Let E be compact and nonempty. Prove that E is bounded and that sup E and inf E both belong to E.
For the first part can we use the fact that if E is compact iff it is closed and bounded? So since its compact than its closed and bounded..or is do I have to actually prove that its bounded? And also for the sup and inf, i'm not sure but would we set E=the interval and show that they are the inf and sup?
Hey this is what i've been working on. Can you let me know if i'm on the right track?
If E is compact then E is closed and bounded. Since E is nonempty, supE and infE both exist. To show supE∈ E, suppose not. Then supE ∈ Ec(compliment) which is an open set. Hence there exists δ > 0 such that
(supE − δ, supE + δ) ⊆ Ec(compliment). This is a contradiction because there must exist an element of E greater than
supE − δ....Then I just have to show it for inf..
well I was refering to the Heine-Borel Theorem which states Let E C R. then E is compact iff it is closed and bounded. So I thought since that was already proved as a theorem I could just state that it was closed and bounded b/c its compact. So instead of proving it I could refer to the theorem (at least my prof let us do somthing like that on one of our workshops)..
That is true. As you well know any compact subspace of a Hausdorff space is closed and any compact subspace of a metric space is bounded.
That said, these ideas in most analysis courses are not pursued and so the Heine-Borel theorem (in some of my books) is an iff statement.
I agree, you could have equally well defined a compact metric space to be one that is complete and totally bounded, but that doesn't change the fact that in a beginning analysis course there is only one "topology". They don't consider anything but the usual topology on