7
$\begingroup$

I am trying to show that a function is upper semicontinuous if and only if the preimage of any open ray $(-\infty, a)$ is open.

The definition given for upper semicontinuity is that $\lim\limits_{k \to \infty} x_k = x \implies \limsup\limits_{k\to \infty} f(x_k) \leq f(x)$.

I find this definition hard to work with, as I have never been comfortable with $\limsup$ and $\liminf$.

Can anyone give me a hint as to how to approach this? Thank you!

$\endgroup$
3
$\begingroup$

The intuitive idea, I believe, is that $\cup_{\{x_k\le x\}}f^{-1}(-\infty,x_k)=\limsup_{\{x_k \le x\}} f^{-1}(-\infty,x_k) $ (in fact actually equal to the limit, since any subsequence would be monotone), thus if

$f^{-1}(-\infty,x)\supset \limsup_{\{x_k \le x\}} f^{-1}(-\infty,x_k)$

this corresponds to the condition

$\limsup_{k→∞}f(x_k)≤f(x)$.

see: lim sup and lim inf of sequence of sets.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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