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Let be $v_1<\cdots<v_n$ and $\mu\in(v_1,v_n)$ real numbers. Show that set $$X=\left\{(p_1,\ldots,p_n)\in[0,1]^n\ |\ \sum_{i=1}^np_i=1,\ \sum_{i=1}^np_iv_i=\mu\right\}$$ is compact.

Obviously, $X$ is bounded from the definition, but what about the closeness?

My attempt – assume a sequence $X\ni\boldsymbol{p}^m\to\boldsymbol{p}\in\mathbb{R}^n$ and I want to show that $\boldsymbol{p}\in X.$ We know $$\lim_{m\to\infty}\boldsymbol{p}^m=\boldsymbol{p}\ \Longleftrightarrow\ \forall i\in\{1,\ldots,n\}:\lim_{m\to\infty}p_i^m=p_i,$$ therefore for every index $i$ and every (let us say) $\varepsilon_i\in(0,1)$ we have some positive integer $M_i$ such that for every integer $m_i>M_i$ $$p_i^{m_i}-\varepsilon_i<p_i<p_i^{m_i}+\varepsilon_i$$ holds. Summing these inequalities oves $i$ we obtain $$1-\sum_i\varepsilon_i<\sum_i p_i<1+\sum_i\varepsilon_i,$$ but how to proceed? And the other equality is even messier since some $v_j$ can be possibly negative.

I don't want a full solution, I'd like to solve it myself, but right now I am stuck...

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I suppose that the simples way is to consider the map$$\begin{array}{rccc}F\colon&[0,1]^n&\longrightarrow&\mathbb{R}^2\\&(p_1,\ldots,p_n)&\mapsto&\displaystyle\left(\sum_{k=1}^np_k,\sum_{k=1}^np_kv_k\right).\end{array}$$Then $F$ is continuous and your set is $F^{-1}\bigl(\{(1,\mu)\}\bigr)$.

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  • $\begingroup$ Really elegant. :) $\endgroup$ – byk7 Mar 5 at 20:34

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