The following images are part of a proof of a theorem in the Ferguson's seminal book "Mathematical Statistics-A decision theoretic approach".
I did not get some parts which uses certain concepts from Analysis.
The information given in the 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 $\left(\mathbf{y^{(n)}}\right)$ 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 $\mathbf{y^{0}}$ is a limit point of the sequence $\left(\mathbf{y^{(n)}}\right)$. But why the last step $\sum p_j y_j^0 =b_0$ ?
Since $b_0=\inf B$, there exists a sequence $(\beta_n)_{n\geq 1}$ of elements of $B$ such that $\lim_n\beta_n = b_0$. As in the book, write $\beta_n = \sum_{j=1}^k p_jy_j^{(n)}$ where $(y_1^{(n)},\ldots,y_k^{(n)})\in S$.
Since $(\beta_n)_{n\geq 1}$ converges, it is bounded above by some $B$, hence for fixed $j$, we have $p_jy_j^{(n)}\leq \sum_{j=1}^k p_jy_j^{(n)}\leq B$. Since $p_j>0$, this yields $\forall n\geq 1,\; y_j^{(n)}\leq \frac B{p_j}$. Consequently, $(\mathbf y^{(n)})_{n\geq 1}$ is bounded coordinate-wise, hence bounded in $\mathbb R^k$. Applying Bolzano-Weierstrass gives some limit point $\mathbf y^{0}$ and a subsequence $(\mathbf y^{(n_m)})_{m\geq 1}$ such that $\lim_m\mathbf y^{(n_m)} = \mathbf y^{0}$. This implies convergence w.r.t each coordinate: $\lim_m y_j^{(n_m)} = y_j^{0}$.
Letting $m\to \infty$ in $\displaystyle \sum_{j=1}^k p_jy_j^{(n_m)} =\beta_{n_m}$ yields $$\sum_{j=1}^k p_jy_j^{0} =b_0$$