Show that if $f$ is lower continuous then $f^{-1}((\alpha,\infty))$ is open Let a function $f:X\to\Bbb R$, where $X$ is a metric space.
Then $f$ is lower continuous if for all $a\in X$ we have that $f(a)\le \liminf f(x_n)$ for every $(x_n)\to a$. Alternatively we can say that $f$ is lower continuous if for all $\epsilon >0$ exists some $\delta>0$ such that
$$x\in\Bbb B(a,\delta)\implies f(a)-f(x)<\epsilon$$
Now I must prove that for any $\alpha\in\Bbb R$ the preimage of $(\alpha,\infty)$ is open. What I tried is set $f(a)=\alpha$, but from this approach I cant conclude that the preimage of $(\alpha,\infty)$ is open.
At most I can conclude that for any $\epsilon>0$ exists a ball $\Bbb B(a,\delta)$ where some of it images belong to some set of the kind $(\alpha,\beta)$.
Some hint or solution will be appreciated, thank you.
 A: Suppose $f$ is lower semicontinuous at each $x \in \Bbb R$, i.e., for each $\epsilon > 0$, there is some $\delta > 0$ so that when $y \in B(x,\delta)$, then $f(y) > f(x) - \epsilon$.
We want to show $f^{-1}(a, \infty)$ is open for each $a \in \Bbb R$.  Then let $z \in f^{-1}(a,\infty)$.  We want to find $\delta > 0$ so that $B(z,\delta) \subseteq f^{-1}(a,\infty)$ (this is just the definition of being open).
Since $z \in f^{-1}(a, \infty)$, this implies $f(z) > a$.  Since $f(z) > a$, we can say $a = f(z) - \epsilon$ for some positive number $\epsilon$, right?  So we have $z \in f^{-1}( f(z) - \epsilon, \infty)$.  Since $f$ is lower semicontinuous everywhere, it's lower semicontinuous at $z$, so there is some $\delta > 0$ such that $y \in B(z,\delta)$ implies $f(y) > f(z) - \epsilon$, i.e., $B(z,\delta) \subseteq f^{-1}(f(z) - \epsilon, \infty) = f^{-1}(a,\infty)$, showing that $f^{-1}(a,\infty)$ is open.
A: Let's use second: 
$f$ LSC if for every $a\in X $ and $\epsilon>0$, $\exists \delta$ such that $f(a)-f(x)<\epsilon$ whenever $x\in B_\delta(a)$
For $\alpha\in {\mathbb R}$ let $I_\alpha=f^{-1}((\alpha,\infty))$. 


*

*Suppose $f$ is LSC. Fix $\alpha\in \mathbb R$. Then for every $a\in I_\alpha$ and $0<\epsilon <f(a)-\alpha$, there exists $\delta>0$ such that whenever $x\in B_\delta(a)$,  $f(a)-f(x)<\epsilon<f(a)-\alpha$, giving  $f(x)>\alpha$. Hence  $B_\delta(a)\subset I_\alpha$, and as a result, $I_\alpha$ is open. 

*Conversely, suppose that for every $\alpha\in {\mathbb R}$, $I_\alpha$ is open. Fix $a\in X$ and $\epsilon>0$. Let $\alpha = f(a)-\epsilon$. Then $I_\alpha$ is open and contains $a$. In particular there exists $\delta>0$ such that $B_\delta(a)\subset I_\alpha$, that is, for all $x\in B_\delta(a)$, $f(x)>\alpha=f(a)-\epsilon$, or $f(a)-f(x)<\epsilon$. 
A: Hint: use your second definition. Fix $a\in f^{-1}(\alpha,\infty)$ and $\epsilon>0$ first, then you get $\delta$, and meanwhile you also get that for all $y\in B_{\delta/2}(a)$, $f(B_{\delta/2}(y))\subset(f(a)-\epsilon,\infty)\subset (f(y)-2\epsilon,\infty)$. 
