What guarantees the existence of smallest infinite sets? I have seen definitions like this in several places

Define $\Phi$ to be the smallest set that such that 
  
  
*
  
*Whenever $p \in \mathcal{P}$, we have $p \in \Phi$
  
*Whenever $\alpha \in \Phi$, we have $\neg \alpha \in \Phi$
  
*Whenever $\alpha, \beta \in \Phi$, we have $(\alpha \implies \beta) \in \Phi$
  

What guarantees the existence of a smallest set? Why can't there be a family of sets satisfying all the conditions such that given any set of the family, there exists a set smaller than it? 
 A: In general, just saying "the smallest set that satisfies such and such condition" can be dangerous, but in your particular case, $\Phi$ is well defined, because if $\{A_i\}_{i\in I}$ is any family of sets that satisfies the three conditions, then
$$A=\bigcap_{i\in I} A_i$$
also satisfies all three conditions, and $A$ is smaller than all the sets in the family.

To prove the claim I made:
Say $A_i$ satisfies all three conditions for all $i\in I$. Then, let's say that $p\in \mathcal P$. 
Let $i\in I$. Then, because $p\in \mathcal P$, we have $p\in A_i$ (because $A_i$ satisfies condition 1). But because $i$ was arbitrary, we know that $$\forall i\in I: p\in A_i$$
which is the same as $p\in A$. So, $A$ satisfies condition 1.

Now let's say $\alpha\in A$. Let $i\in I$. 
From $\alpha\in \bigcap A_i$, we know that $\alpha \in A_i$. Since $A_i$ satisfies condition 2, we know $\neg \alpha \in A_i$. Therefore, since $i$ was arbitrary, we have $$\forall i \in I: \neg \alpha \in A$$ or, in other words, $\neg\alpha\in A$. So, $A$ satisfies condition 2.

Say $\alpha, \beta\in\Phi$. Then again, $\alpha, \beta\in A_i$, therefore $\alpha\implies\beta\in A_i$, and therefore $\alpha\implies\beta\in A$, so $A$ satisfies condition 3.
