Question

Must every compact and connected metric space be locally connected at at least one of its points?

Answer 1

There is a "folklore" counterexample. Peter Nyikos gives the construction here (see the last paragraph for the compactness)

Answer 2

One more picture: Brouwer--Janiszewski--Knaster continuum:

Answer 3

In the product of the closet unit interval $I$ with the Cantor set $C$, identify $(0,x)$ and $\big(1,f(x)\big)$ where $f(x):=3x \mod 1$.

The resulting space $X$ is the mapping cylinder of the continuous map $f:C\to C$. It is a compact metric space locally homeomorphic to $]0,1[\times C\ ,$ thus not locally connected at any point. End-points of $I$ have not a special role; we may equivalently obtain $X$ with a larger quotient, $(\mathbb{R}\times C) / \{ (t,x)=\big (t+1,f(x)\big) \} $. The important feature of the map $f:C\to C$ is that it has a dense orbit $f^n( x_0)$. This is easily seen as it is conjugate to the left-shift map on binary strings on the space ${\bf 2}^\mathbb{N}$, $(c_1,c_2,\dots)\mapsto(c_2,c_3,\dots)$ which is just how we see $f$ on the 2-digits representations of points of the Cantor set. As a consequence, the image of $\mathbb{R}\times \{x_0\}$ in the latter quotient is a path-connected dense subset of $X$, which is therefore connected.

edit. Actually, such spaces are quite common in dynamical systems; an other example is the Smale-Williams Solenoid and several strange attractors.

Answer 4

The two examples given so far do not contain an illustration of the set. So here is one drawn with a "Cantor-pen":

Answer 5

No, examples abound, e.g., the $\sin\frac1x$-curve, i.e., the closure of $\lbrace \sin\frac1x : 0 < x \le 1\rbrace$ in the plane. As noted below, this is not a good example (I misread the question).

However, every indecomposable continuum, such as Knaster's bucket-handle continuum, is an example because every proper subcontinuum is nowhere dense. The pseudo-arc is, of course the ultimate example.