Mara Cook

2022-06-29

statement of the proof was that $S$ was closed from below, and bounded from below. It is fine when they say that $B$ is bounded below which is also obvious from the definition. So, there should be a g.l.b. for the set $B$, which is denoted by ${b}_{0}$.
Now why there should exist a sequence of points $\left({\mathbf{y}}^{\mathbf{\left(}\mathbf{n}\mathbf{\right)}}\right)$ in the risk set $S$ such that $\sum {p}_{j}{y}_{j}\to {b}_{0}$? Is ${b}_{0}$ a limit point of $B$? Is $B$ closed ? Even if this happen why ${b}_{0}$ which is greatest lower bound of $B$ should belong to $B$?
Next, I guess they apply Bolzano Weierstrass theorem to say that ${\mathbf{y}}^{\mathbf{0}}$ is a limit point of the sequence $\left({\mathbf{y}}^{\mathbf{\left(}\mathbf{n}\mathbf{\right)}}\right)$. But why the last step $\sum {p}_{j}{y}_{j}^{0}={b}_{0}$?

trajeronls

Since ${b}_{0}=infB$, there exists a sequence $\left({\beta }_{n}{\right)}_{n\ge 1}$ of elements of $B$ such that $\underset{n}{lim}{\beta }_{n}={b}_{0}$. As in the book, write ${\beta }_{n}=\sum _{j=1}^{k}{p}_{j}{y}_{j}^{\left(n\right)}$ where $\left({y}_{1}^{\left(n\right)},\dots ,{y}_{k}^{\left(n\right)}\right)\in S$.
Since $\left({\beta }_{n}{\right)}_{n\ge 1}$ converges, it is bounded above by some $B$, hence for fixed $j$, we have ${p}_{j}{y}_{j}^{\left(n\right)}\le \sum _{j=1}^{k}{p}_{j}{y}_{j}^{\left(n\right)}\le B$. Since ${p}_{j}>0$, this yields $\mathrm{\forall }n\ge 1,\phantom{\rule{thickmathspace}{0ex}}{y}_{j}^{\left(n\right)}\le \frac{B}{{p}_{j}}$. Consequently, $\left({\mathbf{y}}^{\left(n\right)}{\right)}_{n\ge 1}$ is bounded coordinate-wise, hence bounded in ${\mathbb{R}}^{k}$. Applying Bolzano-Weierstrass gives some limit point ${\mathbf{y}}^{0}$ and a subsequence $\left({\mathbf{y}}^{\left({n}_{m}\right)}{\right)}_{m\ge 1}$ such that $\underset{m}{lim}{\mathbf{y}}^{\left({n}_{m}\right)}={\mathbf{y}}^{0}$. This implies convergence w.r.t each coordinate: $\underset{m}{lim}{y}_{j}^{\left({n}_{m}\right)}={y}_{j}^{0}$.
Letting $m\to \mathrm{\infty }$ in $\sum _{j=1}^{k}{p}_{j}{y}_{j}^{\left({n}_{m}\right)}={\beta }_{{n}_{m}}$ yields
$\sum _{j=1}^{k}{p}_{j}{y}_{j}^{0}={b}_{0}$

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