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Can you think of any spaces that are connected but not path connected? Apart form the topologists sine curve?

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Here are a whole bunch from Counterexamples in Topology. You should be able to find more information online about any of these.

A Pseudo-Arc
An Altered Long Line
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 the Reals
Indiscrete Rational Extension of the Reals
Irrational Slope Topology
Lexicographic Ordering on the Unit Square
Nested Angles
One Point Compactification fo the Rationals
Pointed Irrational Extension of the Reals
Pointed Rational Extension of the Reals
Prime Integer Topology
Relatively Prime Integer Topology
Roy’s Lattice Space
Smirnov’s Deleted Sequence Topoogy
The Extended Long Line
The Infinite Broom
The Infinite Cage
Topologist’s Sine Curve

There is a searchable version of the Counterexamples in Topology reference chart where you can ask other questions of this sort.

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Sorry for the bump of an old question, but I'm still receiving votes on it, so I thought I'd improve the answer. –  Austin Mohr May 29 '12 at 19:14
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Spacebook's great, thanks a lot for doing this! (I noticed a small typo in your database: in the list of properties you have "Extrmally Disconnected".) –  t.b. Jun 27 '12 at 10:56
    
@t.b. Thanks for your keen eyes. I've fixed the typo. –  Austin Mohr Jun 27 '12 at 22:14
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Another standard example is the extended long line. Counterexamples in Topology will have more, but my copy isn't to hand right now.

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Nice example!___ –  George Lowther Mar 26 '12 at 20:53
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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.

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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 $[0, \infty)$ 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.

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The long line is path connected. All locally path connected and connected spaces are path connected. –  George Lowther Mar 26 '12 at 20:47
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@George: The space I described is not locally path-connected in a neighborhood of the extra point $p$. –  MJD Mar 26 '12 at 20:49
    
I think I misunderstood what you meant. Shouldn't this be called the extended long line (or ray)? I don't think the long line has an extra point added at the end. –  George Lowther Mar 26 '12 at 20:56
    
@George: That is my fault. I will edit the reply. Thanks for pointing out that I misspoke. –  MJD Mar 26 '12 at 20:58
    
Ah, ok, I agree that this is a valid example now. –  George Lowther Mar 26 '12 at 21:02
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