I've got a question about semicontinous functions and their hypographs. On wikipedia (https://en.wikipedia.org/wiki/Hypograph_(mathematics)) it is claimed that a function is upper-semicontious iff its hypograph is closed.
So I considered the function $f\colon (0,1)\to\mathbb{R}$ with $f(x)=1$. This function is continous and so especially upper-semicontinous. But the hypograph doesn't seem to be a closed subset of the $\mathbb{R}^2$ because "the right and left" bounderies do not belong to the hypograph (because $(0,1)$ is open). Can someone help me explain where my fault is, please?
Thank you!