some more supremums and infimums (real analysis)

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• Jan 15th 2008, 08:53 PM
alexmin
some more supremums and infimums (real analysis)
For any two non-empty subsets of R let us write P<=Q if , for each x in P, there is a y in Q satisfying x<=y

(a) Show that if P<=Q then supP<= supQ (sup is short for supremum)

(b) Give an Example to show that if P<=Q then it does not follow that infP<=infQ (inf is short for infimum)

(c) Give an example to show that if P<=Q and if Q<=P then it does not follow that P=Q
• Jan 15th 2008, 10:37 PM
Jhevon
Quote:

Originally Posted by alexmin
For any two non-empty subsets of R let us write P<=Q if , for each x in P, there is a y in Q satisfying x<=y

(a) Show that if P<=Q then supP<= supQ (sup is short for supremum)

Proof:

Let $x$ be an arbitrary element of $P$ and $y$ be the element of $Q$ such that $x \le y$. Now we know that $y \le \sup Q$ for all $y \in Q$, by the definition of supremum. So we have $x \le y \le \sup Q$. In particular, we have $x \le \sup Q$ for all $x \in P$. But this means that $\sup Q$ is an upper bound for the set $P$. Now, by the defintion of the supremum, $x \le \sup P$ for all $x \in P$ AND $\sup P$ is less than or equal to any other upper bound for $P$. Thus we have $\sup P \le \sup Q$, as desired.

QED
• Jan 15th 2008, 10:43 PM
Jhevon
Quote:

Originally Posted by alexmin
(c) Give an example to show that if P<=Q and if Q<=P then it does not follow that P=Q

let P be the interval [0,1) and Q be the interval (0,1)
• Jan 15th 2008, 10:46 PM
Jhevon
Quote:

Originally Posted by alexmin
(b) Give an Example to show that if P<=Q then it does not follow that infP<=infQ (inf is short for infimum)

Let P be the interval (1,2) and let Q be the interval (0,2)