I want to prove that the function
is upperly semicontinuos in $R$.
Upper semicontinuity on my textbook has the following definition: $f: X \to R$ is upperly semicontinous in a point $x_0$ if $$\displaystyle\overline{\lim_{x \to xo}}f(x)\le f(x_0)$$ So far I wrote: $\forall \epsilon > 0 ~ \exists ~\delta $ s.t $~~~$$ |x-x_0| < \delta $ $\to $ $[x]-[x_0]<\epsilon$
using the fact that the definition of upperly semicontinuos implies that
$f(x)-f(x_0)< \epsilon$ in a neightborhood of $x_0$.
Now I'm stack since I can't find a proper $\delta$ that satisfies the last disequation. It must be really obvious (observing the graph of the function I guess it has something to do with the fact that the function whole side has discontinuitis regularly and always of the same module) but I'm not able to write it down, nor to obtain it algebrically somehow. Could someone please clarify this last step to me?


If a in R - N, then $f^{-1}(-\infty,a) = (-\infty, [a]+1)$ is open.
If a in N, then $f^{-1}(-\infty,a) = (-\infty,a)$ is open.
Thus f is upper semicontinuous.


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