# Definition of $S_1(A,B)$

The definition of first selection principle is well known: $S_1(A,B)$. Let $A$ and $B$ be families of sets. The symbol $S_1(A, B)$ denotes the statement: for each sequence $(A_n : n \in \omega)$ of elements of $A$ there is a sequence $(x_n : n\in \omega)$ such that each $x_n$ is an element of $A_n$, and $\{x_n : n\in \omega\}$ is an element of $B$.

Can we assume that $x_{n_k}=\emptyset$ for $k\in \omega$ ?

• You say $x_n$ is an element of $A_n$. I would think that in most cases $\emptyset$ is not an element of $A_n$. – Gerald Edgar Jul 22 '17 at 17:53
• Thank you. It seems that it is so. It is interesting that in the definition $S_{fin}(A,B)$ we can choose empty subsets $X_n\subset A_n$. – Alexander Osipov Jul 22 '17 at 18:22
• If $A$ and $B$ are families of subsets of a set $S$ and $B$ is "closed upward", then I think $S_1(A,B)$ is equivalent to the following: for each sequence $(A_n:n\in\omega)$ of elements of $A$ there is a sequence $(B_n:n\in\omega)$ such that $B_n\subset A_n$, $|B_n|\leq 1$ and $\bigcup_{n\in\omega}B_n\in B$. – Renan Maneli Mezabarba Jul 22 '17 at 18:35
• Yes, in this interpretation we get that the answer is positive.... Since $B_{n_k}$ can be empty .... – Alexander Osipov Jul 22 '17 at 18:46
• If I understood @GeraldEdgar's point properly, he was pointing out that you were asking whether you could take the elements $x_n$ to be empty, which seems "ill typed" (a priori they are just elements, not sets). Also, you have only defined $x_n$ in your statement; what is $x_{n_k}$? – LSpice Jul 22 '17 at 19:21

## 1 Answer

I understand your question to mean "must we chose an element $x_n$ for all $n$?". (This is in accordance with the comments.)

Answer. Officially, the answer is "Yes", you must chose an element for each $n$. However, as pointed out in the comments, often you can prove this makes no difference if you are allowed to not chose elements. This is true for all classic selective covering properties of type $S_1(A,B)$.

For some properties this is not entirely trivial, but still provable. For example, $S_1(\Gamma,\Gamma)=S_{\le 1}(\Gamma,\Gamma)$ because WLOG the covers get finer with $n$. If you think about it, this is the essence of the now-classic proof that $S_1(\Gamma,\Gamma)=S_\text{fin}(\Gamma,\Gamma)$.

• Thanks! I was more interested the principle $S_1(\Omega, \Gamma)$, but i find Gerlits-Nagy's result that $S_1(\Omega, \Gamma)={ \Omega\choose \Gamma}$. – Alexander Osipov Jul 23 '17 at 6:48
• If I understood correctly, then this is true for any principle in The Scheepers Diagram ( Including the Borel covers ) ? – Alexander Osipov Jul 23 '17 at 6:54
• @AlexanderOsipov Yes, the proof for $S_1(\Omega,\Gamma)$ is the same. Indeed, in courses I prove for $S_1(\Gamma,\Gamma)$ and then deduce for $S_1(\Omega,\Gamma)$ as an immediate consequence. – Boaz Tsaban Jul 23 '17 at 10:54
• @AlexanderOsipov Yes, this holds for all of the Scheepers diagram, and also in the Borel case. Same proofs work. – Boaz Tsaban Jul 23 '17 at 10:56