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

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

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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?