Isn't the given characterisation of recursively enumerable subsets of the class of all recursively enumerable languages?

Question

$S$ is a subset of the class of all recursively enumerable languages over some finite symbols then $S$ is recursively enumerable iff

1. If $L$ is in $S$ and $L'$ is a language such that $L ⊆ L'$ and $L'$ is recursively enumerable, then $L'$ is in $S$
2. If $L$ is an infinite language in $S$, then there exists at least one finite subset of $L$ that is in $S$
3. The set of all finite languages in $S$ is enumerable, i.e. a Turing machine can list all the finite languages in $S$

Source of the statement: https://cs.stackexchange.com/q/2322 and some online notes

Isn't the 3rd one contradictory to the 1st one?

Answer 1

No, it is not contradictory. Both conditions are positive in the sense that they "put things in $S$", so they cannot be at odds with each other.

There is an equivalent formulation of $S$: there is an r.e. set $B$ of finite languages such that $$S = \{L \mid \text{$L$ is r.e. and $\exists L_0 \in B \,.\, L_0 \subseteq L$}\}.$$ In words: $S$ is generated by an r.e. set of finite language $B$ in the following way: $S$ contains precisely all r.e. languages which contain an element of $B$.