(I have bolded my principal question $\downarrow$).
Reading through Algebraic Curves by W. Fulton I've stumbled upon a couple of problems on proposition three from section 4.3 (pg. 48).
A short recap of the definitions:
For algebraic $V\subseteq \mathbb{A}^n$ define $I^*=\langle\{f^*\in k[X_1,\ldots,X_{n+1}]\mid f\in I(V)\}\rangle$.
For algebraic $V\subseteq \mathbb{P}^n$ define $I_*=\{F_*\in \mid F\in I(V)\}$.
$H_\infty = \{[x_1,\ldots,x_{n+1}]\in \mathbb{P}^n\mid x_{n+1}=0\}$
Where $I(V)$ represents the polyonimials that vanish on $V$, and $V(I)$ the zeroes of the ideal $I$; and $f^*$ and $F_*$ represent homogenization and dehomogenization respectively.
Now, point 7 states that
If no irreducible component of $V\subseteq \mathbb{P}^n$ contains or is contained in $H_\infty$, then $V_*\subsetneq \mathbb{A}^n$ and $(V_*)^*=V$.
In this argument, I don't understand why it's only necessary to prove the result for irreducible $V$ (the proof with this assumption I undersntad). Even if the result is true for the irreducible componentes of a set $V=\cup_i V_i$, it could happen that every $V_i$ is proper but their union is $\mathbb{A}^n$. I don't see how to fill this gap.
Edit: @math54321's comment has solved the $V\subsetneq\mathbb{A}^n$ part for me, I'm still stuck on $(V_*)^*=V$.
It would be nice to see an example where $V$ contains $H_\infty$ but none of it's irreducible componets
On point 6, which states
If $V\subseteq \mathbb{A}^n$ is non empty, then no component of $V^*$ lies in or contains $H_\infty$
I can't find the point in the proof where the hypothesis of irreducibility is used. To me it seems to be proving that the projective closure of any proper algebraic set neither contains nor is contained in $H_\infty$.
I haven't been able to proof the supposedly very simple $(V^*)_*=V$ from point 1.
Finally, if this were true it would solve a couple of my problems:
$$V=\bigcup^N V_i \implies V_* = \bigcup^N {V_i}_*$$
but I can't prove left $\subseteq$ right.
I'll be really grateful for any help, even very partial. Thanks!