Tuesday, March 29, 2016

Mathematics of Ringiana and Ringsanity III: Improving Infinities with Sets

Previous entry II: Solving x + x = x, first entry I: Moves of self-similarity.

To make sense out of our solution x = ∞ and upgrade it to a statement about the object X with the measurement x, we introduce sets. Rigorous axiomatic treatment of sets is long; instead, we follow our intuition and define a set as a collection of elements. Here is a set of 3 people


   S = { Ann, Bob, Carl }.

We list elements of a set inside parenthesis, separated by a comma. Order does not matter, and we can also write 

S = { Carl, Ann, Bob }.

 Exercise: In how many ways can you order elements of this set S? How many ways are there to list elements of a set of 5 people? Can you find a general formula if there are n elements in a set, for n = 1, 2, 3, 4, ... ? 

The number of elements in a set is called its cardinality (it's ok to use "size" if you think "cardinality" is too long). We write cardinality of S as | S |. For our set S above, | S | = 3. 

The empty set with no elements is allowed and denoted ∅.  Listing its elements does not take much space:

  = { }

and there is a pleasant formula 

     | | = 0

saying that the cardinality (size) of the empty set is zero. We won't go  below zero - there are no sets of size -1, -2, and so on. (To interpret the notion of a set of negative size, one needs to do much more - first linearize the concept of a set, then linearize the notion of the boundary of a shape. Some day we'll get to that story.)

Sets can contain various objects; consider this set T of three fruits T = { apple, melon, strawberry }. Set T has cardinality three, | T | = 3, just like S. Let's write S and T side by side: 


S = {Ann, Bob, Carl}   T = {apple, melon, strawberry} 

S and T have the same number of elements. We can match their elements, for instance, by giving apple to Ann, melon to Bob, and strawberry to Carl. A matching between elements of S and T is called an isomorphism between S and T; we denote isomorphism property by the congruence sign 

    S ≅ T 

and say "S is congruent to T" or  "S and T are isomorphic".  
An isomorphism is also called a bijection between elements of S and T. To each element of S we assign an element of T so that different elements of T are assigned to different elements of S (we don't want Bob and Carl to fight over a single strawberry), and each element of T is assigned to some (and unique) element of S. 

If two sets are isomorphic, S ≅ T, they have the same size: 

| S | = | T |  
  
On a deeper level, if two sets are isomorphic, there are many isomorphisms between them unless the sets are really really small (in which case there is only one isomorphism). For our sets S and T,  we can instead choose an isomorphism that gives melon to Ann, apple to Carl, and strawberry to Bob. And there a few other isomorphisms. By the way, the number of isomorphisms between two sets with n elements each is also the number of ways to order elements of either set (this is just me reminding you about the exercise above). 

There is a difference between saying that two sets are isomorphic and choosing a particular isomorphism between them. Ann, Bob, and Carl, the good inhabitants of S, would surely prefer to choose an isomorphism, any isomorphism, between S and T and start eating, rather than spend time staring at elements of T knowing that an isomorphism between them and the set of fruits exists, and more than one.   

In our more advanced examples and manipulations on sets, we will downplay the emotional side of set theory and for the most part will stick to sets of numbers, sequences and points. 


Mathematicians allow infinite sets. In an earlier post we've encountered the infinite set of natural numbers 

N = { 0, 1, 2, 3, ... } .

If a set T is finite, and we create a bigger set U by adding more elements to T, then sets T and U cannot be isomorphic since U has bigger cardinality that T. Interestingly, that is not the  case with infinite sets. 

Infinite sets are strange - we can add elements to them without changing their size. For instance, let's enlarge natural numbers N by adding -1 and calling the resulting set M: 

M = { -1, 0, 1, 2, ... }. 

