help

[vernal tree]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

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[soft grail]

basically we're trying to prove A=I

[cosmic blaze]

Answer to 1: Weierstraß themrum

@soft grail

Maybe you learned it under a different name

Do you need to give formal proofs or is a "loose" proof enough

https://en.m.wikipedia.org/wiki/Extreme_value_theorem

Answer to all the question i think

[soft grail]

Loose is fine

[cosmic blaze]

then you can propably find al of your answers here https://en.m.wikipedia.org/wiki/Extreme_value_theorem

[soft grail]

won't work since the f isn't necessarily continuous

[cosmic blaze]

ah yeah i missed that

[granite oar]

do u still need help

1 is easy because A is bounded above by b and nonempty (a is in A)

2 and 3 can be done by say contradiction. Suppose sup(A)<b. But I'm locally bounded at sup(A), so then I can extend the interval on which I am bounded.

finally, partition your interval in 2: [a, b-epsilon] and (b-epsilon, b]. Bound f on both these intervals for appropriate epsilon

[soft grail]

Huh

[granite oar]

by part b, f is bounded on any interval of form [a, b-epsilon] because sup(A)=b

but f is locally bounded, so also bounded on (b-epsilon, b]