Path Connectedness
students, imagine walking from one point to another without ever lifting your pencil off the page ✏️. In topology, that idea has a precise meaning called path connectedness. It is one of the most important ways to understand whether a space is “all in one piece” in a strong, geometric sense. This lesson will help you recognize path connected spaces, use the definition correctly, and connect the idea to the larger topic of connectedness in Midterm 1 and Connectedness.
What Path Connectedness Means
A topological space $X$ is path connected if for every pair of points $x,y \in X$, there exists a continuous function $f:[0,1]\to X$ such that $f(0)=x$ and $f(1)=y$. The function $f$ is called a path from $x$ to $y$.
The interval $[0,1]$ is important because it represents time. You can think of $f(t)$ as the location of a moving point at time $t$. At $t=0$, you start at $x$, and at $t=1$, you finish at $y$. The requirement that $f$ is continuous means there are no sudden jumps 🚶♂️.
Path connectedness is stronger than just saying a space is connected. If a space is path connected, then it is connected. The converse is not always true, which makes this topic especially useful in topology.
Example: A Line Segment
The interval $[0,1]$ in the real line is path connected. If you choose any two points $a,b \in [0,1]$, you can use the path
$$f(t)=(1-t)a+tb$$
for $t \in [0,1]$.
This formula gives a straight-line motion from $a$ to $b$. At $t=0$, $f(0)=a$, and at $t=1$, $f(1)=b$. Because $f$ is a polynomial in $t$, it is continuous.
How to Build Paths in Familiar Spaces
Many common spaces are path connected because you can draw a continuous route between any two points. In geometry, this is often easy to see. In topology, you still need a clear argument.
In Euclidean Space
The space $\mathbb{R}^n$ is path connected for every positive integer $n$. Given points $x,y \in \mathbb{R}^n$, define
$$f(t)=(1-t)x+ty.$$
This is a straight line path from $x$ to $y$. Since each coordinate changes continuously, the whole map is continuous.
This idea also works for any convex subset of $\mathbb{R}^n$. A set $A$ is convex if for every $x,y \in A$, the entire line segment between them lies in $A$. For such a set, the same formula above gives a path staying inside the set.
Example: An Open Disk
The open disk in $\mathbb{R}^2$ is path connected. If you pick any two points inside the disk, the straight line between them stays inside the disk because the disk is convex. So the line segment itself is a path.
Example: A Circle
The circle $S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}$ is path connected. If you want a path between two points on the circle, you can travel along the circle itself. A continuous path may be given using angles. For example, if the points are written as $e^{i\alpha}$ and $e^{i\beta}$ in the complex plane, then one path is
$$f(t)=e^{i((1-t)\alpha+t\beta)}$$
provided the chosen angle interval is handled consistently. This shows that connected-looking shapes often have path connections that follow their geometry.
Why Continuity Matters
Path connectedness is not just about choosing a route; the route must be continuous. A function $f:[0,1]\to X$ is continuous if nearby values of $t$ produce nearby points in $X$. This rules out teleporting between separated parts of the space.
Continuity is what turns a simple idea into a strong mathematical property. For example, if a space has two pieces that are far apart, then no continuous path can move from one piece to the other without leaving the space or jumping across a gap.
Disconnected Example: Two Separate Intervals
Consider $X=[0,1]\cup[2,3]$ inside $\mathbb{R}$. This space is not path connected. If $x\in[0,1]$ and $y\in[2,3]$, any continuous path from $x$ to $y$ would have to pass through numbers between $1$ and $2$, but those numbers are not in $X$. So such a path cannot exist.
This example shows that path connectedness depends on the space itself, not just on the points you choose.
Path Connectedness and Connectedness
A major fact in topology is: every path connected space is connected. This is a one-way implication.
Why is this true? Intuitively, if any two points can be joined by a path, then the space cannot be split into two separated open parts. A path from one point to another would have to cross from one part to the other, which would force a contradiction with continuity.
However, a connected space need not be path connected. This is a very important distinction for Midterm 1 and Connectedness.
A Famous Example: The Topologist’s Sine Curve
One standard example of a connected space that is not path connected is the topologist’s sine curve. It is the set made from the graph of $y=\sin(1/x)$ for $x>0$, together with the vertical segment at $x=0$ between $-1$ and $1$. This space is connected, but there is no path joining a point on the vertical segment to a point on the oscillating curve.
The reason is that the graph keeps waving faster and faster as $x\to 0^+$. Although the set stays in one connected piece, the geometry is too wild to allow a continuous path between certain points.
This example is powerful because it proves that connectedness and path connectedness are different ideas.
How to Prove a Space Is Path Connected
When solving problems, students, you should look for a direct path formula or a known theorem.
Strategy 1: Use a Straight-Line Path
If the space is convex, use
$$f(t)=(1-t)x+ty.$$
This is the easiest method in subsets of $\mathbb{R}^n$.
Strategy 2: Break the Space into Easy Pieces
Sometimes a space is built from path connected pieces that overlap. If two path connected subsets have a point in common, then their union is path connected. More generally, if a collection of path connected sets overlaps in a chain-like way, you can combine paths across the overlaps.
For example, if two disks in the plane overlap, their union is path connected because you can move within the first disk to a point in the overlap, then move within the second disk to the destination.
Strategy 3: Use Known Path Connected Spaces
You can also rely on standard examples:
- $\mathbb{R}^n$ is path connected.
- Any convex subset of $\mathbb{R}^n$ is path connected.
- Any interval in $\mathbb{R}$ is path connected.
- The circle $S^1$ is path connected.
- Any connected open interval in the real line is path connected.
These facts are often enough for exam-style reasoning.
How to Prove a Space Is Not Path Connected
To show a space is not path connected, you need to find two points that cannot be joined by any continuous path.
A common method is to show the space has two separate pieces with no route between them. For example, if a space is the union of two disjoint open sets, then no path can cross from one to the other.
Another useful idea is to consider a function that would have to be continuous on $[0,1]$ and would force an impossible jump. Since continuous images of intervals in $\mathbb{R}$ are connected, a path cannot land in two isolated pieces without crossing the space in between.
Conclusion
Path connectedness is the idea that any two points in a space can be joined by a continuous path. It is a strong way to describe spaces that feel “unbroken” in a geometric sense 🌟. In Midterm 1 and Connectedness, it is essential to remember that path connectedness implies connectedness, but not every connected space is path connected. That difference is one of the key lessons of topology.
When working with examples, students, look for a clear path formula, use convexity when possible, and remember that continuity is the heart of the definition. These tools will help you recognize path connected spaces and explain why other spaces fail to have this property.
Study Notes
- A space $X$ is path connected if for every $x,y \in X$, there exists a continuous map $f:[0,1]\to X$ with $f(0)=x$ and $f(1)=y$.
- A path is a continuous map from $[0,1]$ into the space.
- The formula $f(t)=(1-t)x+ty$ gives a straight-line path in $\mathbb{R}^n$.
- Every path connected space is connected.
- The converse is false: some connected spaces are not path connected.
- Convex subsets of $\mathbb{R}^n$ are path connected.
- The circle $S^1$ and Euclidean space $\mathbb{R}^n$ are path connected.
- The topologist’s sine curve is connected but not path connected.
- To prove path connectedness, try to construct an explicit continuous path.
- To prove a space is not path connected, find two points that cannot be joined by any path.