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.

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 $y\in B_{\delta/2}(a)$, $f(B_{\delta/2}(y))\subset(f(a)-\epsilon,\infty)\subset (f(y)-2\epsilon,\infty)$. 
