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Is it true that a finite set of points in $\mathbb{A}_k^3$ always an algebraic set is?

Take for example $\{(a,b,c)\}\subset\mathbb{A}^3$. Is there a polynomial $f(x,y,z)$ with $f(x,y,z)=0$ exactly in $(a,b,c)$? $(x-a)(y-b)(z-c)$ doesn't work since $(a,0,0)$ is also a zero.

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    What exactly is your definition of an algebraic set? – 2017-02-28
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    You have to take the three functions $x-a$, $x-b$ and $x-c$. – 2017-02-28
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    The definition is: There is a set of polynomials $J$ with $V=V(J)$. What if there is more than one point? – 2017-02-28

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Let $X = \{x_i \mid 1 \leq i \leq n\}$ be a finite set in $\Bbb A^3(k)$, with $x_i=(a_i,b_i,c_i)$.

Consider the ideals $J_i = (X-a_i,Y-b_i,Z-c_i)$ and $J = J_1 \cap \cdots \cap J_n$ of $k[X,Y,Z]$. Then $$V(J) = \bigcup_{i=1}^n V(J_i) = \bigcup_{i=1}^n \{x_i\} = X$$ is an affine algebraic set.

This actually shows that Zariski's topology on $\Bbb A^r(k)$ is a T1 space, for any $r \geq 1$.