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How to deal with open sets whose boundary is never null I am trying to understand theconcept of open sets. I have, for instance, found that the unit open ball B is closed, whichmakes sense, because the boundary is the unit sphere S. However, I found a paper which
defines the boundary to be the empty set if it's closed. The problem is that a function
does not necessarily have a zero: $$f(t) = \begin{cases} t^2 \sin(\pi x) & \text{ if } t > 0\\ 0 & \text{ if } t \le 0 \end{cases}$$ Here $x$ is a parameter. The set of points where$f$ is not differentiable is $t = 0$. If $t > 0$ the value of the derivative is not defined, so
it's point is in the boundary. However, if $t \le 0$ the value is zero, so the point is outside
the set. So if the empty set is considered as boundary there is a null point of the function.
I understand that the definition of the boundary may depend on the function, but I
believe that the unit open ball is closed, no matter what $f$ is. A: The issue here is that$f(0) = 0$. So, one can't consider $f$ open because the definition of 'open' is that afunction has open image. But here the f678ea9f9e
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