Order by inclusion

Question

Given the set $E = \{a,b,c\}$, order the power set of $E$ (all subsets) by inclusion.

I think the order would be $\varnothing$, $\{a\}$, $\{b\}$, $\{c\}$, $\{a,b\}$, $\{a,c\}$, $\{b,c\}$, $\{a,b,c\}$.

Is that correct?

Answer 1

Hint:

• You have correctly identified the elements of the power set

• Arrange them in a natural way with one at the top, three in the next row, three in the row below that, and one at the bottom, and then show the order by inclusion with arrows.

Answer 2

It would be partially ordered set, because you can't compare 2 arbitrary elements in the power set, as $\{ a \} \not\subseteq \{ b\}$. Check out the Hasse diagram

Answer 3

Hint 2: Not all elements in the powerset of $E$ are comparable by set inclusion, so your diagram will consist of the union of all the linear orders of the subsets of $E$, where $E$ is the maximal element and $\emptyset$ is the minimal element.