supremum and infimum

**flower3** · October 22nd 2009, 10:11 AM

Let A be a bounded subset of R "REAL NUMBERS" B a nonempty subset of A .show that :
$inf A \leq inf B \leq supB \leq sup A$

**aman_cc** · October 22nd 2009, 10:19 AM

Originally Posted by flower3

Let A be a bounded subset of R "REAL NUMBERS" B a nonempty subset of A .show that :
$inf A \leq inf B \leq supB \leq sup A$

First let's prove inf(B) <= sup(b)
For any x in B, by defintion
inf(B)<=x<=sup(B)

All you need to observe now is inf(A) is lower bound of B and sup(A) upper bound of B. Now invoke the definitions of inf,sup

**proscientia** · October 22nd 2009, 10:20 AM

For all $x\in B,$ $x\in A$ (since $B\subseteq A)$ and so $x\leqslant\sup A$ (as $\sup A$ is an upper bound for $A).$ Hence $\sup A$ is an upper bound for $B$ and so $\sup B\leqslant\sup A$ (since $\sup B$ is the least of the upper bounds for $B).$

A similar argument shows that $\inf A\leqslant\inf B.$

Finally, as $B\ne\O,$ let $b\in B.$ Then $\inf B\leqslant b\leqslant\sup B$ (since $\inf B$ and $\sup B$ are a lower and an upper bound for $B$ respectively).

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