Heads, Tails And Beans |S|=3

Doc Mike Finnley, Going Postal

Subsets

Let’s suppose that set \(S\) contains a full deck of playing cards. Therefore, \(|S|=52\). We can split the cards into four suits: \(♠,♣,♥,♦\). Let’s make a set \(T\) that contains only the clubs suit from that deck:

\[T=\{A♣,2♣,3♣,4♣,5♣,6♣,7♣,8♣,9♣,10♣,J♣,Q♣,K♣\}\]

The number of elements inside set \(T\) is \(|T|=13\). We call set \(T\) a subset of the set \(S\). Every element inside of set \(T\) is also an element inside of set \(S\). It’s important to note that set \(S\) still contains all 52 cards. Set \(T\) is simply showing you what is inside set \(S\) – it hasn’t removed those cards. Set \(T\) could contain any of the 52 playing cards and it would be a valid subset, for example:

\[T=\{4♦,J♣,8♠,4♠,A♦,Q♥\}\]

The combinatorical way of writing that \(T\) is a subset of \(S\):

\[T \subseteq S\]

We can also say that \(S\) is the super-set of \(T\):

\[S \supseteq T\]

A subset is a set in its own right and obeys the same rules as for sets. A subset can also contain every element that is in its super-set. Therefore, \(T=S\) iff (if and only if) \(T \subseteq S\) and \(S \subseteq T\).



The Power Set

Here’s a set of two playing cards:

\[S=\{K♠,Q♦\}\]

Remember that sets are unordered, so \(\{K♠,Q♦\}\) is the same as \(\{Q♦,K♠\}\). How many subsets can we make from set \(S\)? A direct enumeration gives us:

\[\{K♠\},\{Q♦\},\{K♠,Q♦\}\]

There appears to be three, but this answer is only partly true.

We can count all of the subsets by the use of the power set rule. It looks like this:

\[\wp(S)\]

Which represents all of the possible subsets of the set \(S\). We want to count the size (all of the groups of elements) of the power set which will count all of the subsets, thus:

\[|\wp(S)|=2^{|S|}\]



The Empty Set

I said that the number of subsets for our card set is three and that’s partly true. To see why, we need to calculate the size of the power-set. First, \(|S|=2\) therefore:

\[|\wp(S)|=2^2=4\]

(Remember, in mathematics a superscript n \(x^n\) means to raise x to the power of n. Or, multiply x by itself n number of times. In this case \(x=2,n=2\), therefore \(x*x=2*2=4\)).

According to the power-set rule we should have counted four subsets and not three.

Where’s the missing subset?

The answer lies with something called the empty set. As the name obviously implies, it’s a set with nothing in it:

\[S=\{\}\]

Hold on. How can I count nothing? An easy way to think of it is as something that didn’t happen but possibly could have happened.

We might decide to not place a card on the table, for example. A lottery ticket is easier to think about. Suppose I don’t match any numbers on my ticket (happens all too often!)? It’s still a lottery ticket. Does it not count?

I can count all of the tickets that match one, two, three etc. numbers. What about all of the tickets that match none?

The power-set of the card set \(S=\{K♠,Q♦\}\) looks like this:

\[\wp(S)=\{\emptyset,\{K♠\},\{Q♦\},\{K♠,Q♦\}\}\]

Where \(\emptyset=\{\}\). The size of the empty set (its cardinality) is zero, \(|\emptyset|=0\). But it does count as one subset. So the number of card subsets is now four.

If we take the cardinality of the power-set given the empty-set, we get this:

\[|\wp(\emptyset)|=2^{|\emptyset|}=1\]

Which means it is a subset of itself. And just to make matters more confusing, all sets have an empty subset and every set is a subset of itself!

To appreciate the potential size of all the subsets from a given set, consider four cards:

\[S=\{J♠,Q♣,K♥,A♦\}\]

According to the power-set rule there should be \(2^{|S|}=2^4=16\) subsets.

Let’s directly enumerate them by placing all of the card subsets on a virtual table (remember, the order does not count):

First, we can put down just one card at a time: \(\{J♠\},\{Q♣\},\{K♥\},\{A♦\}\). That’s four subsets.

Second, we can put them down in pairs: \(\{J♠,Q♣\},\{J♠,K♥\},\{J♠,A♦\},\{Q♣,K♥\},\{Q♣,A♦\},\{K♥,A♦\}\). That’s six subsets.

Third, we can put them down in triplets: \(\{J♠,Q♣,K♥\},\{J♠,Q♣,A♦\},\{J♠,K♥,A♦\},\{Q♣,K♥,A♦\}\). That’s four subsets.

Forth, we can put all of our cards on the table: \(\{J♠,Q♣,K♥,A♦\}\). That’s one subset.

Lastly, we can choose not to put any cards on the table, the empty-set \(\emptyset\). That’s one subset. This gives us a total of sixteen subsets, \(4+6+4+1+1=16\). We’ll see later that these numbers form a part of something called Pascal’s triangle.

As a challenge, try to directly enumerate (count) all 52 playing cards’ subsets. The power-set rule states that you’d need quite a few lifetimes to do so:

\[2^{52}=4,503,599,627,000,000\]

In English – Four quadrillion, five-hundred-and-three-trillion, five-hundred-and-ninety-nine-billion and six-hundred-and-twenty-seven-million different subsets.

Happy counting.
 

© Doc Mike Finnley 2019