6

Could you provide me some examples of sets, which are not based on Cantor's construction, that satisfy the property $\partial A=A$, that is the boundary of a set is the set itself?

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
Salech Alhasov
  • 6,780
  • 2
  • 29
  • 47
  • 6
    The circle $x^2+y^2=1$ in the plane is such an example. The set $\mathbb Z$ of all integers within the line $\mathbb R$ is another. $\qquad$ – Michael Hardy May 15 '16 at 23:20
  • 3
    Any subset of $\mathbb{R}^n$ viewed as a subspace of $\mathbb{R}^m$ for $m>n$ will work. E.g., a line in $\mathbb{R}^2$, a plane or surface in $\mathbb{R}^3$, ... – kccu May 15 '16 at 23:22
  • Generally Jordan curves, compact sets of dimension less than the space. – AJY May 15 '16 at 23:36
  • 1
    @kccu Don't you mean, any closed subset of $\mathbb R^n$? – bof May 15 '16 at 23:42
  • @bof Yes, thank you. It's too late for me to edit. – kccu May 15 '16 at 23:46

1 Answers1

11

Any closed set with an empty interior is its own boundary. @MichaelHardy gave you two examples in his comment: the circle $x^2 + y^2 = 1$ in the plane and the set of $\mathbb{Z}$ of all integers within the line $\mathbb{R}$. Also @kccu noted that you can take any subset of $\mathbb{R}^n$ viewed as a subset of $\mathbb{R}^m$ for $m > n$, like a line in the plane, or a surface in $\mathbb{R}^3$.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
  • 5
    Since comments are ephemeral but answers are eternal, I think it would be a good idea to quote the comment in your answer rather than just refer to it. – bof May 15 '16 at 23:45