Paths Depend on Both Endpoints and Topology
A path is a continuous map from the interval \(I=[0,1]\). Its initial and final values are prescribed, so constructing a path requires both a formula and a check that the formula stays in the space. For familiar spaces, straight-line interpolation often supplies the formula. For a discrete space, in contrast, the topology can rule out every path between distinct points.
Unless specified otherwise, \(\mathbb{R}\) has its standard topology, \(I\) has the subspace topology inherited from \(\mathbb{R}\), and subsets of \(\mathbb{R}\) have the subspace topology. The space \(\mathbb{R}^n\) has its usual Euclidean topology. The circle \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\) has the subspace topology from \(\mathbb{R}^2\). A discrete space is a set \(X\) with topology \(\mathcal{P}(X)\), so every subset is open.
When \(x=y\), the constant map \(\gamma(t)=x\) is a path from \(x\) to \(x\). When \(x\ne y\), a constant map cannot satisfy both endpoint conditions. These endpoint constraints matter in each example below.
Intervals and Euclidean Spaces
Proof. Suppose \(J\) is nonempty and choose any \(x,y\in J\). Define \(\gamma:I\to\mathbb{R}\) by \(\gamma(t)=(1-t)x+ty\). This is continuous because it is an affine function of \(t\). For every \(t\in I\), the number \(\gamma(t)\) lies between \(x\) and \(y\): if \(x\le y\), then \(x\le (1-t)x+ty\le y\), and if \(y\le x\), then \(y\le (1-t)x+ty\le x\). By the defining property of an interval, every number between \(x\) and \(y\) belongs to \(J\). Thus \(\gamma\) takes values in \(J\), and its restriction is continuous as a map to the subspace \(J\). Finally, \(\gamma(0)=x\) and \(\gamma(1)=y\). Since \(x,y\) were arbitrary, \(J\) is path-connected. If \(J=\varnothing\), the defining condition is vacuous. \(\square\)
Worked Example: Joining Points in a Half-Open Interval
Consider \(J=[-2,3)\) and join \(x=-2\) to \(y=2\). The straight-line formula gives \(\gamma(t)=(1-t)(-2)+2t=-2+4t\).
At the endpoints, \(\gamma(0)=-2+4(0)=-2\) and \(\gamma(1)=-2+4(1)=2\). For \(0\le t\le1\), multiplication by \(4\) gives \(0\le4t\le4\), so \(-2\le-2+4t\le2<3\). Hence every value is in \([-2,3)\). For example, \(\gamma(1/4)=-2+4(1/4)=-1\) and \(\gamma(3/4)=-2+4(3/4)=1\), both in the interval. The formula is continuous, so it is a path in \(J\).
Check: substituting \(t=0\) and \(t=1\) recovers \(-2\) and \(2\), respectively. The excluded endpoint \(3\) causes no problem: the entire path has values at most \(2\).
Proof. Let \(x,y\in\mathbb{R}^n\) be arbitrary and define \(\gamma(t)=(1-t)x+ty\), with vector addition and scalar multiplication taken coordinatewise. Each coordinate of \(\gamma(t)\) is an affine, hence continuous, real-valued function of \(t\); therefore \(\gamma:I\to\mathbb{R}^n\) is continuous. Its endpoints are \(\gamma(0)=x\) and \(\gamma(1)=y\), and its values lie in \(\mathbb{R}^n\) because that space contains every vector with \(n\) real coordinates. Thus every pair of points can be joined by a path. \(\square\)
Worked Example: A Segment in the Plane
In \(\mathbb{R}^2\), join \(x=(1,-2)\) to \(y=(5,4)\). Coordinatewise interpolation gives \(\gamma(t)=((1-t)1+t5,(1-t)(-2)+t4)=(1+4t,-2+6t)\).
At \(t=0\), \(\gamma(0)=(1+4(0),-2+6(0))=(1,-2)\). At \(t=1\), \(\gamma(1)=(1+4(1),-2+6(1))=(5,4)\). At \(t=1/2\), \(\gamma(1/2)=(1+4(1/2),-2+6(1/2))=(3,1)\). Every coordinate is an affine function, so the path is continuous, and each value is a point of \(\mathbb{R}^2\).
Check: the displacement from the initial point to the final point is \((5-1,4-(-2))=(4,6)\). The formula satisfies \(\gamma(t)=(1,-2)+t(4,6)\), confirming that it starts at the stated point and moves by the full displacement when \(t=1\).
Paths Around the Circle
A straight segment between two points of \(S^1\) need not stay on \(S^1\). Instead, describe the points using angles and move continuously through the angle. We use the standard parametrization \((\cos\theta,\sin\theta)\) of the unit circle: every point of \(S^1\) has this form for some real \(\theta\).
