About the definition of a path connected set.

453 Views Asked by At

I will start by the usual definition of a path connected set:

A subset $A$ of a topological space $X$ is path connected when, for every pair of points $a,b\in A$, there exists a continuous function $f:[0,1]\rightarrow A$ such that $f(0)=a$ and $f(1)=b$.

This question may be a bit naive, but why the existence of path $f':A\rightarrow [0,1]$ s.t. $f(a)=0$ and $f(b)=1$, for every $a,b\in A$, does not imply that $A$ is a path connected set?

If it actually does, why do people mostly use the first definition?

Thank you.

2

There are 2 best solutions below

0
On BEST ANSWER

By definition, a path in $X$ is a continuous map from the unit interval into $X$. So I’ll address why this is the definition of a path rather than the map in the opposite direction. When you have a map $f:X \to Y$, you can think of it as a ‘copy of $X$’ sitting inside $Y$ that has been contorted by $f$, but since $f$ is continuous, it still somewhat retains the shape of $X$. Most people’s intuitive idea of a path is a curve from one place to another i.e. a contorted unit interval. So we would consider a map of the unit interval into the space.

You could also appeal to the fact that the continuous image of a connected space is connected (but the preimage of a connected space isn’t nevessarily connected). Since $[0,1]$ is connected, this implies that the image under $f$ is a connected subspace that ‘joins’ the endpoints. So the standard definition of path connected implies that there for any two points, there is a connected subspace that joins them. A map $X \to [0,1]$ would not imply the same thing.

Both of these intuitions rely on the idea that continuous maps do something special for you but only in one direction.

0
On

Suppose $X=A=\{0,1\}.$ Then the function $f:A\to [0,1]$ given by $f(0)=0,f(1)=1$ is continuous on $A,$ but $A$ is not path connected, or even connected.