# Thread: Compactness: over the top?

1. ## Compactness: over the top?

If $S = \{ x \in \mathbb{R}: x(x-3) \leq 0 \}$ and $T = \{x \in \mathbb{R}: x \geq 3 \}$, then every element of $T$ is an upper bound for the set $S$.

So $S = [0,3]$ which is a compact set, and so closed and bounded. Consequently there exists a least upper bound, and $\sup S = 3$ and so $T$ is a set of upper bounds for $S$.

Is this correct?

2. Originally Posted by shilz222
If $S = \{ x \in \mathbb{R}: x(x-3) \leq 0 \}$ and $T = \{x \in \mathbb{R}: x \geq 3 \}$, then every element of $T$ is an upper bound for the set $S$.

So $S = [0,3]$ which is a compact set, and so closed and bounded. Consequently there exists a least upper bound, and $\sup S = 3$ and so $T$ is a set of upper bounds for $S$.

Is this correct?
the proof is ok, but i don't think compactness is needed here, it's a bit of an overkill. the completeness axiom which can be applied to any half-bounded set of real numbers could be used to draw the same conclusion.

3. Oh ok, thanks. I just thought it would sort of nifty and neat.

4. Originally Posted by shilz222
Oh ok, thanks. I just thought it would sort of nifty and neat.
yes, it is nifty. i just learnt about compactness the other day myself, cool stuff. but sometimes we want to be simple and economical. using compactness is fine, your proof will stand out from the rest of the pack. but it depends on what type of professor you have that will determine if that is good or bad

5. Show that the infimum of T is larger then the supremum of S. Now since T is closed it contains it own infimum.