How is this function closed?

Question

I am reading appendix of convex optimization book. In example A.1 it is written that $$f(x)=\array{x\log(x)\quad \text{if}\quad x>0 \quad \text{and}\quad 0 \quad \text{if} \quad x=0}$$ Its domain is $[0,\infty)$. Now function is continuous. The boundary of the domain is empty set (please point out if I am wrong in this). So since the empty set is subset of every set therefore the domain of $f$ is closed. But at the same time since the intersection of every set and empty set is an empty set therefore the domain of $f$ is open. So if the domain is closed then the function $f$ is closed but in this case the domain is also open so I do not understand how $f$ is closed. Any help in this regard will be much appreciated. Thanks in advance.

Answer 1

The boundary of your domain is the boundary of $[0, \infty ]$ which is ${\text { {0}}}$ which is a closed set.

Your domain is closed because it includes its boundary. It is not open because there is no open interval centered at $0$ which is contained entirely in the domain.

Answer 2

Nope. The boundary of the domain is the set $\{0\}$. Every open ball centered at zero intersects $[0,\infty)$ and $(-\infty,0)=\mathbb{R}\setminus [0,\infty)$. Moreover, the set $[0,\infty)$ is closed. Take a point $x\in\mathbb{R}\setminus [0,\infty)$, then, if $\varepsilon=\dfrac{|x|}{2}$ then $B_\varepsilon(x)\subseteq\mathbb{R}\setminus [0,\infty)$, i.e., $\mathbb{R}\setminus [0,\infty)$ is open, thus, $[0,\infty)$ is closed.

The previous arguments holds if your whole space is $\mathbb{R}$ with the usual topology. But if your whole space is $[0,\infty)$ then you're correct: the boundary is the empty set and $[0,\infty)$ is open and closed.

Finally, the closed-ness of the open-ness of your function doesn't follows from the properties of the domain (to be closed, for example). Do you have the correct definitions of closed and open function?