Proving the set $C = \{\,x \in \mathbb R^n : \sum x_i = 1, x_i \in [0,1]\,\}$ is compact. Proving the set $C = \{\,x \in \mathbb R^n : \sum_{1}^n x_i = 1, x_i \in [0,1]\,\} \subseteq \mathbb R^n$ is compact.
Alright: I can use the Heine-Borel theorem to prove this, therefore all I need to show is that the set is bounded and closed. Since $x_i \in [0,1]$ I know $C$ is bounded. Therefore I need to show its closed. Any ideas?
 A: As you said, it is bounded since $C\subseteq [0,1]^n$. To see it's closed notice that 
$$
F\colon\ \mathbb R^n\to\mathbb R,\ (x_1,\dots,x_n)\mapsto \sum_{i=1}^n x_i
$$
is a continuous map. Since $\{1\}\subseteq \mathbb R$ is closed it follows that $F^{-1}(1)\subseteq \mathbb R^n$ is closed as well. Now you have
$$
C = F^{-1}(1) \cap [0,1]^n
$$
which is closed as an intersection of closed sets.
A: For the closedness it is enough to check that, for each $x \in C^c$, there exists $\delta > 0$ such that $B(x,\delta) \cap C = \emptyset$. For that purpose, just consider $d:=d(x,C)$, which is not zero because, for each $z \in C$, we have that $\left| \sum_{i \leq n} z_i - \sum_{i \leq n} x_i \right| = \left|\sum_{i \leq n}z_i - x_i\right| > 0,$ as the first sum is one and the second is not.
In particular, there is some $1\leq i \leq n$ such that $x_i \neq z_i$, so $d(x,z)>0$. Now, if we take $\delta = d-\varepsilon$, $\varepsilon > 0$, then for each $y \in B(x,\delta)$ we have $d(y,C) \geq \varepsilon > 0$, thus $y \notin C$ and therefore $B(x,\delta) \cap C = \emptyset$.
