Exact definition of algebraic set Let $k$ be a (algebraically closed) field, as far as I know, the algebraic sets in $k^n$ are the sets $S\subset k^n$ sustaining $V(I(S))=S$,
but I also heard that algebraic sets are the sets sustaining $S=V(T)$ for some $T\triangleleft k[x_1,x_2,...x_n]$;
It is obvious that the first definition implies the second one, but for the second one implies the first one I'm not sure if my proof is correct:
$$V(I(S))=V(I(V(T)))=V(\sqrt{T})$$
It is known that $I:\{\text{algebraic sets (by the second definition)}\}\to\{\text{radical ideals of } k[x_1,...,x_n]\}$ is injective, hence because:
$$I(V(T))=\sqrt{T}=I(V(\sqrt{T}))$$
We get $V(T)=V(\sqrt{T})$, therefore
$$V(I(S))=V(I(V(T)))=V(\sqrt{T})=V(T)=S$$
Therefore $S$ is algebraic by the first definition.
$\blacksquare$
 A: I don't think you need you should need to use the nullstellensatz to prove these two definitions are equivalent.
Given sets in $k^n$ the Zariski topology meaning the closed sets $Z$ are exactly the vanishing locus of some set of polynomials $V(I)$ i.e. an algebraic set according to your second definition. Notice that we can take $I$ to be an ideal because for any set of polynomials $H \subset k[x_1, \dots, x_n]$ we have $V(H) = V((H))$ where $(H)$ is the ideal generated by $H$.
Now I claim that for any set $S \subset k^n$ the following formula holds,
$$ V(I(S)) = \overline{S} $$
where $\overline{S}$ is the closure of $S$ in the Zariski topology. Let's prove this. By definition, $V(I(S))$ is closed and $V(I(S)) \supset S$ so $V(I(S)) \supset \overline{S}$. Conversely, let $C$ be a Zariski closed subset containing $S$. Then, $C = V(T)$ and $S \subset V(T)$ means by definition that every $f \in T$ vanishes on $S$ so $T \subset I(S)$. Therefore,
$$ V(I(S)) \subset V(T) = C $$
because if $x \in V(I(S))$ then every $f \in I(S)$ has $f(x) = 0$ and in particular for $f \in T \subset I(S)$ we have $f(x) = 0$ so $x \in V(T) = C$. Therefore, $V(I(S)) \subset \overline{S}$ because it is contained in every Zariski closed set containing $S$.
Now this formula immediately implies what you want. Indeed, let $S$ be Zariski-closed or equivalently an algebraic set according to the second definition. Then $\overline{S} = S$ so,
$$ V(I(S)) = S $$
proving that $S$ is algebraic according to the first definition.
This does not even require that the field is algebraically closed while the correspondence you cite (using the nullstellensatz) most certainly does.
