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.

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?
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OR
If I change the mapping to $f: \mathbb{R}_{+}^{n} \rightarrow \mathbb{R}$, does $f^{-1} (1)= A$?
