Convexity Gives a Direct Path
In Euclidean space, a straight segment is the simplest candidate for a path between two points. The key condition is that the entire segment—not just its endpoints—must remain in the set. Convexity is exactly the condition that guarantees this. It therefore supplies an explicit path formula and a direct proof of path-connectedness.
Throughout, \(\mathbb{R}^n\) has its Euclidean topology, and every subset \(C\subseteq\mathbb{R}^n\) has the subspace topology. The Euclidean distance between \(x,y\in\mathbb{R}^n\) is \(\|x-y\|\). The parameter interval \(I=[0,1]\) has the subspace topology inherited from \(\mathbb{R}\).
The parameter \(t\) moves from \(0\) to \(1\). At \(t=0\), the expression \((1-t)x+ty\) is \(x\); at \(t=1\), it is \(y\). For intermediate \(t\), it gives the point at the corresponding fraction of the segment from \(x\) to \(y\). Convexity says precisely that all these points stay in \(C\).
The Straight-Line Path
Proof. The endpoint computations are \(\gamma(0)=(1-0)x+0y=x\) and \(\gamma(1)=(1-1)x+1y=y\). For any \(s,t\in I\), subtract the two values and factor: $$ \gamma(t)-\gamma(s)=((1-t)-(1-s))x+(t-s)y=(t-s)(y-x). $$ Taking Euclidean norms gives $$ \|\gamma(t)-\gamma(s)\|=|t-s|\,\|y-x\|. $$ If \(x=y\), this distance is \(0\) for all \(s,t\), so \(\gamma\) is constant and continuous. If \(x\ne y\), then \(\|y-x\|>0\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon/\|y-x\|>0\). Whenever \(s,t\in I\) satisfy \(|t-s|<\delta\), the displayed equality gives \(\|\gamma(t)-\gamma(s)\|<\varepsilon\). Thus \(\gamma\) is continuous in the Euclidean metric. \(\square\)
For a path in \(C\), we also need its values to lie in \(C\). The continuity assertion above is about the map into \(\mathbb{R}^n\); when its values lie in \(C\), the same distance estimate proves continuity with the subspace metric on \(C\).
Proof. If \(C=\varnothing\), there are no pairs \(x,y\in C\), so the defining condition for path-connectedness holds vacuously. Now suppose \(C\ne\varnothing\), and let \(x,y\in C\) be arbitrary. Define \(\gamma:I\to\mathbb{R}^n\) by \(\gamma(t)=(1-t)x+ty\). Since \(C\) is convex, \(\gamma(t)\in C\) for every \(t\in I\). We can therefore regard \(\gamma\) as a map into \(C\). The lemma gives continuity and the endpoint equalities \(\gamma(0)=x\) and \(\gamma(1)=y\). Hence \(\gamma\) is a path in \(C\) from \(x\) to \(y\). Since the pair was arbitrary, \(C\) is path-connected. This proof includes the case \(x=y\): the resulting path is constant. \(\square\)
Worked Examples
Worked Example: A Rectangle in the Plane
Let \(C=[-1,4]\times[-2,3]\subseteq\mathbb{R}^2\), with its subspace topology. First verify that \(C\) is convex. Take arbitrary \((a,b),(c,d)\in C\) and \(t\in[0,1]\). The coordinates of their weighted point are \((1-t)a+tc\) and \((1-t)b+td\). Since \(-1\le a,c\le4\), and both weights \(1-t,t\) are nonnegative with sum \(1\), we have $$ -1=(1-t)(-1)+t(-1)\le(1-t)a+tc\le(1-t)4+t4=4. $$ Likewise, \(-2\le b,d\le3\) gives $$ -2=(1-t)(-2)+t(-2)\le(1-t)b+td\le(1-t)3+t3=3. $$ Thus the weighted point lies in \(C\), proving convexity.
Now take \(x=(1,2)\) and \(y=(4,-1)\), both of which belong to \(C\). The straight-line path is $$ \gamma(t)=(1-t)(1,2)+t(4,-1)=(1+3t,\,2-3t),\qquad 0\le t\le1. $$ At \(t=0\), \(\gamma(0)=(1,2)=x\); at \(t=1\), \(\gamma(1)=(4,-1)=y\). For \(0\le t\le1\), \(1\le1+3t\le4\) and \(-1\le2-3t\le2\), so \(\gamma(t)\in C\). At the midpoint, $$ \gamma(1/2)=(1+3/2,\,2-3/2)=(5/2,\,1/2), $$ which is also in \(C\). The lemma verifies continuity, and the formula and bounds verify that this is a path in the rectangle.
