# 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.

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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 = $$\aleph$$0

5- From 3 and 4, $$\aleph$$L $$\geq$$ $$\aleph$$0

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 $$\geq$$ ( $$\aleph$$0 * $$\aleph$$0)

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 = $$\aleph$$0

10- from 9, $$\aleph$$L $$\geq$$ ( $$\aleph$$0 * $$\aleph$$0 * $$\aleph$$0 )

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

12- $$\aleph$$L $$\geq$$ ( $$\aleph$$0 * $$\aleph$$0 * $$\aleph$$0 * $$\aleph$$0 ...)

13- $$\aleph$$L $$\geq$$ ( $$\aleph$$0 ^ $$\aleph$$0 )

14- ( $$\aleph$$0 ^ $$\aleph$$0 ) $$\geq$$ ( 2 ^ $$\aleph$$0 )

( 2 ^ $$\aleph$$0 ) = $$\aleph$$1 which is the continum , cardinality of all mathematical points on a line segment.

15- from 13,14 $$\aleph$$L $$\geq$$ $$\aleph$$1

16- from 15, there are at least $$\aleph$$1 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 $$\geq$$ $$\aleph$$1 * $$\aleph$$1

18- the same process happen again and again which leads to $$\aleph$$L $$\geq$$ ( $$\aleph$$1 * $$\aleph$$1 * $$\aleph$$1 ... )

19- $$\aleph$$L $$\geq$$ ( $$\aleph$$1 ^ $$\aleph$$0 )

20- from 19, there are at least ( $$\aleph$$1 ^ $$\aleph$$0 ) 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 ( $$\aleph$$1 ^ $$\aleph$$0 )

21- from 20, $$\aleph$$L $$\geq$$ [ ( $$\aleph$$1 ^ $$\aleph$$0 ) * ( $$\aleph$$1 ^ $$\aleph$$0 ) ]

22- the same process happen again and again which leads to $$\aleph$$L $$\geq$$ [ ( $$\aleph$$1 ^ $$\aleph$$0 ) * ( $$\aleph$$1 ^ $$\aleph$$0 ) * ( $$\aleph$$1 ^ $$\aleph$$0 ) ... ]

23- $$\aleph$$L $$\geq$$ ( $$\aleph$$1 ^ $$\aleph$$1 )

24-( $$\aleph$$1 ^ $$\aleph$$1 ) $$\geq$$ ( 2 ^ $$\aleph$$1 )

25- from 23 and 24, $$\aleph$$L $$\geq$$ ( 2 ^ $$\aleph$$1 )

26- from 25, $$\aleph$$L > $$\aleph$$1

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.

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

Homework Helper
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

Staff Emeritus
Homework Helper
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 $$\geq$$ [ ( $$\aleph$$1 ^ $$\aleph$$0 ) * ( $$\aleph$$1 ^ $$\aleph$$0 ) * ( $$\aleph$$1 ^ $$\aleph$$0 ) ... ]

23- $$\aleph$$L $$\geq$$ ( $$\aleph$$1 ^ $$\aleph$$1 )

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...

Staff Emeritus
Gold Member
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.