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 ( y ( n ) ) in the risk set S such that ∑ p j y j → 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 y 0 is a limit point of the sequence ( y ( n ) ) . But why the last step ∑ p j y j 0 = b 0 ?

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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       (            y              (        n        )              )   in the risk set   S such that   ∑      p    j        y    j    →      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       y      0   is a limit point of the sequence       (            y              (        n        )              )  . But why the last step   ∑      p    j        y    j    0    =      b    0  ?

trajeronls · Accepted Answer

Since       b    0    =  inf  B, there exists a sequence   (      β    n        )          n      ≥      1       of elements of   B such that       lim    n        β    n    =      b    0  . As in the book, write       β    n    =      ∑          j      =      1        k        p    j        y    j          (      n      )       where   (      y    1          (      n      )        ,  …  ,      y    k          (      n      )        )  ∈  S.Since   (      β    n        )          n      ≥      1       converges, it is bounded above by some   B, hence for fixed   j, we have       p    j        y    j          (      n      )        ≤      ∑          j      =      1        k        p    j        y    j          (      n      )        ≤  B. Since       p    j    &amp;gt;  0, this yields   ∀  n  ≥  1  ,        y    j          (      n      )        ≤      B          p      j      . Consequently,   (        y          (      n      )            )          n      ≥      1       is bounded coordinate-wise, hence bounded in         R    k  . Applying Bolzano-Weierstrass gives some limit point         y      0   and a subsequence   (        y          (              n        m            )            )          m      ≥      1       such that       lim    m          y          (              n        m            )        =        y      0  . This implies convergence w.r.t each coordinate:       lim    m        y    j          (              n        m            )        =      y    j      0  .Letting   m  →  ∞ in             ∑              j        =        1            k              p      j              y      j              (                  n          m                )              =          β                  n        m             yields      ∑          j      =      1        k        p    j        y    j      0    =      b    0

statement of the proof was that S was closed from below, and bounded from below. It is fine wh

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