In my lecture notes, my professor mentioned that the set of all vectors in $\mathbb{R}^n$ whose components are nonnegative and sum to $1$ is compact.
I am trying to prove this. Since the vectors are in $\mathbb{R}^n$, we just need to prove that the set is closed and bounded.
The set is closed is intuitive to me, but boundedness isn't.
Clearly, the vectors are bounded along every axis (choose one component to be $1$, and all other components to be $0$.) But that is only bouned in $n$ directions, how do I show that the set is bounded in all directions?
Best Answer
The map $f(x_1,\dots, x_n)=x_1+\dots+x_n$ is continuous, hence $f^{-1}(\{1\})$ is closed (since $\{1\}$ is closed). Your set in question $S$ is $f^{-1}(\{1\})\bigcap_i \{x_i\geq 0\} $ which is closed because is an intersection of closed sets.
It is also easy to see that $S$ is bounded in fact, $S\subset B(0,1)$. Since if $(x_1,\dots,x_n)\in S$, we have $x_1^2+\dots+x_n^2\leq (x_1+x_2+\dots+x_n)^2=1$.
You may think your set as a simplex (in 3D is a triangle with vertices at $(1,0,0),(0,1,0),(0,0,1)$ )