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Question:

$(x_n \rightarrow x \implies \limsup_{n \rightarrow \infty } f(x_n) \leq f(x) ) \implies \{x \in X |f(x) < c\} \,\text{is open for any}\, c \in R ?$

I know related questions have been touched open in upper semi-continuity functions , but most of them proved the opposite direction, and I haven't managed to find a satisfying proof for this one.

Can anyone help me with a simple but rigorous proof?

Related MSE Post:

Equivalence of definitions for upper semicontinuity

Show that if $f^{-1}((\alpha, \infty))$ is open for any $\alpha \in \mathbb{R}$, then $f$ is lower-semicontinuous.

Thanks.

Eric Wofsey
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numpynp
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2 Answers2

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It can be easily proved by contradiction.

If $W_c=\{x\in X: f(x)<c\}$ is not open, for some $c\in\mathbb R$, then there exist a $\xi\in W_c$, i.e. $f(\xi)<c$, such that $B_\varepsilon(\xi)\not\subset W_c$, for every $\varepsilon>0$. In partiucular, for $\varepsilon=1/n$, there would exist an $\xi_n\in X$, such that $d(\xi_n,\xi)<1/n$, and $\xi_n\not\in W_c$, i.e. $f(\xi_n)\ge c$. But clearly, $\xi_n\to\xi$, and due to the given condition, we would have that $$ c\le \limsup f(\xi_n)\le f(\xi)<c. $$ Contradiction.

Yiorgos S. Smyrlis
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To show $\{x\in X : f(x) <c\}$ is open, it is enough to show that $B=\{x\in X : f(x) \ge c\}$ is closed.

Proof : Let $\alpha \in B'$ i.e $\alpha$ is a limit point of $B$ .

To show $\alpha\in B$ .

Since $\alpha\in B'$, $\exists (x_n) \in B$ such that $x_n\to \alpha$

Then by the hypothesis, $\limsup f(x_n)\le f(\alpha)$

As $(x_n) \in B$ , $f(x_n) \ge c$

$c\le \limsup f(x_n)$

Then $c\le \limsup f(x_n)\le f(\alpha)$

Hence $f(\alpha) \ge c$ implies $\alpha \in B$

Thus the set $B$ closed. (Proved!)

Sourav Ghosh
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