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## Main Question or Discussion Point

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

[tex]\aleph[/tex]

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 +.... +1/2

Every term 1/2

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 = [tex]\aleph[/tex]

5- From 3 and 4, [tex]\aleph[/tex]

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- [tex]\aleph[/tex]

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 = [tex]\aleph[/tex]

10- from 9, [tex]\aleph[/tex]

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

12- [tex]\aleph[/tex]

13- [tex]\aleph[/tex]

14- ( [tex]\aleph[/tex]

( 2 ^ [tex]\aleph[/tex]

15- from 13,14 [tex]\aleph[/tex]

16- from 15, there are at least [tex]\aleph[/tex]

17- from 16 [tex]\aleph[/tex]

18- the same process happen again and again which leads to [tex]\aleph[/tex]

19- [tex]\aleph[/tex]

20- from 19, there are at least ( [tex]\aleph[/tex]

21- from 20, [tex]\aleph[/tex]

22- the same process happen again and again which leads to [tex]\aleph[/tex]

23- [tex]\aleph[/tex]

24-( [tex]\aleph[/tex]

25- from 23 and 24, [tex]\aleph[/tex]

26- from 25, [tex]\aleph[/tex]

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

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**[tex]\aleph[/tex]

_{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 +.... +1/2

^{n}) where n ---> [tex]\infty[/tex]Every term 1/2

^{n}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 = [tex]\aleph[/tex]

_{0}5- From 3 and 4, [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] [tex]\aleph[/tex]_{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- [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] ( [tex]\aleph[/tex]_{0}* [tex]\aleph[/tex]_{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 = [tex]\aleph[/tex]

_{0}10- from 9, [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] ( [tex]\aleph[/tex]_{0}* [tex]\aleph[/tex]_{0}* [tex]\aleph[/tex]_{0})11- Steps 9 and 10 repeat again and again, infinitely

12- [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] ( [tex]\aleph[/tex]_{0}* [tex]\aleph[/tex]_{0}* [tex]\aleph[/tex]_{0}* [tex]\aleph[/tex]_{0}...)13- [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] ( [tex]\aleph[/tex]_{0}^ [tex]\aleph[/tex]_{0})14- ( [tex]\aleph[/tex]

_{0}^ [tex]\aleph[/tex]_{0}) [tex]\geq[/tex] ( 2 ^ [tex]\aleph[/tex]_{0})( 2 ^ [tex]\aleph[/tex]

_{0}) = [tex]\aleph[/tex]_{1}which is the continum , cardinality of all mathematical points on a line segment.15- from 13,14 [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] [tex]\aleph[/tex]_{1}16- from 15, there are at least [tex]\aleph[/tex]

_{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-intervals17- from 16 [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] [tex]\aleph[/tex]_{1}* [tex]\aleph[/tex]_{1}18- the same process happen again and again which leads to [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] ( [tex]\aleph[/tex]_{1}* [tex]\aleph[/tex]_{1}* [tex]\aleph[/tex]_{1}... )19- [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] ( [tex]\aleph[/tex]_{1}^ [tex]\aleph[/tex]_{0})20- from 19, there are at least ( [tex]\aleph[/tex]

_{1}^ [tex]\aleph[/tex]_{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 ( [tex]\aleph[/tex]_{1}^ [tex]\aleph[/tex]_{0})21- from 20, [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] [ ( [tex]\aleph[/tex]_{1}^ [tex]\aleph[/tex]_{0}) * ( [tex]\aleph[/tex]_{1}^ [tex]\aleph[/tex]_{0}) ]22- the same process happen again and again which leads to [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] [ ( [tex]\aleph[/tex]_{1}^ [tex]\aleph[/tex]_{0}) * ( [tex]\aleph[/tex]_{1}^ [tex]\aleph[/tex]_{0}) * ( [tex]\aleph[/tex]_{1}^ [tex]\aleph[/tex]_{0}) ... ]23- [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] ( [tex]\aleph[/tex]_{1}^ [tex]\aleph[/tex]_{1})24-( [tex]\aleph[/tex]

_{1}^ [tex]\aleph[/tex]_{1}) [tex]\geq[/tex] ( 2 ^ [tex]\aleph[/tex]_{1})25- from 23 and 24, [tex]\aleph[/tex]

_{L}[tex]\geq[/tex] ( 2 ^ [tex]\aleph[/tex]_{1})26- from 25, [tex]\aleph[/tex]

_{L}> [tex]\aleph[/tex]_{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.

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