Topological interior of a standard $n$-simplex A topological standard $n$-simplex is a subset $\Delta^n = \{ (x_0,x_1,...,x_n) \in \mathbb{R}^{n+1} \mid x_i \geq 0, \sum_{i = 0}^n x_i = 1 \}$ of $\mathbb{R}^{n+1}$ endowed with the subspace topology. Simplices have their own definition of boundaries and interiors, but little is said about the topological interior and the topological boundary of $\Delta^n \subseteq \mathbb{R}^{n+1}$ (the closure is, of course, $\Delta^n$ itself since it is closed in $\mathbb{R}^{n+1}$).
Accepted answers to this question (and many similar proofs) implicitly assumes that it is nonempty since a theorem they invoke requires a convex subset of $\mathbb{R}^{n+1}$ to have a nonempty interior.
This question and the answer to it suggest that the interior of $\Delta^n$ to be the subset of those $(x_0,...,x_n)$ for which $x_i > 0$, though I'm not sure whether they are talking about a topological or a simplicial interior.
I've also heard that it is empty, but then it would render all proofs I've seens of the fact that $\Delta^n$ is homeomorphic to $D^n$ invalid, which I find unlikely since it appears to be a folk theorem, which rely on the following theorem:

Let $K \subseteq \mathbb{R}^n$ be a compact convex subset with nonempty interior. Then, for any interior point $p$ of $K$, there is a relative homeomorphism $(D^n,S^{n-1}) \to (K,\partial K)$ which maps $0$ to $p$.

About a year ago I tried to understand the issue, to no avail: googling provides little to none information on the matter. This time I hope to get to the bottom of this, and to provide this answer as a reference to all in my situation.
 A: Let us begin with notation needed below: By $e_i^m$, $i = 0,\ldots,m-1$, we denote the standard basis vectors of $\mathbb R^m$.
The topological interior of $\Delta^n$ in $\mathbb R^{n+1}$ is empty. To see that, take any $x \in \Delta^n$ and any $r > 0$. Let $y = x + \frac{r}{2}e_0$. Then $y \notin \Delta^n$ and $\lVert y - x \rVert =  \frac{r}{2}$. This shows that $B_r(x) = \{ y \in \mathbb R^{n+1} \mid \lVert y - x \rVert < r \} \not\subset \Delta^n$.
The $i$-th face of  $\Delta^n$ is defined as
$$\partial_i \Delta^n = \{ x \in \Delta^n \mid x_i = 0 \} .$$
Let $p_i : \mathbb R^{n+1} \to \mathbb R^n$ denote the projection omitting the $i$-the coordinate from $x = (x_0,\ldots,x_n)$ which is clearly continuous. It is easy to see that $p_i$ maps $\partial_i \Delta^n$ bijectively onto $\Delta^{n-1}$, thus establishes a homeomorphism $\bar p_i : \partial_i \Delta^n \to \Delta^{n-1}$.
The simplicial boundary of $\Delta^n$ is
$$\partial \Delta^n = \bigcup_{i=0}^n \partial_i \Delta^n = \{ x \in \Delta^n \mid x_i = 0 \text{ for some } i \} .$$
You should draw pictures for $n =1, 2$ to see what is going on here. The simplicial interior of $\Delta^n$ is
$$\mathring{\Delta}^n = \Delta^n \setminus \partial \Delta^n = \{ x \in \mathbb R^{n+1} \mid \text{ All } x_i > 0 , \sum_{i=0}^n x_i = 1 \} .$$
You are right, the proofs in your links do not really work.
However, it is easy to see that $\mathring{\Delta}^n$ is the topological interior of $\Delta^n$ in the hyperplane $H \subset \mathbb R^{n+1}$ described by the equation $\sum_{i=0}^n x_i = 1$, and thus $\partial \Delta^n$ is the topological boundary of $\Delta^n$ in $H$. To see this, observe that $O  = \{ x \in \mathbb R^{n+1} \mid \text{ All } x_i > 0 \} = (0,\infty)^{n+1}$ is open in $\mathbb R^{n+1}$ and therefore  $\mathring{\Delta}^n = O \cap H$ is open in $H$. Moreover, no $x \in \partial \Delta^n$ is an interior point of $\Delta^n$ in $H$. In fact, $x \in \partial \Delta^n$ means that some $x_j = 0$. Pick any $k \in \{0,\ldots,n\} \setminus \{j\}$. For any $r > 0$ the point $y = x - \frac{r}{2}e_i^{n+1} + \frac{r}{2}e_k^{n+1}$ is contained in $H$. We have $y_i < 0$, thus $y \notin \Delta^n$. Also $\lVert y - x \rVert = \sqrt{2(\frac{r}{2})^2} = \frac{r}{\sqrt 2} < r$ so that $B_r(x) \cap H \not\subset \Delta^n$.
