Can you give an example of a topological space that's connected but not pathconnected?

Can you give an example of a topological space that's connected but not pathconnected?
http://en.wikipedia.org/wiki/Connect..._connectedness
The extended long line and the topologists's sine curve are examples listed on this page.
Yeah, I looked at this question, but the only counterexamples I could come up with with were somewhat pathological. Like the topologists' sine curve that serpicojr mentioned.
Can I explain why it's a counterexample? . Um, no, not without a lot of work....
P.S; And even then I wonder if this is true if our space has a stricter separation axiom than that of Hausdorff, i.e. Tn > T2?
Anyway, I would say it is safe to assume, for all useful purposes, that, in topological terms, connected means pathconnected.
Now this is not to say a (path) connected space is always arcwise connected, this is a much stronger requirement. That is, a pathconnected space may have "holes", an arcwise connected space may not
How to prove that the topologist's sine curve is connected but not pathconnected? That's what I wanna know!
Actually, the proof is not as hard as I had at first thought. I'm not going to do it for you (at least not yet). Try it using the following heavy hints and show us what you get.
1) Construct the standard topology on R<sup>2</sup>.
2) Explain why this means this space is connected
3) Identify a subset of R<sup>2</sup> that will give you the curve we seek. Well, OK, I'll give you this; noting that the open sets in R<sup>2</sup> are the open intervals (a, b),...., you will want A = (0, sin 1/x) as a subset of R<sup>2</sup>. (What restrictions do you need to place on this set?)
4) Note that any subset of a connected space is connected
5) Find the closure B of A; identify the boundary of B; give an algebraic interpretation of this guy
6) Note that the closure of a connected set is connected
7) finally argue that, even though B is connected, it cannot be path connected (easily seen if you get (3) right!)
8) Good luck!
So the disjoint union of the intervals [0,1] and [2,3] is connected because it sits inside of R?Originally Posted by Guitarist
Yeak, Ok, clever clogs, I was wrong. Scrub (4), and use this
if A is connected, then if A ⊆ B and B⊆ A<sup></sup> (this is the closure of A), then B is connected, and A<sup></sup> is connected.
Rotter. Not talking to you again........
PS, point (4) was not essential for the proof, only the usual definition of connectedness
Just trying to help.
Okay folks, this is how you show connectedness.
Suppose (contrariwise) that the set S = {(0,0)} ∪ {(x,sin(1⁄x))*:*x ∊ ℝ<sup>+</sup>} is disconnected. Let A and B nonempty disjoint open (relative to S) sets such that A ∪ B = S with (0,0) ∊ A.
Now let t = inf{x > 0*:*(x,sin(1⁄x)) ∊ B} (which exists). Since (0,0) ∊ A and A is open and disjoint with B, there is an ε > 0 such that the intersection of B<sub>ε</sub>(0,0) (the open ball with centre (0,0) and radius ε) with S is a subset of A and thus disjoint with B. Moreover, B<sub>ε</sub>(0,0) ∩ S contains points in S other than (0,0) due to the behaviour of the sine curve in the vicinity of (0,0). It follows that t > 0. Hence the point s = (t,sin(1⁄t)) ∊ S.
So either s ∊ A or s ∊ B. If A, then the openness of A means there is an ε<sub>1</sub> > 0 such that B<sub>ε1</sub>(s) ∩ S ⊆ A. But since t+ε<sub>1</sub> is not a lower bound, there must be a t<sub>1</sub> with t < t<sub>1</sub> < t+ε<sub>1</sub> such that (t<sub>1</sub>,sin(1⁄t<sub>1</sub>)) ∊ B – which is a contradiction. OTOH if s ∊ B, we have an ε<sub>2</sub> > 0 such that B<sub>ε2</sub>(s) ∩ S ⊆ B – meaning that there are points t<sub>2</sub> between t−ε<sub>2</sub> and t such that (t<sub>2</sub>,sin(1⁄t<sub>2</sub>)) ∊ B and contradicting the fact that t is a lower bound.
So we get a contradiction either way. Hence S cannot be disconnected.
As for S not being path connected, sorry, I don’t know how to prove that.
Hint: Suppose a path exists between (0,0) and a point (x,sin(1/x)), i.e. some continuous map f(t) from [0,1] to the curve so that f(0) = (0,0) and f(1) = (x,sin(1/x)). First, show that the path must go through every point (x',sin(1/x')) for each 0 < x' < x. Then, say, project the path to the yaxisi.e., if f(t) = (x(t),sin(1/x(t)), then the projection is the function sin(1/x(t)). Finally, show that this can't be continuous at t = 0.Originally Posted by JaneBennet
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