Is it true that for any set $X,Y \subseteq \Bbb A^n(k)$, $I(X) = I(Y )$ iff $X = Y$ . Is it true that for any set $X,Y \subseteq \Bbb A^n(k)$, $I(X) = I(Y )$ iff $X = Y$ . Because I know $I(X) = I(Y )$ implies $\overline X=\overline Y$ But I am thinking that this is not true because if I take $Y=\overline X$ then $I(X) = I(\overline X )$ but $X \neq \overline X$ always.
But in this paper http://www.math.uchicago.edu/~may/VIGRE/VIGRE2010/REUPapers/Bloom.pdf 
I am getting one proof The following are the first basic properties of these objects, and these properties
are easy to verify.
(1) If $X ⊂ Y$ , then $I(X) ⊃ I(Y )$, so this correspondence is inclusion-reversing.
Moreover, $I(X) = I(Y ) ⇐⇒ X = Y $.
We give a construction which will be useful later:
Proposition 1.9. Let $P_1, P_2, . . . , P_n$ be distinct points in $A^
n$. Then, $\exists$
polynomials $F_i ∈ k[X_1, X_2, . . . , X_n]$ such that $F_i(P_j ) = δ_{ij} =0$ if $i \neq j$
and $1$ if $i = j$.
Proof. For each $i$, let $V_i = \{P_1, . . . , P_{i−1}, P_{i+1}, . . . , P_n\}$. Then, from the first
property above, $V_i \subsetneq V_i ∪ \{P_i\}$ implies $I(V_i) \supsetneq I (V_i ∪ \{P_i\})$. Pick $G_i ∈
I (V_i ∪ {P_i}) - I(V_i)$. $G_i$
is zero on $V_i$
, but is non-zero on $P_i$
. Thus, our desired
polynomials are $F_i =\frac 1{G_i(P_i)}Gi$.
So this whole proposition is false right?

The correspondence you mentioned, $X = Y$ if and only if $I(X) = I(Y)$, is only valid when $X, Y$ are assumed to be closed subsets of $\mathbb{A}^n(k)$.  In the proposition you mentioned, all the sets which they were dealing with were closed, so there is no trouble there.
For an example when $I(X) = I(Y)$ holds, but $X \neq Y$, take $n = 1, X = \mathbb{A}^1(k)$, and $Y$ any proper subset of $X$ which has infinitely many elements.  Since $X$ and $Y$ are both infinite, $I(X) = I(Y) = (0)$, because a nonzero polynomial over a field in one variable cannot have infinitely many roots.
