Solution verification: the minimal/maximal elements of a relation on $\mathcal{P}(\{1,2,3,4\})$

Question

Consider the power set $\mathcal P ( \{ 1 , 2 , 3 , 4 \} ) $ and the partial order $ S $ on $ $ such that $ A \mathrel S B $ is defined as

$$A \; S \; B \iff A \cap \{ 1 , 2 \} \subset B \cap \{ 1 , 2 \} $$

$S$ is a partial order and I`m trying to find the minimum and maximum elements of it. Are my results correct?

Minimum elements: $$ \{\text{Ø}\},\{3\},\{4\},\{3,4\} $$ Maximum elements: $$\{1,2\},\{1,2,3\},\{1,2,4\},\{1,2,3,4\} $$

Answer 1

$ \newcommand{\set}[1]{ \left\{ #1 \right\} } \newcommand{\P}{ \mathcal{P}( \set{1,2} ) } \newcommand{\PP}{ \mathcal{P}( \set{1,2,3,4} ) } $ Part of the problem with this, well, problem is that it's hard to intuitively take on except under brute force. However, one thing you might be able to notice off the bat is that the relation $S$ on $\PP$ is the same as the subset order $\subseteq$ on $\P$ in some sense, due to the intersections with $\{1,2\}$. Any $X \in \PP$ satisfies $X \cap \set{1,2} \in \P$, in other words.

In that light, we can reduce this problem somewhat. If we can find the maximal/minimal elements of $\P$ under $\subseteq$, then the sets that generate those extrema in $\PP$ are its maximal/minimal elements under $S$.

Of course, the extreme elements of $\P$ are easily determined. We can even examine the situation visually with a graph like below, where $X$ points to $Y$ if and only if $X \subseteq Y$:

From this, it is clear that the maximal element of $\P$ is $\set{1,2}$, and the minimal element of $\P$ is $\varnothing$.

So the question remains: what sets of $\PP$ generate these under intersection with $\set{1,2}$? Well,

• $X \cap \set{1,2} = \set{1,2}$ if and only if $\set{1,2} \subseteq X$. That is, $1,2 \in X$.
• $X \cap \set{1,2} = \varnothing$ if and only if, similarly, $1,2 \not \in X$.

From these and an enumeration of $\PP$, it is clear that

$$X \cap \set{1,2} = \set{1,2} \iff X \in \Big\{ \set{1,2},\set{1,2,3},\set{1,2,4},\set{1,2,3,4} \Big\}$$

and

$$X \cap \set{1,2} = \varnothing \iff X \in \Big\{ \varnothing, \set{3},\set{4},\set{3,4} \Big\}$$

Thus, we have that

\begin{align*} \text{the set of maxima of } S &= \Big\{ \set{1,2},\set{1,2,3},\set{1,2,4},\set{1,2,3,4} \Big\} \\ \text{the set of minima of } S &= \Big\{ \varnothing, \set{3},\set{4},\set{3,4} \Big\} \end{align*}

This is mostly in line with your answer. However, be cognizant of the difference between $\set{\varnothing}$ and $\varnothing$ -- in particular, $\varnothing \in \mathcal{P}(X)$ for any set $X$, but we can't always say that of $\set \varnothing$.

But otherwise, your answer is correct.

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...granted, this question is a fair bit old (over a month), so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.