Why These Errors Matter
A path argument can fail even when its formulas look plausible. The most common problems are mismatched endpoints, an incorrect change of parameter during concatenation, and an unnoticed departure from the subspace in which the path is supposed to lie. A different error is to treat connectedness as if it guaranteed paths. These are distinct issues: checking a path’s formula does not establish that its image lies in a specified subset, and knowing that a set is connected does not establish that it is path-connected.
Throughout, \(I=[0,1]\) has its usual subspace topology from \(\mathbb{R}\). A subset \(A\) of a space \(X\) has the subspace topology. A path in \(X\) means a continuous map from \(I\) to \(X\); its endpoints are the images of 0 and 1. We will use the standard result that path-connected spaces are connected, proved in the earlier material on why path-connected spaces are connected.
Endpoint Checks and Concatenation
To concatenate paths \(\alpha,\beta:I\to X\), the endpoint of the first path must equal the starting point of the second: \(\alpha(1)=\beta(0)\). When this condition holds, the concatenation follows \(\alpha\) on the first half of the parameter interval and \(\beta\) on the second half.
Proof. On \([0,\frac12]\), the map \(t\mapsto\alpha(2t)\) is continuous as a composition of continuous maps. On \([\frac12,1]\), the map \(t\mapsto\beta(2t-1)\) is also continuous. At the shared parameter value \(t=\frac12\), the first formula gives \(\alpha(1)\), and the second gives \(\beta(0)\). These values agree by hypothesis. The two intervals are closed subsets of \(I\) and cover \(I\), so the pasting lemma for continuous functions on a finite closed cover gives continuity on all of \(I\). Finally, substitution gives \((\alpha*\beta)(0)=\alpha(0)\) and \((\alpha*\beta)(1)=\beta(1)\). \(\square\)
The endpoint hypothesis is essential: it ensures the two pieces give the same value at the join. The reversal of a path is another useful operation. If \(\alpha\) goes from \(x\) to \(y\), define \(\overline{\alpha}(t)=\alpha(1-t)\). This is continuous, starts at \(y\), and ends at \(x\). Reversing changes the direction; it does not change the image of the path.
Worked Example: Concatenate Only After Matching Endpoints
Let \(X=\mathbb{R}\) with its standard topology. Take \(\alpha(t)=t\), which goes from 0 to 1, and \(\beta(t)=1+t\), which goes from 1 to 2. Since \(\alpha(1)=1=\beta(0)\), the endpoint condition holds.
For \(0\le t\le\frac12\), the concatenation is \(\alpha(2t)=2t\). For \(\frac12\le t\le1\), it is \(\beta(2t-1)=1+(2t-1)=2t\). Thus the complete formula is \((\alpha*\beta)(t)=2t\). At the joining parameter, the first piece gives \(\alpha(1)=1\), while the second gives \(\beta(0)=1\). The path starts at \((\alpha*\beta)(0)=0\) and ends at \((\alpha*\beta)(1)=2\).
Check: the formula \(2t\) is continuous on \(I\), and its values at the endpoints are \(2(0)=0\) and \(2(1)=2\), as required. The two pieces agree at \(t=\frac12\), where \(2t=2(\frac12)=1\).
Now consider the tempting but invalid pair \(\alpha(t)=t\) and \(\eta(t)=2+t\). Here \(\alpha(1)=1\), whereas \(\eta(0)=2\). Using the concatenation formula anyway would give \(2t\) on the first half and \(2+(2t-1)=1+2t\) on the second half. At \(t=\frac12\), the first formula gives 1 and the second gives 2. The proposed map is not even well-defined by these two formulas at the join, and the pieces do not match. It is not a continuous concatenation.
Common mistake: Writing down the piecewise formula does not make two paths concatenable. First check the endpoint equality; then substitute \(t=\frac12\) into both pieces to verify that the values match. If they do not, the proposed map cannot be continuous at the join.
A Path Must Stay in the Space Where It Is Claimed to Live
Suppose \(A\subseteq X\). A path in \(A\) is a continuous map \(I\to A\), where \(A\) carries the subspace topology. Composing it with the inclusion \(A\to X\) gives a path in \(X\) whose entire image lies in \(A\). Conversely, a path \(\gamma:I\to X\) whose image is contained in \(A\) can be regarded as a path into \(A\): for each open set \(A\cap U\) of \(A\), its preimage is \(\gamma^{-1}(U)\), which is open in \(I\). Thus the crucial condition is not just that the endpoints lie in \(A\), but that every point of the path does.
Worked Example: An Ambient Path That Is Not a Subspace Path
Let \(X=\mathbb{R}\) with its standard topology, and let \(A=\{0,1\}\) have the subspace topology. The formula \(\gamma(t)=t\) defines a path in \(X\) from 0 to 1. But \(\gamma(\frac12)=\frac12\notin A\), so this particular path is not a path in \(A\).
