Prove, that $A^+\subseteq A^*$ where $A$ is a formal language Prove, that $A^+\subseteq A^*$, where $A$ is a formal language.


The definition of $A^+$ is $\bigcup_{i\in\mathbb{N}\setminus \{0\}}A^i$,  which would be $A^1 \cup A^2 \cup \dots \cup A^i$. Likewise, $A^*$ is $\bigcup_{i\in\mathbb{N}}A^i$ and therefore $A^0 \cup A^1 \cup \dots \cup A^i$. 

Suppose, $x\in A^+$. By definition $x\in A^+ \land x\in  A^*$ for all $x\neq A^0$, which is the set of the empty word. We can conclude, that $A^1\cup A^2\cup \dots \cup A^i \subseteq A^0\cup A^1\cup \dots \cup A^i$ and $\bigcup_{i\in\mathbb{N}\setminus \{0\}}A^i \subseteq \bigcup_{i\in\mathbb{N}}A^i$, that's why $A^+\subseteq A^*$.

Is this "proof" ok?
It seems you have hit the reasoning correctly:
A set $S$ is a subset of another set $A$ if $w\in S\implies w\in A$.
Since you are trying to show that $\forall w \in A^+ \implies w \in A^*$, you are proving that $ A^+ \subseteq A^*$.

One more legant way to formally do this is by finding a bijection between the two sets.
Since $|A^*|$ = $|\Bbb N|$ you need to prove that $|A^+|$ is a subset of  $|\Bbb N|$. So you need to find a Bijection (left to you since we dont do Homework for the OP's) from these two sets.
Note that there exists $2^{\aleph_0}$ such functions, so 
 you have a large selection space :)