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Here it is stated that the interior of the standard $n$-simpliex is $\{ (x_1,...,x_{n + 1}) \in \mathbb{R}^{n + 1} \mid x_i > 0, \sum_{i = 1}^n x_i = 1 \}$. However, I can't find or guess the proof of this fact, hence I ask for help. I know it is customary to post your thoughts or attempts on the matter, but I can't think of anything nontrivial.

Jxt921
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1 Answers1

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The set you describe is the intersection of the open subsets $\pi_i^{-1}[(0,\infty)]$ (for $i=1,\ldots, n+1$) with $\Delta^n$ so an open subset of $\Delta^n$. You only have to show that the complement consists of non-interior points. I.e. if one $x_i=0$ every open ball around it sticks out of the simplex. This makes interior the maximal open subset.

Henno Brandsma
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  • Interesting. Regarding the last part, this is what I came up with: suppose that $x \in \Delta^n$ such that $x_j = 0$. If $y \in B(x,\epsilon)$ such that $y_j \neq 0$ (since $|x - y| = \sum_{i \neq j} (y_i - x_i) + y^2_j$, then our claim is proven. But how to prove such $y$ exists? – Jxt921 Feb 20 '20 at 12:51
  • Oh wait, it seems $y = (x_1,...,\sqrt{\delta},...,x_{n + 1})$ where $\delta \in (0,\epsilon)$ does the trick. Is my logic correct? – Jxt921 Feb 20 '20 at 12:54
  • @Jxt921 I think so, yes. – Henno Brandsma Feb 20 '20 at 12:56
  • Actually, it occured to me that an interior should be the maximal open subset of $\mathbb{R}^{n + 1}$ contained in $\Delta^n$ ratrher that a maximal open subset of $\Delta^n$, which you seem to prove. – Jxt921 Feb 20 '20 at 14:11
  • @Jxt921 Well, $\Delta^n$ is $n$-dimensional in $\Bbb R^{n+1}$ so within the whole space it always has empty interior. I'm talking about the "simplex interior". – Henno Brandsma Feb 20 '20 at 14:25
  • I don't get it: isn't the interior of a subset $X$ in $X$ always $X$t, i.e. the maximal open subset of $X$ contained in $X$? – Jxt921 Feb 20 '20 at 14:29
  • @Jxt921This is a special "affine" notion of interior, not the standard topological one. You could take it as a definition, almost. – Henno Brandsma Feb 20 '20 at 14:31
  • Besides, tom Dieck claims that $\partial \Delta^n = { (x_1,...,x_{n + 1}) \in \mathbb{R}^n \mid x_i = 0$ for some $i }$ which contradicts $\mathrm{int}(\Delta^n) = \varnothing$ since $\partial \Delta^n = \overline{\Delta^n}\setminus\mathrm{int}{\Delta^n}$ and $\Delta^n$ is closed. – Jxt921 Feb 20 '20 at 14:31
  • @Jxt921 He also doesn't use the standard notion of boundary. In $\Bbb R^{n+1}$ the topological boundary of $\Delta^n$ is $\Delta^n$. The "affine" boundary is the set he describes. – Henno Brandsma Feb 20 '20 at 14:33