Prove this set is closed Prove that the following set is convex, closed and bounded:
$$\left\{x: x=(x_{1},\dots , x_{n}) \land (\forall i, x_{i}\geq0) \land \sum_{i=1}^{n}x_{i}=1\right\}.$$  
I've been able to prove it is convex and bounded, but I can't seem to use its compliment to show that the compliment is open. Any suggestions would be very helpful. Thank you.
 A: Hint 
Define $f: \mathbb{R}_+^n \rightarrow \mathbb{R}$ by $f:(x_1,...x_n) \mapsto \sum_{i=1}^{n}x_{i}$
Is $f$ continuous ?
Then try to write this set as $f^{-1}(F)$ where $F$ is a closed set.
A: I don't seem to able to access that account, so please excuse me for using a new account. Is this correct then?
*Thank you for the hints. Hence, is it:
Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}= \sum_{1}^{n} x_{i} $
Define $A=\left\{x: x=(x_{1},\dots , x_{n}) \land (\forall i, x_{i}\geq0) \land \sum_{i=1}^{n}x_{i}=1\right\}.$
$f(A) = \{1\} $
$\{1\}$ is closed because it is a singleton. 
$f(•)$ is continuous because it is linear. 
Hence, $f^{-1} (1)$ is closed. 
adding from hint below
$A= f^{-1} (1) \cap \{x=(x_{1},\dots,x_{n}) | x_{i} \geq 0 \forall i\} $
Hence now I need to prove 
$B = \{x=(x_{1},\dots,x_{n}) | x_{i} \geq 0 \forall i\} $ is closed. 
Do I prove that $B^{c} = \{x=(x_{1},\dots,x_{n}) | x_{i} < 0 \exists i\} $ is open?
----- edit ----
OR
If I change the mapping to $f: \mathbb{R}_{+}^{n} \rightarrow \mathbb{R}$, does $f^{-1} (1)= A$?
A: Referring to the notation used by the OP in this answer (which is really a continuation of the question):
Yes, it's probably easiest to prove that $B^c$ is open, where
$$B = \{(x_1,x_2,\ldots,x_n) \in \mathbb R^n : x_i \geq 0\ \forall i\}$$
Suppose $(x_1,x_2,\ldots,x_n) \in B^c$. We wish to show that there is some $\epsilon$-ball, centered at $(x_1,x_2,\ldots,x_n)$, which is contained in $B^c$.
Since $(x_1,x_2,\ldots,x_n) \in B^c$, there is some $i$ such that $x_i < 0$. Choose any $\epsilon$ satisfying $0 < \epsilon < |x_i| = -x_i$. Let $(y_1,y_2,\ldots,y_n)$ be any point within the open ball of radius $\epsilon$ centered at $(x_1,x_2,\ldots,x_n)$. Then $\sum_{k=1}^{n}(x_k - y_k)^2 < \epsilon^2$, and consequently $(x_i - y_i)^2 < \epsilon^2$. Taking square roots, we obtain $|x_i - y_i| < \epsilon < |x_i| = -x_i$.
This means that
$$y_i = (y_i - x_i) + x_i \leq |y_i - x_i| + x_i < -x_i + x_i = 0$$
which shows that $(y_1,y_2,\ldots,y_n) \in B^c$, so $B^c$ is open, and therefore $B$ is closed.

Now we can finish the argument. As you noted, the function 
$f : \mathbb R^n \to \mathbb R$
defined by $f(x_1,x_2,\ldots,x_n) = x_1+x_2+\ldots+x_n$ is continuous, and $\{1\}$ is a closed subset of $\mathbb R$, so $f^{-1}(\{1\}) = \{(x_1,x_2,\ldots,x_n) \in \mathbb R^n : \sum x_i = 1\}$ is a closed subset of $\mathbb R^n$.

Then $\{(x_1,x_2,\ldots,x_n) \in \mathbb R^n : \sum x_i = 1 \text{ and }x_i \geq 0\ \forall i\} = f^{-1}(\{1\}) \cap B$ is the intersection of two closed subsets of $\mathbb R^n$, so it is closed.
A: Hint, use the first hint and the fact that $\mathbb{R}$ is Hausdorff then the set {1} is closed, so the preimage  $f^{-1}({1})$ of a closed set is closed.
Remains to show that f is continuous.
f is continuous because it is linear and $\mathbb{R}_+^n$ and $\mathbb{R}$ are finite-dimensional normed linear spaces.
