Skip to main content
MathOverflow

Return to Answer

added 4 characters in body
Source Link
Piotr Hajlasz
Piotr Hajlasz
  • 28k
  • 5
  • 85
  • 184

The question has been answered here: https://mathoverflow.net/a/297836/121665. It is connected but not path connected.

The situation is somewhat similar to the topologist's sine curve:topologist's sine curve: the graph of $$ f(x)=\sin\frac{1}{x}, \quad x\in (0,1] $$ is path connected, but its closure (as a subset of $\mathbb{R}^2$) is connected, but not path connected.

The question has been answered here: https://mathoverflow.net/a/297836/121665. It is connected but not path connected.

The situation is somewhat similar to the topologist's sine curve: the graph of $$ f(x)=\sin\frac{1}{x}, \quad x\in (0,1] $$ is path connected, but its closure (as a subset of $\mathbb{R}^2$) is connected, but not path connected.

The question has been answered here: https://mathoverflow.net/a/297836/121665. It is connected but not path connected.

The situation is somewhat similar to the topologist's sine curve: the graph of $$ f(x)=\sin\frac{1}{x}, \quad x\in (0,1] $$ is path connected, but its closure (as a subset of $\mathbb{R}^2$) is connected, but not path connected.

Post Undeleted by Piotr Hajlasz
Post Deleted by Piotr Hajlasz
Source Link
Piotr Hajlasz
Piotr Hajlasz
  • 28k
  • 5
  • 85
  • 184

The question has been answered here: https://mathoverflow.net/a/297836/121665. It is connected but not path connected.

The situation is somewhat similar to the topologist's sine curve: the graph of $$ f(x)=\sin\frac{1}{x}, \quad x\in (0,1] $$ is path connected, but its closure (as a subset of $\mathbb{R}^2$) is connected, but not path connected.