---
tag: 0K3T4A659R
kind: counterexample
name: Topologist's sine curve
aliases: [warsaw-sine-curve]
msc: [54G20]
uses: [0R37DFJKFW, 0K0GE6BCYW, 0D5YHQGDJY]
sources:
  - url: https://en.wikipedia.org/wiki/Topologist%27s_sine_curve
    locator: lead section
  - url: https://en.wikipedia.org/wiki/Topologist%27s_sine_curve
    locator: Properties
same_as:
  wikidata: Q1148367
  wikipedia: Topologist's sine curve
---

## Counterexample

Let
$$T = \left\{ \left( x, \sin \tfrac{1}{x} \right) : 0 < x \le 1 \right\} \cup \{ (0, 0) \} \subseteq \mathbb{R}^2,$$
with the [[subspace topology|0D5YHQGDJY]] of the plane.
Then $T$ is [[connected|0R37DFJKFW]] but not [[path-connected|0K0GE6BCYW]]. Thus connected spaces need not be path-connected.

## Remarks

The graph part $G$ is connected, as the continuous image of $(0, 1]$. The origin is in the closure of $G$, and a set between a connected set and its closure is connected. Thus $T$ is connected.
No path joins the origin to a point of $G$. Near $x = 0$, the curve goes up and down between $-1$ and $1$ infinitely often, and a path cannot follow it continuously.
Some texts use the closed version, which adds the whole segment $\{0\} \times [-1, 1]$. It is compact, connected and not path-connected.
The space $T$ is also not locally connected.

## Explanations

### L1 human

The curve $\sin(1/x)$ oscillates faster and faster as $x$ goes to $0$. The origin is a limit of the curve, so the space is in one piece.
But you cannot walk along the curve to the origin, because the curve never settles down.