Consider Cantor space $2^\omega$ with the standard topology generated by open sets $[\sigma] = \{ \sigma^\frown x: x \in 2^\omega \}$. If $A \subseteq 2^{<\omega}$ and $x \in 2^\omega$, we say $A$ is *dense along $x$* if for every prefix $\sigma \prec x$, there is $\tau \succ \sigma$ such that all finite extensions of $\tau$ are in $A$.


An element $x \in 2^\omega$ is *1-generic* if, for every $\Sigma^0\_1$ (computably enumerable) set $A \subseteq 2^{<\omega}$ which is dense along $x$, we have $x \in [A]$ ($x$ is a path through $A$). I think this is the standard definition (from [here](http://rutgerkuyper.com/pubs/gendiff.pdf)).


Now, suppose $T \subseteq 2^{<\omega}$ is a tree. What conditions can we impose on $T$ that guarantee $[T]$ contains a 1-generic member? Effectively, I'm looking for some type of "generic basis theorem". In particular, if $T$ is infinite and $\Sigma^0\_2$, can we guarantee it contains a 1-generic path?



Finding 1-generic paths through a tree $T \subseteq 2^{<\omega}$

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> What conditions can we impose on $T$ that guarantee $[T]$ contains a 1-generic member?
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An element that is 1-generic *relative to $T$* will not be on $[T]$ unless $[T]$ contains a whole clopen cone $[\sigma]$.
Since "most" 1-generics are 1-generic relative to $T$, I suppose this means the condition to impose is basically that $[\sigma]\subseteq [T]$ for some $\sigma\in 2^{<\omega}$.



> 
> Effectively, I'm looking for some type of "generic basis theorem". In particular, if $T$ is infinite and $\Sigma^0\_2$, can we guarantee it contains a 1-generic path?
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No, if we let $T$ consist of all diagonally non-recursive $\{0,1\}$-valued functions then $T$ contains no 1-generic path. This is because one can show that no 1-generic computes a DNR function.


