Supremum and infimum

**cgiulz** · September 26th 2009, 10:48 AM

Let $S$ and $T$ be non-empty bounded subsets of $\mathbb{R}$ , and suppose that for all $s \in S$ and $t \in T$ , we have $s \leq t$ .

Prove that supremum of $S \leq$ infimum $T$ .

**Plato** · September 26th 2009, 11:06 AM

Originally Posted by cgiulz

Let $S$ and $T$ be non-empty bounded subsets of $\mathbb{R}$ , and suppose that for all $s \in S$ and $t \in T$ , we have $s \leq t$ .
Prove that supremum of $S \leq$ infimum $T$ .

There is no problem knowing that the two exist. Why?
So let $\sigma = \sup (S)\;\& \;\tau = \inf (T)$ .
Working for a contradiction, suppose that $\tau < \sigma$ .
So $\tau$ is not an upper bound of $S$ . Why?
What does that imply about some element of $S$ ?
Where is the contradiction?

**cgiulz** · September 29th 2009, 04:38 PM

Since $S$ and $T$ are bounded there must be an infinimum and supremum.

We can see that $\tau$ is not an upper bound of $S$ since it is smaller than the least upper bound.

So clearly there is some element $t \in T$ and $s \in S$ such that $s > t$ . So we have our contradiction.

Thus, $Sup S \leq Inf T$ .

Beautiful guidelines.

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