Computing Path Components in the Plane
To compute path components, we need two kinds of arguments. To show that two points belong to the same path component, we can construct a continuous path between them that stays inside the set. In many planar examples, a path made from finitely many line segments is enough. To show that two points belong to different path components, we need an obstruction: some feature of every possible path that makes the desired endpoints impossible.
All planar subsets below have the subspace topology inherited from \(\mathbb{R}^2\) with its usual Euclidean topology. A path in a set \(X\subseteq\mathbb{R}^2\) is a continuous map \(\gamma:[0,1]\to X\). Its endpoints are \(\gamma(0)\) and \(\gamma(1)\). A subset is path-connected if each pair of its points can be joined by a path in that subset; a path component is a maximal path-connected subset.
For vertices \(v_0,\ldots,v_n\), the segment from \(v_{i-1}\) to \(v_i\) consists of the points \((1-s)v_{i-1}+sv_i\), where \(0\le s\le1\). Traversing these segments in order gives a continuous path: each parametrized segment is continuous, and consecutive segments agree at their shared vertex. This simple calculation is the basis of the constructions that follow.
Joining Points Through Convex Pieces
Proof. Take any two points \(p,q\in C_1\cup\cdots\cup C_n\). Choose indices \(i,j\) with \(p\in C_i\) and \(q\in C_j\). If \(i=j\), convexity puts the straight segment from \(p\) to \(q\) inside \(C_i\), so that segment is a path in the union. If \(i<j\), choose \(r_k\in C_k\cap C_{k+1}\) for each \(k=i,\ldots,j-1\). The segment from \(p\) to \(r_i\) lies in \(C_i\); for each \(i<k<j\), the segment from \(r_{k-1}\) to \(r_k\) lies in \(C_k\); and the segment from \(r_{j-1}\) to \(q\) lies in \(C_j\). These segments form a finite piecewise-linear path from \(p\) to \(q\) in the union. If \(j<i\), apply the same construction in the reverse order, or reverse the path from \(q\) to \(p\). When \(n=1\), the first case already proves the claim. \(\square\)
Worked Example: A Union of Two Rectangles
Let \(U=(-2,0)\times(-1,1)\), \(V=(-1,2)\times(0,2)\), and \(X=U\cup V\). Both rectangles are convex, and \(U\cap V=(-1,0)\times(0,1)\) is nonempty. The proposition shows that \(X\) is path-connected. Here is an explicit path between \(p=(-3/2,0)\in U\) and \(q=(1,3/2)\in V\).
Use the intermediate vertex \(r=(-1/2,1/2)\), which belongs to \(U\cap V\). For \(0\le t\le1/2\), set \(\gamma(t)=p+2t(r-p)\). For \(1/2\le t\le1\), set \(\gamma(t)=r+(2t-1)(q-r)\). The formulas agree at \(t=1/2\), where both give \(r\).
On the first segment, the \(x\)-coordinate ranges from \(-3/2\) to \(-1/2\), strictly between \(-2\) and \(0\); the \(y\)-coordinate ranges from \(0\) to \(1/2\), strictly between \(-1\) and \(1\). Thus this segment is in \(U\). On the second, \(x\) ranges from \(-1/2\) to \(1\), strictly between \(-1\) and \(2\), and \(y\) ranges from \(1/2\) to \(3/2\), strictly between \(0\) and \(2\). Thus it is in \(V\).
Check: \(\gamma(0)=p+0(r-p)=p\), and \(\gamma(1)=r+(q-r)=q\). The joining point is also correct: \(\gamma(1/2)=p+(r-p)=r\) on the first formula and \(\gamma(1/2)=r+0(q-r)=r\) on the second.
Routing Around a Deleted Segment
A straight segment between two points in a set need not stay in the set. When it meets a deleted obstacle, a path may still exist by taking a different route. The next example shows how to make that route systematically.
Worked Example: The Plane with a Closed Segment Removed
Let \(S=[-1,1]\times\{0\}\) and \(X=\mathbb{R}^2\setminus S\). We show that any two points \(p,q\in X\) can be joined by a piecewise-linear path in \(X\).
First route any \(w=(a,b)\in X\) to \(z=(2,2)\). If \(b\ne0\), use the two segments from \((a,b)\) to \((2,b)\), then from \((2,b)\) to \((2,2)\). The first segment stays at height \(b\ne0\), so it misses \(S\). The second stays on the line \(x=2\), which misses \(S\) because every point of \(S\) has \(x\le1\). If \(b=0\), membership \(w\in X\) means \(a\notin[-1,1]\). Use the segments from \((a,0)\) to \((a,2)\), then from \((a,2)\) to \((2,2)\). The first stays on the line \(x=a\notin[-1,1]\), and the second stays at height \(2\); neither meets \(S\).
For arbitrary \(p,q\in X\), follow the route just described from \(p\) to \(z\), then follow the route from \(q\) to \(z\) in reverse. The resulting finite sequence of segments is a path from \(p\) to \(q\), and every segment avoids \(S\). This also covers points for which a segment has zero length: a constant segment at a point of \(X\) remains in \(X\).
For a numerical instance, take \(p=(-2,0)\) and \(q=(0,1)\). A path with successive vertices \((-2,0),(-2,2),(2,2),(2,1),(0,1)\) joins them. The first segment has \(x=-2\); the second has \(y=2\); the third has \(x=2\); and the fourth has \(y=1\). Each of these values rules out membership in \(S\).
Check: the first vertex is \(p\), and the last is \(q\). The only points of \(S\) have \(y=0\) and \(-1\le x\le1\); each of the four segments has either a fixed \(x\) outside that interval or a fixed nonzero \(y\). Therefore the path does not meet the deleted segment.
