Question about n-simplex and its face I just started to read Elements of Algebraic Topology by Munkres and got question already. On page 5, (Let $\sigma$ be an n-simplex) since Bd $\sigma$ consists of all points $x\in\sigma$ s.t. at least one $t_i(x)=0$, and Int $\sigma$ consists of all pioints $x\in\sigma$ s.t. $t_i(x)>0, \forall i$, it follows:
Given $x\in\sigma$, then there is exactly one face $s$ of $\sigma$ s.t. $x\in$ Int $s$, for $s$ must be the face spanned by those $a_i$ for which $t_i(x)>0$.
My question is, what if $x$ is one of the vertices, say $a_0$. It is not an interior of any faces, so there can't exist a unique face s.t. x contained in the interior of that face. Or am I misunderstood something here. Can someone clarify this. Thanks!
 A: Interior here means the interior with respect to the lowest-dimensional Euclidean space in which the simplex can be embedded. Technically, the standard-$n$-simplex $\Delta^n$ is the convex hull of the unit vectors $e_0,\dots,e_n$ in $\Bbb R^{n+1}$, with the subspace topology. That is 
$$
\textstyle
\Delta^n = \{ (t_0,\dots,t_n) \mid t_i\ge0, \sum_i t_i = 1 \}.
$$
As this set, its interior is empty. However, $\Delta^n$ is also a subset of its affine hull
$$
\textstyle
\text{Aff}(\Delta^n) = \{ (t_0,\dots,t_n) \mid \sum_i t_i = 1 \},
$$
which is a coset of an $n$-dimensional linear subspace of $\Bbb R^{n+1}$ and thus homeomorphic to $\Bbb R^n$. Regarded as a subset of $\Bbb R^n$, $\Delta^n$'s interior is
$$
\textstyle
\mathring\Delta^n = \{ (t_0,\dots,t_n) \mid t_i>0, \sum_i t_i = 1 \}.
$$
That's why we take the interior of a simplex to be the set of all points whose every coordinate is strictly larger than $0$.
If you go down to dimension $0$, then $\Delta^0$ is the subset $\{t_0 \mid t_0=1\}$ of $\Bbb R^1$, and its affine hull is $\Delta^0$ itself, so it is open in there. And this is indeed the set of all elements of the $0$-simplex where every coordinate is $>0$.
