Connected But Not Path-Connected?

Question

Can you think of any spaces that are connected but not path connected apart from the Topologist's Sine Curve?

Answer 1

An example of a connected space that is not path-connected is the deleted comb space: $$ (\{0\} \times \{0,1\}) \cup (K \times [0,1]) \cup ([0,1] \times \{0\})$$ where $K = \{ \frac{1}{n} \mid n \in \mathbb{N} \}$

Taken from here.

Answer 2

Here are a whole bunch from $\pi$-Base, a searchable version of Steen and Seebach's Counterexamples in Topology. You can visit the search result to learn more about any of these spaces.

An Altered Long Line

A Pseudo-Arc

Cantor's Leaky Tent

Closed Topologist's Sine Curve

Countable Complement Extension Topology

Countable Complement Topology

Double Pointed Countable Complement Topology

Finite Complement Topology on a Countable Space

Gustin's Sequence Space

Indiscrete Irrational Extension of $\mathbb{R}$

Indiscrete Rational Extension of $\mathbb{R}$

Irrational Slope Topology

Lexicographic Ordering on the Unit Square

Nested Angles

One Point Compactification of the Rationals

Pointed Irrational Extension of $\mathbb{R}$

Pointed Rational Extension of $\mathbb{R}$

Relatively Prime Integer Topology

Roy's Lattice Space

Smirnov's Deleted Sequence Topology

The Extended Long Line

The Infinite Broom

The Infinite Cage

Topologist's Sine Curve

Answer 3

Another standard example is the extended long line. Counterexamples in Topology will have more, but my copy isn't to hand right now.

Answer 4

The canonical example is the extended long line. You can think of the regular line $[0,\infty)$ as the product of $[0,1)$ and $\omega$ in the dictionary order topology—effectively, a countable number of copies of $[0, 1)$ pasted end-to-end.

The long line is the same way, except that instead of a countable number of copies you use an uncountable number of copies: take $[0, 1)\times\omega_1$ in the dictionary order topology, where $\omega_1$ is the smallest uncountable ordinal. Then to get the extended long line, you add one more point $p$ onto the far end. It's clearly connected, but it isn't path-connected because the path from any finite point, say $(1/2, 1)$, is too far from $p$ for the path between them to be the image of $[0,1]$.

The book Counterexamples in Topology by Seebach and Steen is good for answering questions like this.