<p>The union is not a good idea because you cannot tell, if a number is in the union, whether the number got in because it is an element of <span class="math-container">$0^{(n)}$</span> or <span class="math-container">$0^{(n+1)}$</span>, etc.  Instead, if you take the union of the <span class="math-container">$&lt;n,a&gt;$</span> where <span class="math-container">$a$</span> is an element of <span class="math-container">$0^{(n)}$</span> then you can keep them separate.</p>

CC BY-SA 4.0
2020-05-05T21:18:41.387Z
2020-05-05T21:18:41.387Z
Jim Hefferon
<p>In the book Computability theory (Rebecca Weber) I stumbled about Exercise 7.1.24 with the definition of the omega jump.
The book says:
The omega jump of <span class="math-container">$A$</span>, <span class="math-container">$A^{(\omega)}$</span> is the join of all <span class="math-container">$A^{(n)}$</span> for <span class="math-container">$n \in \mathbb{N}$</span>: <span class="math-container">$A^{(\omega)}=\{\langle x, n \rangle : x \in A^{(n)}\}$</span>. <br/>
I tried to understand this definition but i couldn't. I think I know that the jump is but I dont understand the omega jump. Is it simply <span class="math-container">$A^{(\omega)} = \cup_{n \in \mathbb{N}} A^{(n)}$</span>? Why (not) ? <br/>
Can someone give me an equivalent definition or explain it to me?</p>

<computability>
Definition of Omega Jump
GMiiX
# Definition of Omega Jump

(Asked by: username_0 on 2020-05-04)

In the book Computability theory (Rebecca Weber) I stumbled about Exercise 7.1.24 with the definition of the omega jump.
The book says:
The omega jump of $A$, $A^{(\omega)}$ is the join of all $A^{(n)}$ for $n \in \mathbb{N}$: $A^{(\omega)}=\{\langle x, n \rangle : x \in A^{(n)}\}$.   
I tried to understand this definition but i couldn't. I think I know that the jump is but I dont understand the omega jump. Is it simply $A^{(\omega)} = \cup\_{n \in \mathbb{N}} A^{(n)}$? Why (not) ?   
Can someone give me an equivalent definition or explain it to me?

**Tags:** computability

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## 1 answers

### Answer by username_1 on 2020-05-05. Score: 1

The union is not a good idea because you cannot tell, if a number is in the union, whether the number got in because it is an element of $0^{(n)}$ or $0^{(n+1)}$, etc. Instead, if you take the union of the $$ where $a$ is an element of $0^{(n)}$ then you can keep them separate.


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