From Paths to Connectedness
Connectedness rules out a split into two disjoint, nonempty open pieces. Path-connectedness gives a way to test such a proposed split: choose one point in each piece and follow a path between them. The path begins in one piece and ends in the other, so the preimages of the pieces would split the unit interval. Since the interval is connected, this cannot happen.
Throughout, a separation of a space \(X\) means a pair of disjoint, nonempty open subsets of \(X\) whose union is \(X\). A space is connected if it has no separation. When \(A\subseteq X\), “open in \(A\)” means open in the subspace topology: a set of the form \(A\cap U\), where \(U\) is open in \(X\).
Why the Unit Interval Is Connected
The proof uses the connectedness of \(I\). Here is the interval fact in the form needed: \(I\) cannot be written as a union of two disjoint, nonempty subsets that are open in \(I\). The proof relies on the least-upper-bound property of \(\mathbb{R}\).
Proof. Suppose, to obtain a contradiction, that \(I=P\cup Q\), where \(P\) and \(Q\) are disjoint, nonempty, and open in \(I\). Choose one point from each set. After exchanging the names \(P\) and \(Q\) if necessary, there are \(a\in P\) and \(b\in Q\) with \(a<b\). The set \(S=P\cap[a,b]\) is nonempty because \(a\in S\), and it is bounded above by \(b\). Let \(c=\sup S\). Then \(a\leq c\leq b\).
If \(c\in Q\), then \(c\ne a\), since \(a\in P\) and \(P\cap Q=\varnothing\). Because \(Q\) is open in \(I\), there is \(\delta>0\) such that \((c-\delta,c+\delta)\cap I\subseteq Q\). Choose \(r\) with \(0<r<\min(\delta,c-a)\). The definition of supremum gives some \(s\in S\) with \(c-r<s\leq c\). This \(s\) lies in \((c-\delta,c+\delta)\cap I\), so \(s\in Q\), but \(s\in S\subseteq P\). That contradicts disjointness. Thus \(c\notin Q\), and since \(P\cup Q=I\), we have \(c\in P\).
If \(c=b\), then \(c\in P\cap Q\), contrary to disjointness. Hence \(c<b\). Since \(P\) is open in \(I\) and \(c\in P\), there is \(\eta>0\) such that \((c-\eta,c+\eta)\cap I\subseteq P\). Choose \(t\) with \(c<t<\min(b,c+\eta)\). Then \(t\in P\cap[a,b]=S\), which contradicts that \(c\) is an upper bound for \(S\). Both possibilities for \(c\) have led to contradictions, so no such \(P,Q\) exist. Therefore \(I\) is connected. \(\square\)
The Path-Connectedness Theorem
Proof. Suppose \(X\) is path-connected and, seeking a contradiction, has a separation \(U,V\). Choose \(x\in U\) and \(y\in V\). By path-connectedness, there is a continuous map \(\gamma:I\to X\) with \(\gamma(0)=x\) and \(\gamma(1)=y\). Define \(P=\gamma^{-1}(U)\) and \(Q=\gamma^{-1}(V)\). Continuity of \(\gamma\) means that preimages of open sets are open, so \(P\) and \(Q\) are open in \(I\). They are disjoint because \(U\cap V=\varnothing\), and they cover \(I\) because \(U\cup V=X\) and every value of \(\gamma\) lies in \(X\). Also, \(0\in P\) since \(\gamma(0)=x\in U\), and \(1\in Q\) since \(\gamma(1)=y\in V\). Thus \(P,Q\) are a separation of \(I\), contradicting the proposition. Therefore \(X\) has no separation and is connected. If \(X=\varnothing\), it has no pair of nonempty subsets and is connected directly, so the conclusion includes that case. \(\square\)
The key step is taking preimages. The sets \(U,V\) live in the space \(X\), while connectedness of the interval concerns subsets of \(I\). Continuity transfers openness from \(X\) to \(I\), and the endpoint conditions ensure that both preimages are nonempty.
Worked Example: A Path in the Unit Interval
Take \(X=I\) with its usual topology, and choose \(a=\frac14\) and \(b=\frac34\). Define \(\gamma:I\to I\) by \(\gamma(t)=(1-t)a+tb\). Substituting the chosen endpoints gives \(\gamma(t)=(1-t)\frac14+t\frac34=\frac14+\frac12t\).
At the endpoints, \(\gamma(0)=\frac14+\frac12(0)=\frac14=a\), and \(\gamma(1)=\frac14+\frac12(1)=\frac34=b\). For \(0\leq t\leq1\), the formula gives \(\frac14\leq\gamma(t)\leq\frac34\), so \(\gamma(t)\in I\). The map is continuous because it is an affine function of \(t\). For instance, at \(t=\frac12\), \(\gamma(\frac12)=\frac14+\frac12\cdot\frac12=\frac12\), which lies between the endpoints. Thus any two points of \(I\) can be joined by the same affine construction, and the theorem implies that \(I\) is connected.
