I haven’t  really been active, so please accept my apologies since my first post seemed to promise that I would write quite often than I have done. But it was all for a good cause, so I have decided to share something which I found interesting while reading up some Mathematics this holiday (any corrections, additions of definitions, etc. will be appreciated).

Now, the idea I am going to talk about is the Cardinality of a set. In simple terms:
Cardinality is a “measure  of the size” of a set.

Suppose that we have a set, \(A\), such that \(A={a,b,c,d,e}\). The cardinality of the set, denoted by \(|A|\), is 5, because there are 5 elements.
It is indeed worth noting that unlike ‘lists’, in Mathematics, order and number of elements doesn’t determine much concerning the identity  of a set, so {a,b,c}={a,a,a,b,c}={a,b,b,b,c,c}={a,…,b,…,c, …} as long as you use the same elements, they are all equal, and of course, cardinalities would be the same, because all of them are a representation of the same mathematical entity.

Let’s  talk about something more interesting. Suppose that you have another set with infinitely many elements, like the set of natural numbers, \(N\), or real numbers \(R\). The question  is: is \(|N|=|R|\)? The answer is: NO.

See, of all infinite sets, we have the following “kinds of infinity”:
-the kind where you can keep track of your infinite elements.
-the opposite kind.

A handy definition here is as follows:
-An infinitely  countable set is one whose elements can be arranged in an infinite list. Infinitely countable sets are denoted some ‘weird’ Hebrew (alphabet) letter “aleph”, which I don’t  seem  to have on my keyboard, so I will use Φ, for control. For instance, for \(N\):
{1,2,3,4,5,6,…}
Or \(Z\) (integers): {0,1,-1,2,-2,3,-3,…}. It is obvious that the elements of each of these sets can be arranged in an infinite list, as required by our definition of an infinitely countable set.
This is just a neat way of saying that if anyone had the “time”, they would be able to count the numbers (…but you don’t  want to be that guy–life is too short).  On the other hand, something like \(R\) causes serious problems with this idea. For instance, what is the smallest ‘positive’ real number ? (Say, for instance, you want to list real numbers from zero). With some proof by contradiction, you should be able to convince yourself that there is no such ! Also, it can be proven (with a really neat technique, which I won’t  show here) that \(R\) is uncountable!

Next, let’s  look at what it means for two sets that are infinite  to have equal cardinalities. Refer to the lists of numbers from \(N\) and \(Z\) above, and observe that you can align an integer to every natural number, so we use this basic idea to say whether two sets have equal cardinalities, or not. It is true that |N|=|Z|. “But,” you will say, “we have two integers for one natural number, except zero, so how…,” and I will answer: “Because both sets are equal (in terms of cardinality), you will not run out of natural numbers much like you  won’t  run out of integers!”
So, |Ν|=|Z|=  Φ
Βut a more formal statement is the following:
-Two sets, \(A\) and \(B\), have the same cardinality if and only if there exists a bijection from \(A\) to \(B\).

A bijection is a map that is both injective and surjective.
Injective, in simple terms, means that the map  is “one to one”.
Surjective means that it is “on to”–i.e if you map elements from \(A\) to \(B\), then any \(b\) in \(B\) has an element \(a\) in \(A\) from which it is mapped.

The following idea is what I found  most interesting…

Can you ‘show’ that the cardinality of the interval \((0,1)\) is equal to \(|R+|\) (i.e the cardinality of the set with all positive real numbers)?
That is absurd ! Obviously not.. I mean, imagine how many ‘more’ elements \(R+\) must have,right ? MAYBE… NO.

Let’s  draw an \(xy\) plane, with the following:
-Put a point at \((x,0)\)
-Put another at \((-1;1)\)
-Draw a line from the first to the second point (this should pass through  the y-axis).
-Connect \((-1;1)\) to the x-axis, label this point a (for reference).
-Observe that from x to a the distance is \(x+|-1|=x+1\)
-Label the y-axis interval from zero to the intersection as \(f (x)\).

(here is a sketch)

Observe and convince yourself that the two triangles formed are similar, so from proportionality, we get \(\dfrac {f (x)}{1}=\dfrac{x}{(x+1)}\)
In this case we see that there exists a “bijection”, \(f\), from \((R+)\), to \((0,1)\), so \(|(R+)|=|(0,1)|\)… (and you thought  you could “count”).

Unfortunately, that is it for now. I hope you had a great time reading!