Worked Example: The Open Unit Disk
Let \(D=\{u\in\mathbb{R}^2:\|u\|<1\}\), with its subspace topology. This set is convex. Indeed, let \(u,v\in D\) and \(t\in[0,1]\). The triangle inequality and homogeneity of the Euclidean norm give $$ \|(1-t)u+tv\|\le(1-t)\|u\|+t\|v\|<(1-t)\cdot1+t\cdot1=1. $$ The strict inequality holds because both \(\|u\|<1\) and \(\|v\|<1\), and the nonnegative weights sum to \(1\). Thus \((1-t)u+tv\in D\), including when \(t\) is either endpoint.
For a specific path, take \(x=(0,0)\) and \(y=(3/5,0)\). Both belong to \(D\), since \(\|x\|=0<1\) and \(\|y\|=3/5<1\). The path is $$ \gamma(t)=(1-t)(0,0)+t(3/5,0)=(3t/5,0). $$ Its endpoints are \(\gamma(0)=(0,0)\) and \(\gamma(1)=(3/5,0)\). For every \(t\in[0,1]\), its norm is \(\|\gamma(t)\|=3t/5\le3/5<1\), so every value lies in the open disk. The inequality is strict at every parameter, including \(t=1\), as required for membership in \(D\). Continuity follows from the lemma.
Worked Example: An Affine Plane in Three Dimensions
Consider $$ P=\{(a,b,c)\in\mathbb{R}^3:a+2b-c=4\}, $$ with the subspace topology. This plane is convex. If \(u=(a,b,c)\) and \(v=(d,e,f)\) lie in \(P\), then \(a+2b-c=4\) and \(d+2e-f=4\). For \(t\in[0,1]\), the first coordinate plus twice the second coordinate minus the third coordinate of \((1-t)u+tv\) is $$ (1-t)(a+2b-c)+t(d+2e-f)=(1-t)4+t4=4. $$ So the weighted point still belongs to \(P\).
Choose \(x=(4,0,0)\) and \(y=(0,2,0)\). Substitution gives \(4+2\cdot0-0=4\) and \(0+2\cdot2-0=4\), so both points lie in \(P\). The path formula yields $$ \gamma(t)=(1-t)(4,0,0)+t(0,2,0)=(4-4t,\,2t,\,0). $$ At the endpoints this gives \(\gamma(0)=(4,0,0)\) and \(\gamma(1)=(0,2,0)\). At every parameter \(t\in[0,1]\), the plane equation evaluates to $$ (4-4t)+2(2t)-0=4-4t+4t=4. $$ Therefore the path remains in \(P\); the lemma gives its continuity. This example shows that a convex set need not be open or have full dimension in its ambient Euclidean space.
A Common Pitfall: The Segment Must Stay in the Set
The formula \((1-t)x+ty\) always gives a continuous map into \(\mathbb{R}^n\), but it need not give a map into an arbitrary subset containing \(x\) and \(y\). For example, consider the unit circle \(S^1=\{(a,b)\in\mathbb{R}^2:a^2+b^2=1\}\) with its subspace topology. Both \(x=(1,0)\) and \(y=(-1,0)\) lie in \(S^1\). The straight-line formula is $$ \gamma(t)=(1-t)(1,0)+t(-1,0)=(1-2t,0). $$ At \(t=1/2\), it gives \(\gamma(1/2)=(0,0)\), and \(0^2+0^2=0\ne1\). Thus this formula does not define a path in \(S^1\).
The mistake would be to infer that any set containing two endpoints contains the segment between them, or that the straight-line formula automatically stays in the set. The correction is to check convexity, or to check directly that every value of the proposed map belongs to the target set. Here the computation only shows that the straight segment is not contained in \(S^1\); by itself, it does not rule out a different path between those points.
The usefulness of convexity is that it turns a question about finding a path into a formula that can be checked directly. Once the endpoints are chosen, the path is specified, its continuity follows from the Euclidean distance estimate, and convexity keeps all its values in the set. This applies equally to bounded and unbounded sets, and to sets of any dimension.
Check Your Understanding
For each question, keep track of both requirements for a path: continuity and values in the specified space.
- Write the straight-line path from \((2,-1)\) to \((-1,3)\) in \(\mathbb{R}^2\), and compute its value at \(t=1/4\).
- Prove directly from the definition that every interval in \(\mathbb{R}\), whether open, closed or half-open, is convex.
- For \(C=\{(a,b)\in\mathbb{R}^2:a+b=1\}\), verify that the segment between any two points of \(C\) stays in \(C\).
- Give two points of the unit circle for which the straight-line segment leaves the circle, and identify a parameter value where it leaves.
- Explain why the formula \((1-t)x+ty\) gives a constant path when \(x=y\), and why this handles the equal-endpoint case in the theorem.