Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Restricting us to function of a single real variable, I was used to prove Extreme value theorem via the short way: show that continuous functions preserve compactness, and the job is done.

Now, I want to prove similar results for semicontinuous functions. Wikipedia show me how to do:

This convinces me, however is a bit long and messy.

There is any way to prove the theorem more straightforwardly? For example, proving that an upper semicontinuous function has an image that owns its supremum?


share|cite|improve this question

There is a more straightforward way by using that a function is lower semicontinuous if and only if for each $c \in \mathbb{R}$, the set $\{ x \in X : f(x) \le c \}$ is closed.

For the proof of this and the proof of extreme value theorem (also known as Weierstrass' theorem) using this fact, see pp.43-44 of Infinite Dimensional Analysis: Hitchhiker's Guide, which preview is available in google books.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.