In fact, there is no path in \(A\) from 0 to 1. If such a path existed, it would be a continuous map \(f:I\to A\) with \(f(0)=0\) and \(f(1)=1\). A path-connected space is connected, so its continuous image is connected; hence \(f(I)\) would be a connected subset of \(A\) containing both 0 and 1. But \(\{0\}\) and \(\{1\}\) are open in \(A\), since they equal \(A\cap(-\frac12,\frac12)\) and \(A\cap(\frac12,\frac32)\), respectively. They separate \(A\), so \(A\) is disconnected. This is a contradiction.
Check: the ambient path has the stated endpoints because \(\gamma(0)=0\) and \(\gamma(1)=1\); its intermediate value \(\gamma(\frac12)=\frac12\) verifies that its image is not contained in \(A\). The disconnectedness check uses the disjoint nonempty open sets \(\{0\}\) and \(\{1\}\), whose union is \(A\).
This example distinguishes two claims that are often blurred: there is a path in \(X\) between points of \(A\), and there is a path in \(A\) between those points. The first does not imply the second. In a proof about a subspace, explicitly state the target of the path and verify that its full image lies there.
Connected Components Need Not Be Path Components
Every path component lies inside a connected component, but the two kinds of component need not coincide. The reason for the inclusion is that a path component is path-connected, hence connected; a connected subset that meets a connected component lies within that component. This uses the standard union fact that the union of two connected sets with a common point is connected: any separation of the union would have to place each connected set entirely on one side, and their shared point forces them onto the same side.
Proof. Let \(P\) be a path component and choose \(x\in P\). Let \(C\) be the connected component containing \(x\). The set \(P\) is path-connected by the definition of path components, so it is connected by the standard theorem that path-connected spaces are connected. Since \(P\cap C\) contains \(x\), the union \(P\cup C\) is connected by the union fact stated above. The maximality of \(C\) among connected subsets implies \(P\cup C\subseteq C\), and therefore \(P\subseteq C\). \(\square\)
Worked Example: A Connected Component with Two Path Components
Consider the topologist’s sine curve \(T=\{(x,\sin(1/x)):0<x\le1\}\cup(\{0\}\times[-1,1])\) with the subspace topology inherited from \(\mathbb{R}^2\). Write \(G=\{(x,\sin(1/x)):0<x\le1\}\) for its graph part and \(V=\{0\}\times[-1,1]\) for its vertical segment.
The earlier material on the connectedness and path components of the topologist’s sine curve establishes that \(T\) is connected and that its path components are exactly \(G\) and \(V\). These are the specific facts needed here; they are not conclusions drawn merely from a picture of the curve. Since \(T\) is connected, it is its own connected component when regarded as a space. Since its path components are the two disjoint nonempty sets \(G\) and \(V\), it is not path-connected.
Check: the points \((1,\sin 1)\in G\) and \((0,0)\in V\) belong to the same connected component, namely \(T\), but to different path components. So they cannot be joined by a path in \(T\). This is precisely why “the connected component containing these points” does not by itself provide a path between them.
A reliable component argument therefore keeps its conclusion matched to its hypothesis. Connectedness rules out a separation into two disjoint nonempty relatively open sets. Path-connectedness asks for a continuous map from \(I\) joining each chosen pair of points. The first property implies the second only under additional hypotheses; without such hypotheses, the topologist’s sine curve shows that connected components can contain multiple path components.
A Short Diagnostic Checklist
Before accepting a path or component argument, check the following points:
- Endpoints: For a concatenation, verify that the first path’s endpoint equals the second path’s starting point.
- Join: Substitute the joining parameter into both formulas and compare the values.
- Target space: Verify that every point of the path lies in the claimed subspace, not merely that its endpoints do.
- Conclusion: Do not replace “connected” by “path-connected” without a theorem that supplies that implication.
- Components: A path component is contained in a connected component; the reverse inclusion can fail.
Check Your Understanding
For each question, identify the hypothesis that matters and give a reason for your conclusion.
- Suppose \(\alpha(1)=p\) and \(\beta(0)=q\), where \(p\ne q\). What fails if the usual two-piece concatenation formula is used?
- Let \(A=\{0,1\}\subseteq\mathbb{R}\). Why does the path \(t\mapsto t\) in \(\mathbb{R}\) not show that \(A\) is path-connected?
- For paths \(\alpha,\beta:I\to X\) with \(\alpha(1)=\beta(0)\), evaluate both pieces of the concatenation at \(t=\frac12\).
- What does the topologist’s sine curve show about the relationship between connected components and path components?
- State the image condition that turns a path in \(X\) with endpoints in \(A\subseteq X\) into a path in \(A\).