Understanding topology and arc connectedness

Question

I started learning topology quite a while ago, and I still struggle with the core idea of topology. Today I learned about arc-connected topological spaces and that the unit interval with euclidean topology is path-connected and arc-connected but the unit interval with the trivial topology is path-connected but not arc-connected. I read the proof and understood it, but I can't really see the intuition behind it. Is there any way to visualize it? Can someone please explain it to me intuitively?

Answer 1

Both paths and arcs are continuous maps $u : [0,1] \to X$. However, the concept of a path is much more general because there is no other requirement than continuity. A path is an arc if it is an embedding (i.e. $u : [0,1] \to u([0,1])$ is a homeomorphism, where $u([0,1])$ has the subspace topology inherited from $X$).

You see that arc connectedness is a very strong property: It means that for any two distinct points $x, y\in X$ there exists a subspace $I' \subset X$ which is a homeomorphic copy of $[0,1]$ and contains both $x,y$.

1. Each one-point space is arc connected simply because it does not have two distinct points.

2. An arc connected space $X$ with more than one point contains a copy of $[0,1]$, thus its cardinality must be at least $\mathfrak c$.

3. Any two element subset $\{x,y\}$ of an arc conncted space $X$ has the discrete topology.

The unit interval with the trivial topology does not satisfy 3. Nevertheless it is path connected because any function to it is continuous.

Remark.

A space $X$ satisfies 3. if and only if it is a $T_1$-space.

In fact, if $X$ is $T_1$, then both $\{x\}$ and $\{y\}$ are closed in $X$, thus also closed in the subspace $\{x, y\}$. Hence they are also open in $\{x, y\}$ which means that $\{x, y\}$ is discrete.

Conversely, let 3. be satisfied. Given two distinct points $x,y \in X$, we know that $\{x\}$ is open in $\{x, y\}$. Thus there exists an open $U \subset X$ such that $U \cap \{x, y\} = \{x\}$. Hence $x \in U$ and $y \notin U$. This shows that $X$ is $T_1$.