Using Coordinates to Prove That Paths Do Not Exist
For an impossibility proof, begin with an assumed path and use its coordinates. If \(\gamma(t)=(u(t),v(t))\) is continuous, then \(u\) and \(v\) are continuous real-valued functions, because they are the compositions of \(\gamma\) with the coordinate projections. The intermediate value theorem can then prevent a coordinate from changing sign without taking the value zero.
Worked Example: The Plane with a Line Removed
Let \(L=\{0\}\times\mathbb{R}\), the vertical axis, and \(X=\mathbb{R}^2\setminus L\). The two sets \(H_-=\{(x,y):x<0\}\) and \(H_+=\{(x,y):x>0\}\) cover \(X\). Each is convex, so any two points in either one can be joined by their straight segment, which stays in that half-plane.
No path in \(X\) joins a point of \(H_-\) to a point of \(H_+\). Suppose such a path \(\gamma(t)=(u(t),v(t))\) existed. Its first coordinate \(u\) is continuous, and its endpoint values have opposite signs. By the intermediate value theorem, \(u(t_0)=0\) for some \(t_0\in[0,1]\). Then \(\gamma(t_0)\in L\), contradicting \(\gamma([0,1])\subseteq X\).
Consequently the path components are exactly \(H_-\) and \(H_+\): each is path-connected, and no path can join one to the other. For example, \((-1,0)\) and \((2,0)\) cannot be joined in \(X\), because any continuous first coordinate running from \(-1\) to \(2\) must take the value \(0\).
Check: the segment from \((-1,0)\) to \((2,0)\) has first coordinate \(-1+3t\). It is zero at \(t=1/3\), so the segment meets \(L\), exactly as the general obstruction predicts.
Worked Example: The Coordinate Axes Without the Origin
Let \(X=(\mathbb{R}\times\{0\})\cup(\{0\}\times\mathbb{R})\setminus\{(0,0)\}\). We show that its path components are the four open rays: the positive and negative \(x\)-axis and the positive and negative \(y\)-axis.
Each ray is path-connected: for two points on the same ray, the straight segment between them stays on that ray. Now suppose \(\gamma(t)=(u(t),v(t))\) is any path in \(X\). At every \(t\), exactly one of \(u(t)\) and \(v(t)\) is zero. Define \(E=\{t\in[0,1]:v(t)=0\}\) and \(F=\{t\in[0,1]:u(t)=0\}\). Continuity of \(u\) and \(v\) makes \(E\) and \(F\) closed in \([0,1]\). They are disjoint and cover \([0,1]\), so each is also open relative to \([0,1]\), since its complement is the other closed set.
The interval \([0,1]\) is connected, so it cannot be the union of two disjoint nonempty open subsets. Therefore one of \(E,F\) is empty. If the path starts on the \(x\)-axis, then \(0\in E\), so \(F\) is empty and the path stays on the \(x\)-axis. Its first coordinate never vanishes; by the intermediate value theorem it cannot change sign. Thus it cannot join the positive \(x\)-axis to the negative \(x\)-axis. The same reasoning, with the coordinates exchanged, applies to paths starting on the \(y\)-axis. A path cannot move between the axes either, because that would require both \(E\) and \(F\) to be nonempty.
Hence no path joins points on different rays, while each individual ray is path-connected. These four rays are precisely the path components.
Check: \((1,0)\) and \((-1,0)\) lie on different rays. If a path joined them, it would have to stay on the \(x\)-axis, and its first coordinate would move continuously from \(1\) to \(-1\). The intermediate value theorem would force that coordinate to equal \(0\), placing the path at the deleted origin. This is impossible.
A Common Pitfall: The Straight Segment Is Not Always a Path in the Set
For \(X=\mathbb{R}^2\setminus\{(0,0)\}\), the straight segment from \((-1,0)\) to \((1,0)\) is not a path in \(X\): its parametrization is \(\gamma(t)=(-1+2t,0)\), and \(\gamma(1/2)=(0,0)\). The incorrect conclusion “the set is not path-connected because this segment fails” confuses one particular route with every possible route.
Instead use the two segments from \((-1,0)\) to \((0,1)\), then from \((0,1)\) to \((1,0)\). The first is \((-1+s,s)\), \(0\le s\le1\); it could equal the origin only if \(s=0\) and \(-1+s=0\), which cannot both hold. The second is \((s,1-s)\), \(0\le s\le1\); it could equal the origin only if \(s=0\) and \(1-s=0\), which also cannot both hold. Thus both segments avoid the origin. The correct test is always whether the constructed path stays in the set, not whether the direct segment does.
These examples illustrate a useful division of work: convexity and explicit intermediate vertices build paths, while continuous coordinates and the intermediate value theorem rule out paths. To identify path components completely, establish both sides—points within each proposed component can be joined, and points in different proposed components cannot.
Check Your Understanding
For each question, specify the subset topology and justify your answer by a path construction or an obstruction.
- Let \(X=\mathbb{R}^2\setminus(\{0\}\times\mathbb{R})\). Which path component contains \((-3,2)\), and why can it not contain \((4,2)\)?
- In the plane with the closed segment \([-1,1]\times\{0\}\) removed, construct a piecewise-linear path from \((3,0)\) to \((0,-2)\).
- In the coordinate axes with the origin removed, can a path join \((0,2)\) to \((0,-5)\)? Give the coordinate-based obstruction.
- For two intersecting convex subsets of \(\mathbb{R}^2\), explain how a point in their intersection can be used as an intermediate vertex.
- Why does the failure of the straight segment between two points to lie in a subset not, by itself, prove that the points are in different path components?