M properly contains N. That is, every element of N is also an element of M, and some element of M is not in N. In this sense, M is strictly bigger than N. But actually M and N are isomorphic. To construct an isomorphism, we need to assign elements of M to elements of N in a nice (bijective) way. One such bijection is easy to guess if we put N and M next to each other:  

N = { 0,  1, 2, 3, ... }

M = { -1, 0, 1, 2, ... }.

Simply shift every element of N down by 1, that is, assign -1 to 0, 0 to 1, 1 to 2 and so on. To element n of N we assign element n-1 of M, for n = 0, 1, ... This is a sort of infinite shift, and this neat game can only be played with infinite sets. We do get a bijection between N to M this way, even though M contains all elements of N and an extra element. 

Now we come back to our task from the first post, looking for an object X isomorphic to two copies of itself. Let's take an infinite set as X, and, specifically, the set N of natural numbers. Is there a bijection between N and two copies of N? To construct one, we split N into two infinite sets. Any splitting would do, and to keep it simple split N into sets of even and odd integers: 

Even = { 0, 2, 4, 6, 8, ... }  
Odd = { 1, 3, 5, 7, ... } .  

We can write 
N = Even ∪ Odd


to denote that N is a union of two sets, Even and Odd. 
A brief side trip into difference between the usual union of sets and the disjoint union is needed. The usual union 

  X ∪ Y 
of sets X and Y consists of elements of X and elements of Y. If the same element b appears in both X and Y we list it only once in the union. For instance, 

{ a, b, c }  ∪ { b, c, d, e}  = { a, b, c, d, e}.

b appears in both sets, and then once in the union, ditto for c. An element cannot appear in a set more than one. 

Sometimes, we need to engineer an artificial union of two sets and keep track of both copies of each element that appears in X and Y. It's called the disjoint union of X and Y and denoted 

  X ⨆ Y

To implement the disjoint union, for each b that's in both X and Y we add additional labels to signify that it came from a particular set. Relabel b in X by b0 and b in Y by b1 to create two different copies of b and same for c in X and Y. Then, 
for the disjoint union of X and Y as above we get 

{ a, b, c }  { b, c, d, e}  = { a, b0, b1, c0, c1, d, e}.

The size of the disjoint union X ⨆ Y is the sum of sizes of X and Y 
  | X ⨆ Y | = | X | + | Y | ,
a simpler formula when compared to  the more familiar case of just the union of sets, where an additional term, the size of the intersection X ∩ Y, enters the equation: 

| X ∪ Y | = | X | + | Y | - | X Y | .

Since Even and Odd sets in the decomposition of natural numbers N have no elements in common (no number is both even and odd), we can rewrite the union as the disjoint union 

N = Even  ⨆  Odd

where, recall, 

Even = { 0, 2, 4, 6, 8, ... } 
Odd = { 1, 3, 5, 7, ... }.  

There are many bijections between Even and N and between Odd and N. For a bijection between Even and N we choose the one taking 0 to 0, 2 to 1, 4 to 2, 6 to 3, etc. In general, even number 2n is taken to natural number n. Ditto to get a bijection between Odd and N. Take 1 to 0, 3 to 1, 5 to 2, 7 to 3 and so on. Odd number 2n+1 goes to natural number n, for n = 0, 1, 2, ... These bijections give us isomorphisms between sets
  Even  ≅  N   and  Odd  ≅  N

Combining the decomposition N = Even Odd with these isomorphisms, we get an isomorphism between N and two copies of N :
  ≅  N N   
giving us a desired object X isomorphic to two copies of itself together with a particular isomorphism. What about the measurement x of X  that must equal infinity? That measurement is the cardinality (number of elements) of N. It is indeed infinite, and we write 

| N |  =  ∞ 

Actually, cardinalities of infinite sets constitute a sophisticated hierarchy, but for now we observe that N is an infinite set carrying an appropriate label for its size. 

To summarize the fruits of today's labor, other than those in the set T, we did find a solution to the equation X = X + X  in the world of infinite sets and isomorphisms between them. The next couple of posts will consider ways to prettify and further improve this answer. 

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