---
tag: A2Q4RXD162
kind: counterexample
name: Comb space
msc: [54G20, 54D05]
uses: [0WB6DH499T, FDVAHRJKF3, 1ZVR8BRK65]
sources:
  - url: https://en.wikipedia.org/wiki/Comb_space
same_as:
  wikidata: Q5150587
  wikipedia: Comb space
---

## Counterexample

Let $K = \set{0} \cup \set{1/n : n \ge 1} \subseteq \R$.
The *comb space* is the subspace of $\R^2$
$$C = ([0, 1] \times \set{0}) \cup (K \times [0, 1]).$$
It consists of a base segment, the tooth $\set{0} \times [0, 1]$, and the teeth $\set{1/n} \times [0, 1]$.

The space $C$ is [[path-connected|0WB6DH499T]].
Each point can be joined to $(0, 0)$ by going down its tooth and then along the base.

The space $C$ is not [[locally connected|FDVAHRJKF3]].
Take $p = (0, 1)$ and the open set $U = C \cap B(p, 1/2)$.
For large $n$, the set $U$ contains a piece of the tooth $\set{1/n} \times [0, 1]$ near height $1$.
Any open set $V$ with $p \in V \subseteq U$ contains such pieces, but $U$ contains no part of the base.
The vertical line $x = c$ with $c$ strictly between two teeth splits $V$ into two disjoint open parts, one of them containing $p$.
So no open neighbourhood of $p$ inside $U$ is connected.

## Remarks

So connectedness, even path-connectedness, does not imply local connectedness.
For the same reason, $C$ is not [[locally path-connected|1ZVR8BRK65]].
The comb space is contractible, so it also shows that contractible spaces need not be locally connected.

## Explanations

### L1 human

Picture a hair comb with infinitely many teeth crowding toward the left edge.
You can walk from any point to any other along the teeth and the base, so the comb is in one piece.
But near the top of the left edge, every small window shows many separate bits of teeth.
So the space is connected overall, yet not connected in the small.