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

I've tried looking for this question on this site, but I can't seem to find it. But if anyone can direct me to it, that would be great. But I'll pose my question in the mean time.

As the title states, if $$f^{-1}(\alpha, \infty)$$ is open, then show that $$f$$ is lower-semicontinuous.

Starting with the basics. The definition according to my professor gave on lower-semicontinuous is the following:

Given a function $$f: X \to \mathbb{R}$$ on a topological space $$X$$, $$f$$ is lower-semicontinuous if, for any $$x \in X$$ and for any $$\epsilon > 0$$, there is a neighborhood $$N$$ of $$x$$ such that

$$f(x) - \epsilon < f(x')$$ for all $$x' \in N$$.

I proved the conversed of this statement, but for this direction I seem to be stuck, been thinking for it for an hour or two. To my understanding a neighborhood of a point $$x$$ is a subset of $$X$$ such that it contains an open set which has $$x$$.

To me, it seems that I must consider if $$x \in f^{-1}(\alpha, \infty)$$ or $$x \in f^{-1}(-\infty, \alpha)$$. I started with the consideration of $$x \in f^{-1}(\alpha, \infty)$$. So by definition of pre-image, $$f(x) \in (\alpha, \infty)$$. We have that $$\alpha < f(x) < \infty$$. So it seems to me that I want a neighborhood of $$x$$ such that it satisfies $$f(x) - f(x') < \epsilon$$ for all $$x' \in$$ nieghborhood of $$x$$. Fix $$x \in X$$. Then I went on the path of suppose that $$x \in f^{-1}(\alpha, \infty)$$ and using that using $$f^{-1}(\alpha, b)$$ where $$b = f(x)$$. Since $$f^{-1}(\alpha, \infty)$$ can be written as the union (index starting at $$k$$) of $$f^{-1}(\alpha, k)$$ for $$k > \alpha$$ and $$k \in \mathbb{R}$$, then $$f^{-1}(\alpha, b)$$ must be open. Then the problem is that I can't get a neighborhood of $$x$$ such that $$f(x) - f(x') < \epsilon$$. I get a neighborhood of $$x$$ but it has elements such that $$f(x) - f(x') \not<\epsilon$$. I'll currently will update the progress of my work. Hopefully I can progress somewhere. Thanks.

Suppose we have $x \in X$ and $\epsilon > 0$. Then $f^{-1}(f(x)-\epsilon,\infty)$ is our desired neighbourhood of $x$.
• I did that awhile ago, but I came to some point where I was struck. Let me put what I have so far. I said that $f^{-1}(f(x)- \epsilon, \infty) = N_x$ (a neighborhood of $x$). Now, to show that $f$ is lower-semicontinuous, then we must find an existence of a neighborhood of $x$ such that $f(x) - f(x') < \epsilon$, but I'm thinking that $N_x$ can contain some $x'$ such that $f(x') > f(x)$. – MathNewbie Mar 22 '12 at 19:06
• So what if $f(x')>f(x)$? In that case we have $f(x)-f(x')<0<\epsilon$. – Chris Eagle Mar 22 '12 at 19:26
• I was focusing on the $x'$ so much that I forgot the objective of this proof. Thanks a bunch Chris. – MathNewbie Mar 22 '12 at 19:29
• @Chris Eagle If G = { $(x, y)$ belongs to $R^{2}$ such that $y = f(x)$} is closed in $R ^{2}$.then $f$ is lsc?? – sani May 6 '17 at 14:33