1

Taken from here Boundary/interior of $0$-simplex

The standard-$n$-simplex $\Delta^n$ is the subspace $$ \textstyle \Delta^n = \{x=(x_0,\dots,x_n)\in\Bbb R^{n+1}\mid \sum_0^n x_i=1,\,x_i\ge0\,\forall i \} $$

Question what is the interior of the above simplex?

Originally, I thought it is simply:

$$ \textstyle \text{int}\Delta^n = \{x=(x_0,\dots,x_n)\in\Bbb R^{n+1}\mid \sum_0^n x_i<1,\,x_i\ge0\,\forall i \} $$

But then it seems stuff in the interior do add up to $1$.

So instead it should be:

$$ \textstyle \text{int}\Delta^n = \{x=(x_0,\dots,x_n)\in\Bbb R^{n+1}\mid \sum_0^n x_i=1,\,x_i>0\,\forall i \} $$

Which one is correct?

Community
  • 1
Olórin
  • 5,703
  • 1
    Note that this is true for the combinatorial interior of the simplex. It's topological interior is empty as long as the simplex's dimension does not match the space dimension. – Yaël Mar 24 '21 at 13:52

1 Answers1

5

The second is correct. Let $n=2$. Your standard simplex is then $x+y+z=1, x,y,z \ge 0$. The boundary has at least one coordinate equal to zero, two in the corners. The interior then has all coordinates positive.

With the definition as given, the above is incorrect. If we take $n=1$, for example, the 1-simplex is the open segment from $(1,0)$ to $(0,1)$. As a set in $\Bbb R^2$ the interior of this is empty. The n-simplex as defined, because it is in $\Bbb R^{n+1}$ has empty interior. Thanks to guest1 for the correction while Astaulphe made the same observation in a comment on the original post.

Ross Millikan
  • 383,099
  • 1
    but isnt this the relative interior? I.e., isnt the interior of the simplex actually empty since it is a subset of $\mathbb{R}^n$ but embedded into $\mathbb{R}^{n+1}$? – guest1 Feb 18 '25 at 13:19