Is there a nice way to express the power set of a set $H$ minus the largest subset in the power set?

Question

Given the definition of the power set of a set $H$, $\mathcal P(H)$, as the set of all subsets of $H$, what is $\mathcal P(H) \setminus H$? Ie the power set minus the largest subset, $H$ its self. Something like

$$\mathcal P(H) - \{ H \} = \sum\limits_{i\in H}\mathcal P(H\setminus i) - \sum\limits_{(i,j)\in H, i\neq j}\mathcal P(H\setminus i, j) + \dots$$

Where the sum just denotes adding two sets together in such a way that we keep any repeated elements.

Explanation of the summation: I care about repetitions of set elements, so cannot use the more elegant solution suggested below of using the set union

Answer 1

"Adding two sets together in such a way that we keep any repeated elements" is the same as taking the union of two sets. It would be correct to say $$\mathcal{P}(H)-\{H\} = \bigcup_{i \in H} \mathcal{P}(H -\{ i\}).$$ Why? Well if $A \in \mathcal{P}(H)-\{H\}$, then $A$ is a subset of $H$ but it is not all of $H$. So there exists $i \in H$ such that $i \notin A$... Conversely, if $A \in \bigcup_{i \in H} \mathcal{P}(H-\{i\})$ then there exists $i\in H$ such that $A \in \mathcal{P}(H -\{ i\})$. Hence $A \subseteq H- \{i\} \subseteq H$, but $i \notin A$...

I'm not sure why you are subtracting sets, but that is not needed.

Answer 2

First, instead of using $\Sigma$, write $\bigcup_{i \in H}$, as (if I understood you correctly) you are thinking about the union of subsets of $P(H)$.

$\bigcup_{i \in H} P(H \backslash i)$ then consists of all subsets of $H$ that do not contain all elements of $H$, i.e. $\bigcup_{i \in H} P(H \backslash i) = P(H) - \{H\}$.

If you additionally subtract the set you did in your post, you will instead get the set of subsets of $H$ which are missing exactly one element, but that is not what you wanted.