Set Theory, working with cardinality, subsets and minimal maximal. I am asked:

Let $A =\{n:n$ is an integer and $-6&lt;n&lt;6\}$ and $X=\{B\subset A : 0&lt;|B|&lt;5\}$. Show that $B_1\le B_2$ if and only if $B_1$ is a subset of $B_2$ defines a partial order on $X$. What are the maximal and minimal elements? suppose $Y=\{B\subset A : 0\le|B|\le 9\}$. What are the maximal and minimal elements of $\le$ on $Y$ ?

So far I have found that $A={-5, -4, -3, -2, -1, 0, 1, 3, 4, 5}}. 

I'm not sure how to find $X$ or show that $B_1\le B_2$ if and only if $B_1$ is a subset of B2 defines a partial order on $X$. Can someone point me in the right direction? 

Thanks a lot.
They do indeed tell you what the set $\mathscr{X}$ is: all the subsets of $A$ that contain between $1$ and $4$ elements, inclusive. (Note: I'm using $\mathscr{X}$ instead of $X$ to emphasize the hierarchy we have: $\mathscr{X}$ is a set of sets, things in $\mathscr{X}$ are sets containing elements of $A$, and I would use lower case letters for things in $A$; the numbers $-5$ through $5$.

So, some sets in $\mathscr{X}$ would be $\{-3, 2, 5\}$, or $\{2\}$, and so on: Any set that contains between $1$ and $4$ elements of $A$ (integers from $-5$ to $5$), and nothing else. There are nearly $300$ of these, you do not want to write them all down! Thankfully, we don't need to, in order to verify that we have a partial order.

(As an aside, defining $B_1 \leq B_2$ whenever $B_1 \supseteq B_2$ (that is, $B_2 \subseteq B_1$) is called ordering by reverse inclusion, and is quite natural in certain situations).

Remember what it takes to make a partial order. Our relation  $\leq$ must be 

$\bullet$ reflexive: $B \leq B$ for all $B \in \mathscr{X}$

$\bullet$ antisymmetric: If $B_1 \leq B_2$ and $B_2 \leq B_1$, then $B_1 = B_2$, and 

$\bullet$ transitive: if $B_1 \leq B_2$ and $B_2 \leq B_3$, then $B_1 \leq B_3$.

So, to show that you have a reflexive relation, you need to pick an arbitrary $B \in \mathscr{X}$ and show that $B \leq B$; that is, $B \supseteq B$. Hopefully you do not need help doing this!

Likewise, for antisymmetry we'll pick arbitrary $B_1, B_2 \in \mathscr{X}$ such that $B_1 \leq B_2$, and $B_1 \geq B_2$. In other words, we have $B_1 \supseteq B_2$ and $B_2 \supseteq B_1$. Can you see why it must be the case that $B_1 = B_2$?

OK, have you shown it's a partial order? Good! Now let's think about minimal elements. 

To say a set $B \in \mathscr{X}$ is minimal with respect to $\leq$ means that there is no $B' \in \mathscr{X}$ with $B' \leq B$, other than $B$ itself.

So, we need to find sets $B$ such that there are no $B' \in \mathscr{X}$, aside from $B$ itself, with $B' \leq B$, that is, $B' \supset B$. This amounts to saying that there are no sets in $\mathscr{X}$ that properly contain $B$. So, ordering by reverse inclusion calls the largest sets of $\mathscr{X}$ the smallest sets with respect to $\leq$!

As an example, what about the set $ B = \{-5, -4, -3, -2\}$? Any set $B'$ containing this $B$ must either: be $B$ (which is fine), or be strictly larger meaning it will have to contain more than $4$ elements. But sets in $\mathscr{X}$ contain at most $4$ elements, so there's no $B'$ in $\mathscr{X}$ with $B' \leq B$ other than $B' = B$. What was special about that $B$ that made it minimal? The elements it contained, or how many elements it contained? There are plenty of other sets that are maximal, but they all have something in common with that particular set.

Thinking about maximal sets is very similar (the opposite, in fact!). Since we're ordering by reverse inclusion, 'big' and 'small' gets turned on its head, with respect to $\leq$. For a minimal set, would you expect it to be big or small, in the conventional sense?
They do indeed tell you what the set $\mathscr{X}$ is: all the subsets of $A$ that contain between $1$ and $4$ elements, inclusive. (Note: I'm using $\mathscr{X}$ instead of $X$ to emphasize the hierarchy we have: $\mathscr{X}$ is a set of sets, things in $\mathscr{X}$ are sets containing elements of $A$, and I would use lower case letters for things in $A$; the numbers $-5$ through $5$.

So, some sets in $\mathscr{X}$ would be $\{-3, 2, 5\}$, or $\{2\}$, and so on: Any set that contains between $1$ and $4$ elements of $A$ (integers from $-5$ to $5$), and nothing else. There are nearly $300$ of these, you do not want to write them all down! Thankfully, we don't need to, in order to verify that we have a partial order.

(As an aside, defining $B_1 \leq B_2$ whenever $B_1 \supseteq B_2$ (that is, $B_2 \subseteq B_1$) is called ordering by reverse inclusion, and is quite natural in certain situations).

Remember what it takes to make a partial order. Our relation  $\leq$ must be 

$\bullet$ reflexive: $B \leq B$ for all $B \in \mathscr{X}$

$\bullet$ antisymmetric: If $B_1 \leq B_2$ and $B_2 \leq B_1$, then $B_1 = B_2$, and 

$\bullet$ transitive: if $B_1 \leq B_2$ and $B_2 \leq B_3$, then $B_1 \leq B_3$.

So, to show that you have a reflexive relation, you need to pick an arbitrary $B \in \mathscr{X}$ and show that $B \leq B$; that is, $B \supseteq B$. Hopefully you do not need help doing this!

Likewise, for antisymmetry we'll pick arbitrary $B_1, B_2 \in \mathscr{X}$ such that $B_1 \leq B_2$, and $B_1 \geq B_2$. In other words, we have $B_1 \supseteq B_2$ and $B_2 \supseteq B_1$. Can you see why it must be the case that $B_1 = B_2$?

OK, have you shown it's a partial order? Good! Now let's think about minimal elements. 

To say a set $B \in \mathscr{X}$ is minimal with respect to $\leq$ means that there is no $B' \in \mathscr{X}$ with $B' \leq B$, other than $B$ itself.

So, we need to find sets $B$ such that there are no $B' \in \mathscr{X}$, aside from $B$ itself, with $B' \leq B$, that is, $B' \supset B$. This amounts to saying that there are no sets in $\mathscr{X}$ that properly contain $B$. So, ordering by reverse inclusion calls the largest sets of $\mathscr{X}$ the smallest sets with respect to $\leq$!

As an example, what about the set $ B = \{-5, -4, -3, -2\}$? Any set $B'$ containing this $B$ must either: be $B$ (which is fine), or be strictly larger meaning it will have to contain more than $4$ elements. But sets in $\mathscr{X}$ contain at most $4$ elements, so there's no $B'$ in $\mathscr{X}$ with $B' \leq B$ other than $B' = B$. What was special about that $B$ that made it minimal? The elements it contained, or how many elements it contained? There are plenty of other sets that are maximal, but they all have something in common with that particular set.

Thinking about maximal sets is very similar (the opposite, in fact!). Since we're ordering by reverse inclusion, 'big' and 'small' gets turned on its head, with respect to $\leq$. For a minimal set, would you expect it to be big or small, in the conventional sense?