7
$\begingroup$

Say that a topological space $X$ is locally connected at some point $x$, if it has a local base at that point consisting of connected open sets. Also $X$ is locally connected if it is locally connected at each of its points.

It is well known that a connected space is not necessarily locally connected.

For example, let $X = \bigl([0;1] \times (\mathbb{Q}\cap[0;1])\bigr) \cup (0\times [0;1])$ be the union of horizontal segmets with rational $y$-coordinate together with a vertical segment intersecting all of them. Then $X$ is connected but not locally connected. Namely it is not locally connected at any point except for the points of that vertical segment $0\times[0;1]$. On the other hand, it is still locally connected at each point of $0\times[0;1]$.

In fact, in all such examples which I know, connected spaces are still locally connected at some of their points.

The question is to construct an example of a connected space being not locally connected at any of its point.

Probably this is also a standard example, but I never meet it. I would be grateful for any information about that question.

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ The empty set is an example :) But of course the question is whether there's another one (and maybe you'd like Hausdorff too). $\endgroup$
    – YCor
    Commented May 2, 2023 at 7:32
  • 3
    $\begingroup$ I would do this: let $\{q_i\}_{i\in\mathbb Z}$ a bijection and take $X$ to be the quotient of $\mathbb Q\times [0,1]$ via $(q_i,1)\sim(q_{i+1},0)$ $\endgroup$
    – Pietro Majer
    Commented May 2, 2023 at 7:47
  • 3
    $\begingroup$ Also, a natural example is a dense orbit of a flow on a manifold (of dim≥2) $\endgroup$
    – Pietro Majer
    Commented May 2, 2023 at 8:26
  • 3
    $\begingroup$ @YCor The empty set is not connected. It has zero connected components. $\endgroup$
    – Zhen Lin
    Commented May 2, 2023 at 9:49
  • 2
    $\begingroup$ Yes, but I am saying the other convention is misguided. It is analogous (in a way that can be made precise!) to how the zero ring is not a field or integral domain. The correct definition can be stated in category theoretic language: an object $X$ is connected if and only if, for every coproduct $\coprod_{i \in I} Y_i$ and every morphism $X \to \coprod_{i \in I} Y_i$, there is a unique $j \in I$ such that the morphism factors through $Y_j \to \coprod_{i \in I} Y_i$. $\endgroup$
    – Zhen Lin
    Commented May 2, 2023 at 10:27

3 Answers 3

Reset to default
14
$\begingroup$

There are compact groups that are connected, but not locally connected at any point. For instance, solenoids $(\mathbf{R}\times\mathbf{Z}_p)/\langle (1,1)\rangle$. (It is locally homeomorphic to $\mathbf{R}\times$ Cantor.)

This is not path connected. A path-connected non-empty compact Hausdorff space that is not locally connected is given as follows: start from a Cantor subset $C$ of $[0,1]$ containing $0$, consider the product $C\times [0,2]$ and identify $(0,c)$ to $(c,2)$ for every $c\in C$.

(Not very artistic) illustration

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ I had in mind something similar but this is even better $\endgroup$
    – Pietro Majer
    Commented May 2, 2023 at 7:49
9
$\begingroup$

Here's a metric (and contractible) example, it is a subspace of $\Bbb R^2$. On each segment of the main zig-zagging line there's segments sticking out on a dense set, the idea is the same as the example you mention in the question body, but it is repeated to break local connectivity everywhere.

See this question for an (harder) example of a space which is compact, metric, connected and contractible, but not locally connected at any point.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Actually I first thought of this one and then "closed" it to make it compact. $\endgroup$
    – YCor
    Commented May 2, 2023 at 10:10
  • 2
    $\begingroup$ @Ycor makes sense! I like this example because it is visually obvious that it works $\endgroup$
    – Alessandro Codenotti
    Commented May 2, 2023 at 10:14
  • 1
    $\begingroup$ I was going to suggest $$\{(n+t,mt):n\in\mathbb Z,t\in[0,1],m\in\mathbb Q\cap[0,1]\}$$ but your example is nicer. $\endgroup$
    – bof
    Commented May 2, 2023 at 11:15
1
$\begingroup$

The closed topologist's sine curve is the classic example of a connected but not locally connected space. But local neighborhoods away from the y-axis are copies of $\mathbb R$, which is connected.

To fix this, replace the copy of $(0,1]$ that waves toward the y-axis with a copy of $(0,1]\times 2^\omega$. The result is still connected as it's the union of many copies of the closed topologist's sine curve that intersect on the y-axis. But now local neighborhoods away from the y-axis are copies of $\mathbb R\times 2^\omega$, which is disconnected.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .