# Thread: some more supremums and infimums (real analysis)

1. ## 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

2. 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

3. 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)

4. 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)