Is recursive join of a sequence of low sets also low?

Question

A set $A$ is low when $\deg(A)'\leq 0'$.

Suppose we have a sequence of low sets $(A_i)_{i\in\omega}$ such that for every $n\in\omega$ we have $$\deg(\bigoplus_{i<n}A_i)'\leq 0'$$

Let $A=\bigoplus_{i\in\omega}A_i$.

Is it also true that $\deg(A)'\leq 0'$ i.e. is it true that $A$ is low?

Answer 1

No.

Let $\{A_i\}_{i \in \omega}$. My definition of the infinite join $\bigoplus A_i$ is $\langle i, x \rangle \in \bigoplus A_i$ if and only if $x \in A_i$.

Now let $A_i = \begin{cases} \{1\} & \quad i \in \emptyset' \\ \{0\} & \quad i \notin \emptyset' \end{cases}$

It is clear that for each $i$, $A_i$ is low since $A_i$ is computable. But $\bigoplus A_i \equiv_T \emptyset'$, so the join is not low.

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This example shows that extra information is coded into the sequence. A good question may be if $f : \omega \rightarrow \omega$ is a computable function such that each $W_{f(i)}$ are low c.e. sets. If $A_i = W_{f(i)}$, then is $\bigoplus A_i$ low.