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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!

Martin Sleziak
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Aden Dong
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  • Take a look at http://math.stackexchange.com/questions/123379/show-that-if-f-1-alpha-infty-is-open-for-any-alpha-in-mathbbr?rq=1. Your definition of upper semicontinuity agrees with the corresponding one in the link, if your domain is a metric space. – Thomas Mar 15 '13 at 13:24

1 Answers1

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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.

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Chill2Macht
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