Proof. Choose arbitrary \(p,q\in S^1\). By the circle parametrization, there are real numbers \(\alpha,\beta\) such that \(p=(\cos\alpha,\sin\alpha)\) and \(q=(\cos\beta,\sin\beta)\). Define \(\gamma(t)=(\cos((1-t)\alpha+t\beta),\sin((1-t)\alpha+t\beta))\). The coordinate functions are compositions of continuous functions, so \(\gamma\) is continuous as a map into \(\mathbb{R}^2\). For each \(t\in I\), the identity \(\cos^2\theta+\sin^2\theta=1\), applied to \(\theta=(1-t)\alpha+t\beta\), gives \(\gamma(t)\in S^1\). Thus it is also continuous as a map into the subspace \(S^1\). Substitution gives \(\gamma(0)=(\cos\alpha,\sin\alpha)=p\) and \(\gamma(1)=(\cos\beta,\sin\beta)=q\). Therefore \(p\) and \(q\) are joined by a path. \(\square\)
Worked Example: A Quarter-Circle Path
Join \((1,0)\) to \((0,-1)\) on \(S^1\). Take \(\gamma(t)=(\cos(\pi t/2),-\sin(\pi t/2))\). This is the angle formula for an angle moving from \(0\) to \(-\pi/2\), since \(\cos(-\theta)=\cos\theta\) and \(\sin(-\theta)=-\sin\theta\).
At \(t=0\), \(\gamma(0)=(\cos 0,-\sin 0)=(1,0)\). At \(t=1/2\), \(\gamma(1/2)=(\cos(\pi/4),-\sin(\pi/4))=(\sqrt{2}/2,-\sqrt{2}/2)\). At \(t=1\), \(\gamma(1)=(\cos(\pi/2),-\sin(\pi/2))=(0,-1)\), since \(\cos(\pi/2)=0\) and \(\sin(\pi/2)=1\).
For every \(t\), the squared distance of \(\gamma(t)\) from the origin is \(\cos^2(\pi t/2)+(-\sin(\pi t/2))^2=\cos^2(\pi t/2)+\sin^2(\pi t/2)=1\). So the path stays on \(S^1\). The endpoint substitutions verify that it has the required direction as well as the correct trace.
Discrete Spaces: Only Constant Paths
Proof. If \(X\) is empty, it is path-connected vacuously. If \(X=\{x\}\), the constant map \(\gamma(t)=x\) joins its only pair of points. Conversely, suppose \(X\) contains distinct points \(x\) and \(y\), and assume for contradiction that a path \(\gamma:I\to X\) satisfies \(\gamma(0)=x\) and \(\gamma(1)=y\). In a discrete space, \(\{x\}\) and \(X\setminus\{x\}\) are both open. Their preimages under the continuous map \(\gamma\) are therefore open in \(I\). They are disjoint, their union is \(I\), and both are nonempty: \(0\in\gamma^{-1}(\{x\})\), while \(1\in\gamma^{-1}(X\setminus\{x\})\). This separates \(I\) into two nonempty disjoint open sets, contradicting the standard connectedness of the interval \(I\). Thus no path joins distinct points of a discrete space. If \(X\) has at least two points, it is not path-connected, proving the equivalence. \(\square\)
Worked Example: A Two-Point Discrete Space
Let \(X=\{a,b\}\) have the discrete topology. Suppose a path \(\gamma:I\to X\) joined \(a\) to \(b\). The set \(\gamma^{-1}(\{a\})\) would be open because \(\{a\}\) is open and \(\gamma\) is continuous. Its complement \(\gamma^{-1}(\{b\})\) would also be open. The first contains \(0\), and the second contains \(1\); together they would separate \(I\), which is impossible because \(I\) is connected. So no such path exists. The constant maps at \(a\) and at \(b\) are paths from each point to itself, but they do not connect the two distinct points.
Check: this space has two points, so it fails the “at most one point” condition in Theorem 4; the endpoint argument confirms that it is not path-connected.
A Common Endpoint Mistake
A path formula can look plausible while its endpoint calculation is wrong. For the quarter-circle formula above, a mistaken calculation might claim \(\gamma(1)=(0,1)\) by treating \(\sin(\pi/2)\) as \(-1\). The correct trigonometric value is \(\sin(\pi/2)=1\), so \(\gamma(1)=(\cos(\pi/2),-\sin(\pi/2))=(0,-1)\). In general, substitute \(t=0\) and \(t=1\) into the actual formula, and evaluate any special function values before concluding which endpoints the path joins.
The four settings illustrate two distinct checks. For intervals and Euclidean spaces, interpolation works because the whole segment stays in the space. For the circle, the angle formula stays on the circle by the identity \(\cos^2\theta+\sin^2\theta=1\). For a discrete space, continuity and connectedness of \(I\) prohibit a path from moving between distinct points. A path-connectedness claim is therefore about more than whether two points can be joined by a curve in a picture: the entire map must be continuous and take values in the given topological space.
Check Your Understanding
For each question, identify the space topology and check both the endpoint conditions and whether the proposed path stays in the space.
- Write an explicit path from \(3\) to \(-1\) in the interval \((-2,4]\), and verify its endpoints.
- Give a path in \(\mathbb{R}^3\) from \((0,1,2)\) to \((2,-1,0)\) using linear interpolation.
- Why does the straight segment between \((1,0)\) and \((0,1)\) fail to lie entirely in \(S^1\), and what angle formula gives a path on \(S^1\) instead?
- Which discrete spaces are path-connected: the empty set, a singleton, and a three-point set?
- Explain why a continuous map from \(I\) into a discrete space cannot take two distinct values.