Every open interval $(a,b)\subseteq \mathbb R$ is indeed homeomorphic to $\mathbb R$. The one-point compactification of $\mathbb R$ is homeomorphic to the unit circle $\mathcal S^1$.
Now, $\mathbb R$ and $\mathcal S^1$ are not homeomorphic because the latter is compact (and the real line isn't). Is that what you mean by "compact circumference"?
The two-point compactification of $\mathbb R$ is homeomorphic to the closed interval $[0,1]$ (or any other closed real interval).
So it looks like you got it right.
[Also, here's a post where you can find a definition of the two-point compactification and also an interesting discussion: Question on compactification ]
"We start with the fact that between any two real numbers there is a rational number."
True. But not just one; there are an infinite number of rational numbers between two real numbers.
"There is an infinite amount of real numbers between any two real numbers."
True. There are an infinite amount of real numbers including an infinite amount of rational numbers between two real numbers.
" Hence any real interval can accommodate the whole set of rational numbers which is also infinite."
Well, it can contain a set of the same cardinality as the whole set of rational numbers. We'll call that "accomodating". (You can't actually put 1/5 into [37.5, 37.8]. But you can put $\phi(1/5)$ into [37.5,37.8] when $\phi$ is a bijection from Q to $Q \cap [37.5,37.8]$.) It can also accommodate the whole set of real numbers between to real numbers. (That is we can make a bijection, $\rho$ where $\rho(x) \in [37.5,37.8]$ for all $x \in \mathbb R$.)
"We know there is a bijection between the set of natural numbers and the set of rational numbers."
True.
" Hence if there were a "bijection" between the natural and real numbers the domain (the natural numbers) would be mapped entirely within this interval of real numbers to all the rational numbers within."
Yes. And what's wrong with that?
Between x and y there are an infinite number of rationals. There is a bijection, $\phi$, between $Q$ and the rationals between x and y. And There is a bijection, $\chi$ between $N$ and $Q$. So $\phi\circ\chi$ is a bijection between $N$ and the rationals between x and y. Nothing wrong with that.
"So there cannot be any natural number mapping to something outside of this real number interval so there cannot be a bijection so the set of real numbers is uncountable. "
Sure there can. That would be just a different mapping. Having one map map to somewhere doesn't "use up" the ability to have another map mapping elsewhere.
Consider $f(n) = 2n + 1$. This is a 1-1 map from the natural numbers to the odd numbers. There is "nothing left" to map to any even number. Yet $g(n) = n$ maps to all the even numbers including the odd numbers. This isn't a contradiction.
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There is nothing in this proof that pertains to the real numbers that doesn't pertain to the rationals as well. So whatever this proves about the reals it also proves about the rationals. As the rations are countable this can not prove the reals are countable so there must be an error.
The error seems to be that you assume if you have a bijection $\phi: N \rightarrow S$ then you can not also have a bijection $\rho: N \rightarrow T$ where $S$ is a proper subset of $T$.
That is simply not the case on infinite sets. We have bijections from that Natural numbers to the integers, the rationals, the even numbers, etc. many of which are subset and supersets of each other.
In particular we can and do have bijections between the natural numbers and the rationals in any interval.
Let [x,y] be an interval. And let x< q < r < y so that q and r are rational. Let $f(t) = (t/t+1)(1/2(r - q)) + q + (r-q)/2$. $f$ is just such a injection from all rationals to the rationals in [x,y].
Best Answer
Well, the point being talked about is called denseness/density and it applies to rational numbers also. The case of rational numbers is simpler and ideally should be in the mind of a seventh grader. It is rather unfortunate that most textbooks don't emphasize this concept of density at the right time and later on students have to struggle while studying calculus.
We start with the following:
This is an immediate consequence of the following:
This is easy to understand and prove as well. If $m/n$ is a positive rational number then $m/(n+1)$ is a positive and smaller rational number. So the whole thing is ultimately dependent on the existence of a larger positive integer $n+1$ given a positive integer $n$. Moreover this also shows that there is no least positive rational number.
Next we can use theorem 2 to prove theorem 1. If $a, b$ are two rationals with $a<b$ then the number $d=b-a>0$ and we just need to find a smaller positive rational number $d'$ (which exists via theorem 2) and we can take $c=a+d'$ as our rational number lying between $a, b$.
Both the theorems above can also be proved using midpoint technique. Thus if $d$ is a positive rational number $d/2$ is a smaller one. And clearly $(a+b) /2$ is a number which lies between $a, b$. It is important to note both the approaches towards these theorems. Note also that the result in theorem 1 can be repeated to get as many rational numbers as we please between any two given rationals.
Now the idea of a neighbor (successor or predecessor for the case of integers) breaks down for rational numbers. To put it more precisely given any rational number $r$, there is no least rational number which exceeds $r$ and there is no greatest rational number which is exceeded by $r$.
This is expressed informally by saying that given any rational number $r$ we can find a rational number as close (near) to $r$ and less (or greater) than $r$ as we please.
The definitions of key concepts (limits) in calculus use the ideas mentioned above and are crucially dependent on the fact that there is no least positive number and there is no largest positive integer. However the fact that there is no next neighbor for a given rational number is not a drawback. It's a feature which is used everywhere in the definitions in calculus. A consequence of these facts is that we have an infinite supply of smaller and smaller positive numbers and greater and greater positive integers. And calculus in general deals with and therefore needs such infinite things.
Another key and slightly difficult ingredient in calculus is what we call completeness and that deals with the fact that although rationals are dense, they do lack something and this inadequacy is fulfilled by creating real numbers. The idea of completeness is not given sufficient focus in most textbooks of calculus (like in this answer too, but that's only to keep the length of the answer in control and can be discussed in another answer if you wish) but one can summarize the picture as follows: