I'm a computer science student and during the lecture, the lecturer prove that a set $\cup_{n=0}^\infty A_n \subseteq B$ by induction on $n$. i.e $\forall n$ $A_n \subseteq B$
I remember from my mathematic for computer science that kind of induction doesn't hold.
Who wrong?
Induction allows you to prove (in a certain formal way that often isn't needed in ordinary mathematical proofs) that the infinitely many statements $A_0 \subseteq B,$ $A_1 \subseteq B,$ $A_2 \subseteq B, \, \ldots$ all hold, and hence that $(\forall n)(A_n \subseteq B)$ holds.
However, once you have this, it follows easily from the definition of union that $\cup_{n=0}^\infty A_n \subseteq B$ holds:
$x \in \cup_{n=0}^\infty A_n$ implies there exists $N$ such that $x \in A_N$ (by definition of union), and hence from the already established $A_N \subseteq B$ we have $x \in B.$ Thus, we have $x \in \cup_{n=0}^\infty A_n$ implies $x \in B,$ and hence (by definition of subset) $\cup_{n=0}^\infty A_n \subseteq B.$
That said, it sounds like the lecturer is confused about induction, because it doesn't establish an "infinite case" from all finite cases. Instead, it simply allows you to "simultaneously in a finite amount of time, so to speak" prove all infinitely many of the finite cases.
