# In the definition of partial derivative, why the function must be defined on an open set?

On the page Partial derivative on Wikipedia, the following formal definition was found:

I am wondering if in this definition, the condition that $$U$$ being open is always necessary. For example, if in $$\mathbb{R}^n$$, a function $$f$$ is defined on the set of union of all axises: $$A=\{(x_1,\ldots,x_n)\mid \text{all but one }x_i \text{'s are zero}\}$$, it seems to me that one can still compute all partial derivatives of $$f$$ at $$(0,\ldots, 0)$$ without requiring the domain of $$f$$ being an open set.

Should we loosen the condition that $$U$$ is open to something like "U contains the intersection $$\{(x_1,\ldots,x_n)+(a_1,\ldots, a_n)\mid \text{all but one }x_i \text{'s are zero}\}\cap B$$, where $$B$$ is an open ball containing $$(a_1,\ldots, a_n)$$"? In this case the domain of $$f$$ is not necessarily an open set but the definition of the partial derivative at $$(a_1,\ldots, a_n)$$ still makes sense to me. Did I miss anything here?