# 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.

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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