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From Counterexamples in Topology by Steen and Seebach (2nd edition) example 129 page 145 we have an example of connected and totally path-disconnected space. It is defined as follow:

Fix $p= (1/2,1/2)$. Let $C$ be the Cantor set in the unit interval $[0,1]$. Let $E \subset C$ be the subset of $C$ that is the endpoints of the removed intervals. Let $F = C \setminus E$. Let $L( c)$ be the line segment connecting the point $c \in C $ to $p$. Define $Y_E=\{(x, y)\in L( c)~|~ c \in E, y \in \mathbb Q \}$ and $Y_F=\{(x, y)\in L( c)~|~ c \in F, y \not\in \mathbb Q \}$ We define Cantor's Leaky Tent or Cantor's Teepee as $Y =Y_E \cup Y_F\subset \mathbb{R}^2$ with the subspace topology.

I have two questions:

  1. Is there another "well-know" example of connected and totally path-disconnected space?
  2. Cantor's Teepee is not compact. Can we construct a non trivial compact, connected and totally path-disconnected space?
Portland
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    "Spacebook" has a sort of "electronic index" of Steen and Seebach; searching for "Is Compact" + "Is Connected" + "Is Totally Path-Disconnected" turns up A Pseudo-Arc, Finite Complement Topology on a Countable Space, and One Point Compactification fo the Rationals. – Nate Eldredge May 24 '14 at 19:00
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    Cantor's toupee? -click- Oh... well, that's disappointing :) – Bruno Stonek May 25 '14 at 17:43

2 Answers2

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I happened to run across an answer to 2. (and therefore 1.) just by clicking on one of the related MO links, finding the same question in a comment by Timothy Gowers under MO16578 that was later answered in a comment by Anton Petrunin, and then googling for more information. The key term you want seems to be "pseudo-arc", apparently a construction very familiar to the point-set topology community.

A pseudo-arc is a continuum $X$ (a connected compact metrizable space) with more than one point such that no subcontinuum $A$ (a subspace that is a continuum) is a union of two proper subcontinua of $A$. Remarkably, all pseudo-arcs are homeomorphic to one another!

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Todd Trimble
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The order topology provides many examples of compact connected ordered sets which are totally path disconnected. A totally ordered set is said to be complete if it complete as a lattice. A totally ordered set $X$ is said to be dense if whenever $x<y$ then there is some $z$ with $x<z<y$. One can get a totally ordered set that is both complete and dense by taking the Dedekind-MacNielle completion of a dense totally ordered set. A totally ordered set is compact in the order topology if and only if it is complete, and a totally ordered set is connected in the order topology if and only if it is dense and every bounded nonempty subset has a least upper bound. Therefore the dense complete total orders are precisely the compact connected total orders. On the other hand, except for the trivial cases, a compact totally ordered set need not have any paths. To be more precise, if $X$ is a compact connected totally ordered set, then there is a path from a point $x$ to a point $y$ with $y>x$ if and only if the interval $[x,y]$ is order isomorphic to the unit interval in the real numbers.

Joseph Van Name
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    It looks like you forgot to bring in density where you say "a totally ordered set is connected in the order topology if and only if every bounded nonempty subset has a least upper bound" (e.g., a finite totally ordered set has this l.u.b. property). For those interested, there's a nice paper by Pete Clark on this theme: http://arxiv.org/pdf/1208.0973v1.pdf. – Todd Trimble May 26 '14 at 13:27
  • A complete dense Aronszajn line gives a concrete example which is very similar to $\mathbb{R}$ in almost every other way. – François G. Dorais May 26 '14 at 16:02