Proving \epsilon_0 is the Smallest Epsilon Number: A Step-by-Step Guide

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In summary, we can prove that \epsilon_0 is an \epsilon number by showing that it satisfies the equation \omega^{\epsilon_0}=\epsilon_0 and that it is the smallest number to do so. To prove this, we can use the fact that \omega^{\epsilon_0}=\lim_{n<\omega}\phi(n) where \phi(n)=\omega^{\omega^{\omega^{...^{\omega}}}} with \omega appearing n times. We can also use the fact that for any \epsilon number, e, e=\omega^e and e must be greater than w, w^w, w^w, w^w^w, etc, which is precisely
  • #1
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prove that [tex]\epsilon_0[/tex] is an [tex]\epsilon[/tex] number and that it's the smallest number.
[tex]\epsilon_0=\lim_{n<\omega}\phi(n)[/tex]
[tex]\phi(n)=\omega^{\omega^{\omega^{...^{\omega}}}}[/tex] where [tex]\omega[/tex] appears n times.
an epsilon number is a number which satisfies the equation [tex]\omega^{\epsilon}=\epsilon[/tex].
for the first part of proving that it's an epsilon number i used the fact that [tex]1+\omega=\omega[/tex], for the second part I am not sure i understand how to prove it:
i mean if we assume there's a number smaller than [tex]\epsilon_0[/tex] that satisfy that it's an epsilon number, then [tex]\omega^{\epsilon^{'}}=\epsilon^{'}<\epsilon_0=\omega^{\epsilon_0}[/tex]
i know that there exists a unique ordinal such that [tex]\epsilon_0=\epsilon^{'}+\alpha[/tex]
if [tex]\alpha[/tex] is a finite ordinal then [tex]\epsilon_0=\epsilon^{'}[/tex] and it's a contradiction, but how to prove it when alpha isn't a finite ordinal?
 
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I don't know if this is at all valid in the theory of ordinals, but surely lims and exponentials commute, hence e_0 is an epsilon number.

Secondly, if e is an epsilon number then e=w^e (=>w) = w^w^e (=>w^w) =... hence e must be greater than w, w^w, w^w, w^w^w,.. and therefore e must be greater thanor equal to the smallest ordinal larger than all of w, w^w, w^w^w, etc which is precisely e_0.

(This is exactly the same as showing that if t is larger than 0.9, 0.99, 0.999,... then t=>1.)
 
  • #3


To prove that \epsilon_0 is an epsilon number, we need to show that it satisfies the equation \omega^{\epsilon_0} = \epsilon_0. We can rewrite this equation as \omega^{\omega^{\epsilon_0}} = \epsilon_0, since \epsilon_0 = \lim_{n<\omega}\phi(n) and \phi(n) = \omega^{\omega^{\omega^{...^{\omega}}}} where \omega appears n times.

Now, let's consider the expression \omega^{\omega^{\omega^{...^{\omega}}}}} where \omega appears \epsilon_0 times. This can be written as \omega^{\omega^{\omega^{...^{\omega}}}}} = \omega^{\omega^{\epsilon_0}} = \epsilon_0. This shows that \epsilon_0 satisfies the equation \omega^{\epsilon_0} = \epsilon_0, and therefore, it is an epsilon number.

To prove that \epsilon_0 is the smallest epsilon number, we need to show that there is no other epsilon number that is smaller than \epsilon_0. Let's assume that there exists an epsilon number \epsilon^{'} such that \epsilon^{'} < \epsilon_0 and \omega^{\epsilon^{'}} = \epsilon^{'}.

Since \epsilon^{'} < \epsilon_0, we know that \epsilon^{'} < \omega^{\epsilon_0}. Using the fact that 1+\omega = \omega, we can rewrite this as \epsilon^{'} < 1 + \omega^{\epsilon_0} = \omega^{\epsilon_0}. This means that \omega^{\epsilon_0} is an upper bound for the set of all epsilon numbers smaller than \epsilon_0.

But, we know that \epsilon_0 = \lim_{n<\omega}\phi(n) = \omega^{\omega^{\epsilon_0}}. This means that \omega^{\epsilon_0} is the smallest upper bound for the set of all epsilon numbers smaller than \epsilon_0. Therefore, \omega^{\epsilon_0} must be the smallest epsilon number, which contradicts our assumption that \epsilon^{'} < \epsilon_0.

Hence, there cannot exist an epsilon number smaller than \epsilon_0, and therefore, \epsilon_0 is the smallest
 

1. What is the significance of proving \epsilon_0 is the smallest epsilon number?

The concept of epsilon numbers is important in the field of set theory and mathematical logic. Proving that \epsilon_0 is the smallest of these numbers helps to establish a foundation for understanding the infinite and transfinite numbers that are used in mathematical proofs.

2. How is the proof of \epsilon_0 being the smallest epsilon number achieved?

The proof involves showing that any candidate for a smaller epsilon number would lead to a contradiction, thus proving that \epsilon_0 is indeed the smallest. This is done by using the definition of an epsilon number and basic principles of set theory.

3. Can you explain the concept of epsilon numbers in more detail?

Epsilon numbers are a type of transfinite number that are used to describe the size of infinite sets. They are defined as the smallest ordinal number that cannot be expressed as the sum of a smaller number of smaller ordinal numbers. Epsilon numbers are important in understanding the hierarchy of infinite numbers.

4. What is the practical application of proving \epsilon_0 is the smallest epsilon number?

While the concept of epsilon numbers may seem abstract, they have practical applications in fields such as computer science and artificial intelligence. Understanding the properties of epsilon numbers can help in the development of algorithms and decision-making processes.

5. Are there any challenges or controversies surrounding this proof?

There have been debates and discussions surrounding the concept of epsilon numbers and the proof of \epsilon_0 being the smallest. Some mathematicians argue that there is a flaw in the proof, while others believe it is sound. Further research and exploration of this topic continue to be a subject of interest in the mathematical community.

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