I have decided to share something which I found interesting while reading up some Mathematics this holiday.

The idea I am going to talk about is that of the cardinality of a set. In simple terms,

Definition 1: Cardinality is a “measure  of the size” of a set.

Example:
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. This means that $\{a,b,c\}=\{a,a,a,b,c\}=\{a,b,b,b,c,c\}=\{a,...,b,...,c, ...\}$

as long as you use the same elements, they are all equal. Of course, cardinalities would be the same, because all of them are a representation of the same mathematical entity.

Let’s  talk about something slightly more interesting. Suppose that you have another set with infinitely many elements, like the set of natural numbers, $N$, or real numbers $\mathbb{R}$. A question we can ask is the following:

Is $|\mathbb{N}|=|\mathbb{R}|$ (i.e do we have as many real numbers as we do natural numbers)? Even though I have gone through the trouble of highlighting this question, the answer is no. However, we can say more than just, “no.”

Of all infinite sets, we have the following types of infinity:
1) The type where you can keep track of your infinite elements.
2) The type where 1) is not possible.

Definition 2: An infinitely  countable set is one whose elements can be arranged in an infinite list.

The cardinality of infinitely countable sets are denoted by $\aleph .$

For instance, for $\mathbb{N}=\{1,2,3,4,5,6,\ldots\},$

or $\mathbb{Z}=\{0,1,-1,2,-2,3,-3,\ldots\}$.

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 life is too short).

On the other hand, something like $\mathbb{R}$ 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 $\mathbb{R}$ is uncountable!

Let’s pause, and look at what it means for two sets that are infinite  to have equal cardinalities. Refer to the lists of numbers from $\mathbb{N}$ and $\mathbb{Z}$ 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 $|\mathbb{N}|=|\mathbb{Z}|$.

“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!”
For this case, we write $|\mathbb{N}|=|\mathbb{Z}|=\aleph .$

A more formal statement is the following.
Definition 3: Two sets, $A$ and $B$, have the same cardinality if and only if there exists a bijection from $A$ to $B$.

Recall:
A bijection is a map (or a function, if you prefer that name better) 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 each $b$ in $B$ has an element $a$ in $A$ from which it is mapped.

The following idea is what I found  most interesting, and the main motivation for the post.

The question is whether one can show that the cardinality of the interval $(0,1)$ is equal to $|\mathbb{R}_{>0}|$ (i.e the cardinality of the set with all positive real numbers)?

That is absurd ! Obviously not.. I mean, imagine how many ‘more’ elements $\mathbb{R}_{>0}$ 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 $f (x)$.

This should be done as is depicted in the following sketch: (Credits go to Richard Hammack: The Book of Proof)

Observe and convince yourself that the two triangles formed are similar, and 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 $\mathbb{R}_{>0}$, to $(0,1)$, so $|\mathbb{R}_{>0}|=|(0,1)|$

… (and we all thought we could “count”)!

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

 How clear is this post?