Check: for arbitrary \(a,b\in I\), \(\gamma(t)=(1-t)a+tb\) has \(\gamma(0)=a\) and \(\gamma(1)=b\). Since \(0\leq t\leq1\), both coefficients are nonnegative and sum to \(1\), so \(\gamma(t)\) lies between \(a\) and \(b\), hence in \(I\). This verifies the path construction for every pair, not only the selected one.
Worked Example: The Unit Square
Let \(X=[0,1]^2\) with the subspace topology from the Euclidean plane, and join \(p=(0,0)\) to \(q=(1,1)\). Define \(\gamma(t)=(t,t)\) for \(t\in I\). Its coordinates satisfy \(0\leq t\leq1\), so \(\gamma(t)\in[0,1]^2\) for every \(t\). The endpoint calculations are \(\gamma(0)=(0,0)=p\) and \(\gamma(1)=(1,1)=q\). At the midpoint, \(\gamma(\frac12)=(\frac12,\frac12)\).
For \(s,t\in I\), the Euclidean distance between the two path values is \(\|\gamma(s)-\gamma(t)\|=\|(s-t,s-t)\|=\sqrt{(s-t)^2+(s-t)^2}=\sqrt{2}\,|s-t|\). Given any \(\varepsilon>0\), choosing \(\delta=\varepsilon/\sqrt{2}\) makes \(|s-t|<\delta\) imply \(\|\gamma(s)-\gamma(t)\|<\varepsilon\); hence this path is continuous. More generally, for any \(p=(p_1,p_2)\) and \(q=(q_1,q_2)\) in the square, the formula \(\gamma(t)=(1-t)p+tq\) has each coordinate between the corresponding coordinates of \(p\) and \(q\), so it stays in the square and joins the endpoints. The same distance calculation, with \(\|q-p\|\) as a constant factor, proves continuity. Thus the square is path-connected and therefore connected.
Check: in the displayed path, the distance formula gives \(0\) when \(s=t\), since \(\sqrt{2}|t-t|=\sqrt{2}\cdot0=0\). This agrees with \(\gamma(t)=\gamma(t)\) and confirms the computed distance expression at equal parameters.
Worked Example: A Two-Point Indiscrete Space
Let \(X=\{a,b\}\) have the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). Define \(\gamma:I\to X\) by \(\gamma(t)=a\) for \(0\leq t<1\), and \(\gamma(1)=b\). Then \(\gamma(0)=a\) and \(\gamma(1)=b\). To check continuity, the only open sets in \(X\) are \(\varnothing\) and \(X\), whose preimages are \(\varnothing\) and \(I\), respectively. Both are open in \(I\), so \(\gamma\) is continuous. A path from \(b\) to \(a\) is obtained by defining \(\eta(t)=b\) for \(0\leq t<1\) and \(\eta(1)=a\); its endpoint values are \(b,a\), and its open-set preimages are again \(\varnothing\) and \(I\). Constant paths handle pairs with equal endpoints. Thus \(X\) is path-connected.
The only open subset of \(X\) other than \(\varnothing\) is \(X\) itself, so there cannot be two disjoint nonempty open subsets covering \(X\). Hence \(X\) is connected as well. This example also shows why paths are defined by continuity in the given topology: the displayed path changes value at its final parameter, but its preimages of all open sets in the indiscrete target are open.
Continuity Is Essential
A common mistake is to argue that the image of a connected interval must be connected without checking that the map is continuous. The correct argument uses continuity precisely to ensure that the preimages of open sets are open. A concrete discontinuous map shows what can go wrong.
Common pitfall. Define \(f:I\to\mathbb{R}\) by \(f(t)=0\) when \(0\leq t<\frac12\), and \(f(t)=1\) when \(\frac12\leq t\leq1\), where \(\mathbb{R}\) has its usual topology. Its image is \(\{0,1\}\), which is disconnected: the subsets \(\{0\}\) and \(\{1\}\) are disjoint, nonempty, open in the subspace \(\{0,1\}\), and cover it. But \(f\) is not continuous. The open interval \((-\frac12,\frac12)\) has preimage \([0,\frac12)\). That preimage is not open in \(I\): every relative neighborhood of \(\frac12\) contains points greater than or equal to \(\frac12\), while \(\frac12\notin[0,\frac12)\). So this example does not contradict the theorem; it demonstrates why continuity cannot be omitted.
The theorem gives a one-way implication: path-connectedness guarantees connectedness. Its proof needs neither a metric on \(X\) nor a separation axiom such as Hausdorffness. The only topological input about the path is continuity, together with its prescribed endpoints. When checking an application, verify the topology on \(X\), construct a path for every pair of points, and then invoke the theorem; a continuous bijection or a picture of a space alone does not establish path-connectedness.
Check Your Understanding
For each question, identify where continuity or the endpoint conditions enter the argument.
- In the proof of the theorem, why are \(\gamma^{-1}(U)\) and \(\gamma^{-1}(V)\) open in \(I\)?
- Why does \(\gamma(0)\in U\) imply that \(\gamma^{-1}(U)\) is nonempty?
- Write an affine path in \(I\) from \(\frac15\) to \(\frac45\), and verify its endpoints and range.
- Why does the step function in the common pitfall not contradict the theorem?
- What does the separation argument say if one of the proposed open pieces is empty?