The Proof: There exist interval [A,B] that cannot have a sub-interval

[said_alyami]

This is a proof that:

There exist interval [A,B] on the real numbers line that is indivisible as a mathematical point is indivisible.

------------------------------
Its a proof by contradiction, starting with premise 1

Proof

\aleph _L represents the cardinality of all possible consecutive closed intervals within the interval [0,1]

1- EVERY closed interval [A,B] on the real numbers line can be divided into infinite consecutive sub-intervals.

2- The interval [0,1] is a line segment on the real numbers line with length = 1
This unity line segment can be expressed as the sum of infinite consecutive smaller line segments between 0 and 1 as in the infinite series ( 1/2 + 1/4 + 1/8 + 1/16 +... +1/2ⁿ ) where n ---> \infty
Every term 1/2ⁿ in the infinite series is a line segment that can be represented as a closed sub-interval within the mother interval [0,1]

3- These infinite consecutive sub-intervals within the interval [0,1] can be put in a (one to one) correspondence with the set of integer numbers Z

4- The set of integer numbers Z has cardinality = \aleph₀

5- From 3 and 4, \aleph_L \geq \aleph₀

6- now, each sub-interval of those infinite sub-intervals within [0,1] is by itself a unique closed interval on the real numbers line that can be represented by a line segment and an infinite series, exactly as done with the mother interval [0,1] in step 2

7- From 6, every sub-interval within [0,1] has infinite sub-intervals that can be put in one-to-one correspondence with the set of integers Z

8- \aleph_L \geq ( \aleph₀ * \aleph₀)

9- now every new sub-sub-interval is again a unique closed interval on the real numbers line and can be represented by a line segment and an infinite series with cardinality = \aleph₀

10- from 9, \aleph_L \geq ( \aleph₀ * \aleph₀ * \aleph₀ )

11- Steps 9 and 10 repeat again and again, infinitely

12- \aleph_L \geq ( \aleph₀ * \aleph₀ * \aleph₀ * \aleph₀ ...)

13- \aleph_L \geq ( \aleph₀ ^ \aleph₀ )

14- ( \aleph₀ ^ \aleph₀ ) \geq ( 2 ^ \aleph₀ )

( 2 ^ \aleph₀ ) = \aleph₁ which is the continum , cardinality of all mathematical points on a line segment.

15- from 13,14 \aleph_L \geq \aleph₁

16- from 15, there are at least \aleph₁ closed sub-intervals within the mother interval [0,1] every one of them can be treated like the mother interval, which means that every one of them can be divided into at least N1 sub-intervals

17- from 16 \aleph_L \geq \aleph₁ * \aleph₁

18- the same process happen again and again which leads to \aleph_L \geq ( \aleph₁ * \aleph₁ * \aleph₁ ... )

19- \aleph_L \geq ( \aleph₁ ^ \aleph₀ )

20- from 19, there are at least ( \aleph₁ ^ \aleph₀ ) sub intervals, every one of them can be treated like the mother interval [0,1] which means every one of them can be divided too into ( \aleph₁ ^ \aleph₀ )

21- from 20, \aleph_L \geq [ ( \aleph₁ ^ \aleph₀ ) * ( \aleph₁ ^ \aleph₀ ) ]

22- the same process happen again and again which leads to \aleph_L \geq [ ( \aleph₁ ^ \aleph₀ ) * ( \aleph₁ ^ \aleph₀ ) * ( \aleph₁ ^ \aleph₀ ) ... ]

23- \aleph_L \geq ( \aleph₁ ^ \aleph₁ )

24-( \aleph₁ ^ \aleph₁ ) \geq ( 2 ^ \aleph₁ )

25- from 23 and 24, \aleph_L \geq ( 2 ^ \aleph₁ )

26- from 25, \aleph_L > \aleph₁

27- from 26, on a closed interval on the real numbers line, there can be more consecutive closed sub-intervals than mathematical points "continuum" on that interval, which is not the case.

28- Contradiction.

29- premise 1 is False

30- So, There exist interval [A,B] that can be divided only into Finite sub-intervals

31- From 30, those finite sub-intervlas cannot be continously devided otherwise that will result into infinite sub-intervals and the contradiction in 27

32- From 31 there exist sub-interval [A,B] that can Not be divided anymore

33- From 32, There exist interval [A,B] that is indivisible as a mathematical point is indivisible.

End

 

[CompuChip]

To be honest, unless I have misunderstood the question, your proof seems a little over-complicated.

The only interval without (proper) subintervals is [A, A]. If you consider [A, B] for A < B, then you can explicitly construct a subinterval such as [A + d/4, B - d/4] where d = B - A.

If A = B then you can easily show that [A, A] = {A}, and any interval [C, D] is either disjoint or a superset.

 

[said_alyami]

To be honest, unless I have misunderstood the question, your proof seems a little over-complicated.

The only interval without (proper) subintervals is [A, A]. If you consider [A, B] for A < B, then you can explicitly construct a subinterval such as [A + d/4, B - d/4] where d = B - A.

If A = B then you can easily show that [A, A] = {A}, and any interval [C, D] is either disjoint or a superset.

the proof shows this is not the case "its counter-intuitive ofcourse, but math does not care as we know"

in a nutshell, the proof shows one important notion:

" when you accept the premise that says: Every interval has a proper sub-interval, when you accept this as true, this leads to the contradiction: within interval [A,B] where A< B there can be more proper sub-intervals than all mathematical points within that interval"

which leads to the conclusion: "There must Exist interval [A,B] that does not have a proper sub-interval and A< B

 

[micromass]

Sadly, there are quite a lot of mistakes in your proof. I'll just mention one:

22- the same process happen again and again which leads to \aleph_L \geq [ ( \aleph₁ ^ \aleph₀ ) * ( \aleph₁ ^ \aleph₀ ) * ( \aleph₁ ^ \aleph₀ ) ... ]

23- \aleph_L \geq ( \aleph₁ ^ \aleph₁ )

I don't see why 23 follows from 22. It is simply not true that

\aleph_1^{\aleph_0}...\aleph_1^{\aleph_0}=\aleph_1^{\aleph_1}.

You make some other (understable mistakes): for example, if \kappa and \lambda are cardinals such that

\kappa\geq \lambda^n for all n

then this DOES NOT imply that

\kappa\geq \lambda^{\aleph_0}.

This sounds tempting to do, but it is incorrect...

If you want more information about cardinals and set theory, you should read the book "introduction to set theory" by Hrbacek and Jech. It's an excellent read!

 

[Fredrik]

the proof shows this is not the case
...
which leads to the conclusion: "There must Exist interval [A,B] that does not have a proper sub-interval and A< B

That doesn't make sense. If A<B, then, as CompuChip already mentioned, [A + d/4, B - d/4] with d=B-A is a proper sub-interval. This is very easy to prove. A (closed) interval is by definition a set [x,y]={r|x≤r≤y} with x≥y. So all we need to do is to prove that B-d/4≥A+d/4, and this isn't hard at all:

B-d/4=B-A+A-(B-A)/4=3(B-A)/4+A>(B-A)/4+A=A+d/4

This means that there must be a flaw somewhere in your proof. I haven't examined it, so I can't tell you where it is.

 

[CompuChip]

I just realized that this even works when you restrict yourself to the rationals, by the way :)

 

[chronon]

Step 11. If you repeat infinitely then you end up with points, not intervals. (and even if all of the intervals have rational endpoints, the points will be all of the reals in [0,1])