# Show that if $A$ and $B$ are recognizable subsets of $\Sigma^*$ then so is $A \cup B$, $A \cap B$ and also $\Sigma^* - A$

I am working through a chapter of a book by Samuel Eilenberg about Automata, Languages and Machines as part of an university course in computer science.

And as an exercise in this chapter I have to show the statement above. I have to use two propositions that were shown in the book earlier.

Let $$A$$ be a subset of $$\Sigma^*$$ then the cardinality of this family of sets: $$\{\ s^{-1}A\ \vert \ s \in \Sigma^*\}$$ is called $$\alpha ^{e}A$$.

$$\ s^{-1}A = \{x \ \vert \ sx \in A \}$$

With $$\varnothing$$ deleted the cardinality of the family of sets is called $$\alpha A$$.

Any subset $$A$$ of $$\Sigma^*$$ is recognizable iff this family of sets is finite.

The first proposition is the following:

Let $$A$$ be a recognizable subset of $$\Sigma^*$$, and let $$\mathfrak A = (Q, i, T)$$ be a deterministic (resp. complete) $$\Sigma$$-Automaton such that $$\vert \mathfrak A \vert = A$$ (Meaning that $$\mathfrak A$$ recognizes $$A$$.), then $$card \ Q \ge \alpha A$$ (resp. $$card \ Q \ge \alpha ^{e}A$$). With equality holding if and only if $$\mathfrak A$$ is minimal (resp. complete minimal).

The second proposition is the following:

If $$f\colon \ \Sigma^* \to \Gamma^*$$ is a morphism and $$B$$ is a recognizable subset of $$\Gamma^*$$, then $$A = Bf^{-1}$$ is a recognizable subset of $$\Sigma^*$$ and $$\alpha A \le \alpha B$$.

I think the first proposition explains it for $$C = A \cup B$$ and $$D = A \cap B$$ since $$\alpha C \le \alpha A + \alpha B$$ and $$\alpha D \le \alpha A$$ or $$\alpha D \le \alpha B$$.

For $$E = \Sigma^* - A$$ I dont understand how to show that it's true. I guess I have to use the second proposition because so far I didn't need it. Anyway somehow I have to derive bounds for $$\alpha E$$ to show that it is recognizable.