to \(p\) stays in \(A\).

The contraction is \(H((u,v),t)=((1-t)u,(1-t)v)\). For example, \(x=(2,1/4)\) lies in the first rectangle. At \(t=1/2\), \(H(x,t)=(1,1/8)\), and \(1\in[-2,2]\) while \(1/8\in[-1/2,1/2]\), so the result lies in that rectangle. At \(t=0\) the formula gives \((2,1/4)\), and at \(t=1\) it gives \((0,0)\).

This set is not convex. The points \((2,0)\) and \((0,2)\) both belong to \(A\), but their midpoint is \((1,1)\). That midpoint is in neither rectangle: its second coordinate has absolute value \(1>1/2\), excluding the first rectangle, and its first coordinate has absolute value \(1>1/2\), excluding the second. So convexity is not necessary for this construction.

Worked Example: Contracting a Line Segment

Let \(A=\{(u,v)\in\mathbb{R}^2:u\ge0,\ v\ge0,\ u+v=1\}\), with the subspace topology, and choose \(p=(1,0)\). This is a convex set: for two points whose coordinates are nonnegative and sum to \(1\), each coordinate of their interpolation is nonnegative, and the sum of the interpolated coordinates is \((1-t)\cdot1+t\cdot1=1\).

The contraction formula is \(H((u,v),t)=((1-t)u+t,(1-t)v)\). For the point \(x=(1/4,3/4)\) at \(t=1/2\), this gives \(((1/2)(1/4)+(1/2),(1/2)(3/4))=(5/8,3/8)\). Both coordinates are nonnegative and their sum is \(5/8+3/8=1\), so the result lies in \(A\). More generally, for any \((u,v)\in A\), both coordinates in the formula are nonnegative and their sum is \((1-t)(u+v)+t=(1-t)+t=1\). At \(t=0\) the point is unchanged, and at \(t=1\) the result is \(p=(1,0)\in A\).

This example also shows why the theorem does not require \(A\) to have interior in \(\mathbb{R}^2\). The segment is one-dimensional, but the formula and the subspace-topology continuity argument still apply.

A Common Pitfall: The Point Must Be a Star Center

For a convex set, any chosen point of the set can serve as the endpoint of the contraction. For a star-shaped set, the definition guarantees this only for a star center. In the cross above, an incorrect choice would be \(q=(2,0)\): using the same interpolation from \((0,2)\) to \(q\) at \(t=1/2\) gives \((1,1)\). The coordinate checks above show \((1,1)\notin A\), so this formula does not even define a map \(A\times I\to A\). The correction is to use the center \(p=(0,0)\); interpolation from \((0,2)\) to \(p\) at \(t=1/2\) gives \((0,1)\), which lies in the vertical rectangle.

Not every subset has a star center. For instance, in \(S^1=\{x\in\mathbb{R}^2:\|x\|=1\}\), with the subspace topology, any proposed center \(p\in S^1\) has its antipode \(-p\in S^1\). The midpoint of the segment between them is \((p+(-p))/2=(0,0)\), whose norm is \(0\), not \(1\). That segment is not contained in \(S^1\), so \(p\) is not a star center. This only shows that the linear star-center construction does not apply; it does not by itself settle whether some other kind of contraction exists.

The practical value of these results is that the homotopy is available as a formula, not merely as an existence claim. Once the chosen point and containment condition are identified, the endpoint equations and continuity can be checked directly. The key distinction is between convexity, which permits interpolation between any two points, and star-shapedness, which permits interpolation only toward a suitable center.

Check Your Understanding

For each question, distinguish the segment-containment condition from the continuity and endpoint checks.

State the definition of a convex subset of \(\mathbb{R}^n\), and explain why every nonempty convex set is star-shaped about each of its points.
For a star-shaped set with center \(p\), write the linear contraction and verify its values at \(t=0\) and \(t=1\).
In the cross-shaped set, why does interpolation from \((0,2)\) to \((2,0)\) fail at the midpoint, while interpolation from \((0,2)\) to the origin stays in the set?
Why does proving continuity of the ambient interpolation map require an intersection with \(A\times I\) when checking preimages for the restricted map?
Does the definition of an arbitrary contraction require that its chosen endpoint be fixed throughout the homotopy? What additional property does the linear formula have?