Corollary, The Proof

Corollary; The extended real number line is the real number line. Proof; I shall attempt a proof with as well as I understand them, some arguments from set theory. Let R={x:–∞‹x‹+∞}, and R c={x:– ∞≤x≤+∞}.Where, R and R c stand for the real and extended real number lines respectively. We want to show the following inclusions; R ⊆ Rc and R c ⊆ R. Obviously the inclusion R ⊆ Rc is self evident. So now we shall attempt to demonstrate the inclusion R c ⊆ R. Consider the set R. Consider two points that we place “at the edges” of the chain R. Call those points “ α ” and “ β ”.With the property that; {y: y‹ α }= Ø ={y: y>b}.Where Ø denotes the empty set. Then, we generate the set Rc: = R U { α , β } ={x:–∞≤x≤+∞}. Haven constructed Rc from R we want to extend our argument to Rc as follows. Lets pick two other points ξ and ξ and place them as done previously with the points a and b for the set R at the edges of the set Rc. So doing we can naturally generate a new set −

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C={x: ξ ≤ x≤ ξ }.this is possible since R is a chain and naturally, so is Rc. Haven done this construction, we observe naturally that; R ⊆ Rc ⊆ C. Since; C\ R c= {ξ , ξ } and R c\ R= { α , β }. Whence, we have that; C\R = {ξ α , β ,ξ }. We thus can, on the basis of the extrapolation of our argument on the set Rc, take the set R c to be the set R (R = R c….1), and take the set C to be the set R c(C= R c….2) as they are “mechanically” equal and hence similar by the above construction algorithm employed for the establishment of these sets. Yet the very definition of the points “ α ” and “ β ”, we have that α =ξ and β =ξ . And consequently the most celebrated equality R = R c=C as a result of 1. And 2.Therefore we have that R = Rc.Certainly, this consequent equality is a stronger condition by which we can safely deduce the inclusion R c ⊆ R. Thus haven shown that R ⊆ Rc and that R c ⊆ R, we can deduce or rather conclude that R = Rc. which is the required result. −

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−

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Q.E.D Just to mention, a beautiful consequence of this corollary is the fact that the entity “infinity” symbolically denoted “∞” is a real and quantifiable number. Since it can be shown that

+∞≤1 and consequently 1=+∞.This result may not to be interpreted as thus, but rather intuitively as to the veritable nature of the quantifiability of the notion of infinity. On a demonstration Corollary; The notion; ‘Infinity’ mathematically denoted ‘∞’ is a real number. Proof: Here, we attempt to establish the equality: 1=+∞, since the number 1 is a real number and as such, establishing such a relationship between them will eventually induce a similar relationship on their mathematical nature. Clearly, 1≤+∞.and in fact 1<+∞.Now we establish the other inequality namely; +∞ ≤ 1. Clearly, 1

a = a ∀ a ≠ 0.

Define the sequence of real

numbers { a n} with a ≥0 properties that: a → 0 1. n∈ℵ

i

∀i ∈ ℵ with

n → +∞

n

2. a

≤

n +1

∞

3.

∑a i =1

i

a. n

=a

Then, 1 Thus;

.

a =a=

, as a ≠ 0 .

the

1

a =a a≠0

=

=

= . For some ζ ∈ ℵ .

≥

Whence, 1 ≥ = +∞. That is, 1≥+∞. We thus have both inequalities established. And consequently the equality 1=+∞. As required… Bizarre enough, lets apply the argument to the denominator and we will obtain a mathematical enigma. Thus; a1

+

+∞

1=

=lim

a →0

a2

∑a ∑a i =1

i

+ ... +

+∞

i =1

i

a1

= lim ∑ ai

→0

i =1

+∞

ai ≥0 ∀i ∈ ℵ , ⇒

∑a

i

∑a i =1

+

+∞

+∞

ak

a2

i

i =1

i

+ ... +

+∞

∑a ∑a i =1

+ ...

+∞

i

ak

+ ...

+∞

∑a i =1

a → 0 →0

i

Since and necessarily, a ∀i ∈ ℵ , an consequently, the following horror; i =1

1=

i

→0

0 0 0 0 + + + ... + + ... 0 0 0 0

Observe that the strong point of this argument is that it can be applied to any real number by simply using the fact that “1”is a multiplicative factor to any real number.