The map $h : H \to \mathbb R^n, h(x_0,\ldots,x_n) = (x_0,\ldots,x_{n-1})$, is a homeomorphism (its inverse being $(x_0,\ldots,x_{n-1}) \mapsto (x_0,\ldots,x_{n-1},1 - \sum_{i=0}^{n-1} x_i)$). Considering the compact convex set $h(\Delta^n) \subset \mathbb R^n$ which has non-empty interior in $\mathbb R^n$ reduces everything to what has been proved in your links.
You can also proceed as follows. $\Delta^n$ is the convex hull of the $n+1$ standard basis vectors $e_i^{n+1}$ of $\mathbb R^{n+1}$. Thus $h(\Delta^n)$ is the convex hull of the $n+1$ vectors $v_i = h(e_i^{n+1})$. But $v_i = e_i^n$ for $i < n$ and $v_n = 0$. Drawing a picture is helpful. Now have a look at my answer to Homeomorphism between $k$-simplex and a product of $k$ unit intervals. There I denoted $h(\Delta^n)$ by $\Delta^n$, so please be not be confused. Anyway, I constructed a homeomorphism $h(\Delta^n) \to I^n$. It should be well-known that $I^n$ is homeomorphic to $D^n$.
A: I shall attempt to answer your first question on the topological interior of the standard $n$-simplex. More specifically, I will show that every point in
\begin{equation}
x\in\mathrm{Int}(\Delta^{n})=\left\{(x_{0},\dots,x_{n})\in\mathbb{R}^{n+1}:\sum_{j=0}^{n}x_{j}=1,x_{j}>0\right\}
\end{equation}
is contained in some open ball $B(x,\delta)$ for some $\delta>0$, and $B(x,\delta)\subseteq\mathrm{Int}(\Delta^n)$. I am in fact working with the above definition (as in the equation above) of the interior of a simplex.
Let $a=(a_{0},\dots,a_{n})\in\mathrm{Int}(\Delta^n)$. Let
\begin{equation*}
2\delta=\inf\left\{\Vert\alpha-a\Vert_{2}:\alpha\in\mathrm{Int}(\Delta^n),\alpha\text{ has exactly one coordinate }0,\text{ and the rest positive}\right\}
\end{equation*}
where $\Vert\cdot\Vert_{2}$ is the usual Euclidean norm. We need to show that $B(x,\delta)\subseteq\mathrm{Int}(\Delta^n)$. Let $y\in B(x,\delta)\cap\Delta^{n}$. Then we see that from the definition of $\delta$, we must have $y\in\mathrm{Int}(\Delta^n)$. This shows that the open set $B(x,\delta)\cap\Delta^n$ is completely contained in $\mathrm{Int}(\Delta^n)$, where $B(x,\delta)\subseteq\mathbb{R}^{n+1}$ is the open ball centered at $x$ of radius $\delta$.
To answer your second question on the topological boundary, it follows that
\begin{equation*}
\partial\Delta^{n}=\Delta^{n}\setminus\mathrm{Int}(\Delta^n)=\left\{(x_{0},\dots,x_{n})\in\mathbb{R}^{n+1}:\sum_{j=0}^{n}x_{j}=1,x_{j}=0\text{ for some }j\right\}
\end{equation*}
Remark: Note that I am working with the topology on $\mathbb{R}^{n+1}$ induced by the usual Euclidean metric, and hence working with the subspace topology on the $n$-simplex $\Delta^n$ induced by the (Euclidean) metric topology on $\mathbb{R}^{n+1}$.
Any comments are welcome.
