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Is there a notation to express "this set is closed for all it's operations", in the sense of, given a set with it's defined operations, after a change in the set, the set retains the "closedness" of the operations it previously had, either by just using the same definitions on the new set, or by creating new definitions for the operations, yet to be said.

For use in the examples, let's say we have the operator $\overset*\setminus$, similar to \setminus, but capable of removing all the instances of something. For example:

$\mathbb C\overset*\setminus(\mathbb Ri) =: \mathbb R$, $\mathbb C$ without all the imaginary numbers becomes $\mathbb R$ (but $\mathbb R$ is not closed under the exponentiation, diferenttly than $\mathbb C$).

$\mathbb C\overset*\setminus\mathbb R$ becomes something where the square of the unity (the imaginary unity) is not present on the set, so it's not closed.

Supposing if $\left[\mathbb E\right]_*^*$ is the standard notation I'm looking for, we would have:

$\left[\mathbb C\overset*\setminus\mathbb R\right]_*^*$ is closed, with something like $i^2 = 0$ (with the $0$ from $\mathbb Ri$).

$\left[\mathbb H\overset*\setminus\mathbb R\right]_*^*$ mighty be defined as something like $\mathbb V^3$, the set of the 3d vectors, with $i^2 = j^2 = k^2 = ijk = 0$ (instead of $- 1$).

Is there a standard notation for this managing of sets?

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  • $\begingroup$ Your description of the operator $\overset*\setminus$ is unintelligible. Please clarify. $\endgroup$
    – Rob Arthan
    Mar 12 '21 at 1:16
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    $\begingroup$ @Rob Arthan It seemed to be an inverse to direct sum of vector spaces/additive groups, except that I don't know of a way to select a privileged inverse. $\endgroup$
    – Mark S.
    Mar 12 '21 at 14:29
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Minus operator

For your examples of $\overset{*}{\setminus}$, it seems like you are positing the existence of an operation that is not well-defined. Treating $\mathbb{R}$ and $\mathbb{C}$ as real vector spaces, it is essentially true that $\mathbb{C}=\mathbb{R}\oplus(\mathbb{R}i)$, where $\oplus$ is the direct sum. This is how I understand your intent for $\mathbb{C}\overset{*}{\setminus}\left(\mathbb{R}i\right)$ to be $\mathbb{R}$ and $\mathbb{C}\overset{*}{\setminus}\mathbb{R}$ to be $\mathbb{R}i$.

However, it is also true that, say, $\mathbb{C}=\left(\mathbb{R}(3+4i)\right)\oplus(\mathbb{R}i)$. Given $a+bi\in\mathbb{C}$, we have $a+bi=\dfrac{a}{3}(3+4i)+\dfrac{3b-4a}{3}(i)$. The notation $\mathbb{C}\overset{*}{\setminus}\left(\mathbb{R}i\right)$ doesn't seem to have any way to know that the answer you wanted is $\mathbb{R}$ instead of $\mathbb{R}(3+4i)$.

There may be a careful way to define this sort of inverse of direct sum (in some context) so that it has a unique result, but it's not obvious to me what that would be, and I'm skeptical that there would be such a definition at all.

Brackets notation

You essentially wrote "$\left[\mathbb{R}i\right]_{*}^{*}$ is closed, with something like $i^{2}=0$". If you redefine multiplication so that $i^{2}=0$, then you have the $\varepsilon$-axis of the "dual numbers" , which is indeed closed under addition and multiplication.

But if you redefine $i^{2}=0$, then the original meaning of $i$ is lost completely. So it seems at first glance that you want $\left[X\right]_{*}^{*}$ to mean something like "make whatever algebraic redefinitions of the operations are necessary to turn $X$ into something closed under addition and multiplication". There may be a careful way to define that (in some context) so that it always exists and has a unique result, but it's not obvious to me what that would be either.

Conclusion

In any case, since even the existence of these ideas as well-defined concepts is not obvious to me, I'm pessimistic that there's a common notation for it that I'm unaware of, but I'd be happy to be proven wrong by another answer.

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