0
$\begingroup$

I am integrating a function over the multidimensional domain $\Omega$, which is a subset of a larger domain $X$. Omega is defined by ALL points $x \in X$ that satisfy some condition, the details of which are not important, that I will denote $\text{cond}\!\left(x\right) = \text{TRUE}$. There are many points throughout $X$ that satisfy this condition (i.e. $\Omega$ is the union of a number of disjoint regions in $X$) due to symmetries. I am having some trouble providing an unambiguous definition of $\Omega$ in set notation. Here is what I've got:

$$\Omega = \left\{x \mid x \in X, \text{cond}\!\left(x\right) = \text{TRUE}\right\}$$

Does this clearly imply ALL points $x \in X$ that satisfy the condition (i.e. even the symmetrically equivalent points--they need to be included in the integration even though it is essentially double counting)? Or is there a better way to make this clear?

$\endgroup$
  • 2
    $\begingroup$ This looks fine to me. You could even shorten it to $$\Omega = \{x\in X | \operatorname{cond}(x) =\textrm{TRUE}\}$$ $\endgroup$ – MPW Feb 1 '16 at 17:49
  • $\begingroup$ Yes, that's what this notation means. It doesn't take into account symmetries in any way. $\endgroup$ – Qiaochu Yuan Feb 1 '16 at 17:49
  • $\begingroup$ @MPW: If you submit your answer as an answer I would be happy to accept it. Thank you. $\endgroup$ – okj Feb 4 '16 at 1:02
0
$\begingroup$

[Converted from comment to answer]

This looks fine to me. You could even shorten it to $$\Omega = \{x\in X \mid \operatorname{cond}(x) =\textrm{TRUE}\}$$

Addendum: Since the condition is evidently truth-valued, you could even write just

$$\Omega = \{x\in X \mid \operatorname{cond}(x)\}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.