First let's try the following lemma:
Lemma: If $ k $ is not algebraically closed field and $n\in{\mathbb{N}}$, then there exists $f\in{k[x_ {1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $V(f)=\{(0,0,\dots,0,0)\}\subseteq{\mathbb{A}_{k}^{n}}$.
Proof: Let $k^{a}$ the algebraic closure of $k$. (Reasoning Inductively) First we note that if $n=1$ then there exists $f(x_{1})=x_{1}$ in $k[x_{1}]$ such that $V(f)=\{0\}\subseteq{\mathbb{A}_{k}^{1}}$. By induction hypothesis suppose there is $g(x)\in{k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $V(g)=\{(0,0,\ dots,0,0)\}\subseteq{\mathbb{A}_{k}^{n}}$. Then consider the following cases:
Case(I). If $k$ is of characteristic zero. In that case are $\alpha\in{k^{a}\smallsetminus{k}}$ and $p_{\alpha,k}(x)$ its minimal polynomial. Then as $ p_{\alpha,k}(x)$ is separable (for $k$ is of characteristic zero), note that $k(\alpha,\alpha_{1},\ldots,\alpha_{r-1},\alpha_{r})\mid{k}$ where $\{\alpha_{0},\alpha_{1},\ldots,\alpha_{r-1},\alpha_{r}\}$ are the roots of $p_{\alpha,k}(x)$ in $k^{a}$ is a Galois extension whose Galois group will be denoted by $G$, then we notice that:
i) $f(x_{1},x_{2},\ldots,x_{n},x_{n+1}):=\displaystyle\prod_{\sigma\in{G}}\sigma(g(x_{1},x_{2},\ldots,x_{n-1},x_{n})+\alpha{x_{n+1}^{r}})$ is a polynomial in $k[x_{1},x_{2},\ldots,x_{n},x_{n +1}]$. In effect it as $\sigma(f)=f$ for all $\sigma\in{G} $ we have that $f$ coefficients are in the fixed field $k^{G}=k$.
ii) $V(f)=\{(0,0,\dots,0,0)\}\subseteq{\mathbb{A}_{k}^{n +1}}$. Indeed: First notice that $f(0,0\ldots,0,0)=\displaystyle\prod_{\sigma\in{G}}\sigma(g(0,0,\ldots,0,0)+\alpha{0^{r}})=0$. On the other hand if $(a_{1},a_{2},\ldots,a_{n},a_{n+1})\in{V(f)}$, there exists $\sigma\in{G}$ such that $\sigma(g(a_{1},a_{2},\ldots,a_{n-1},a_{n})+\alpha{a_{n+1}^{r}})=0$, $g(a_{1},a_{2},\ldots,a_{n-1},a_{n})+\sigma(\alpha){a_{n+1}^{r}}=0$, then $a_{n+1}=0$ (pues si $a_{n+1}\not=0$, $\sigma(\alpha)=-\displaystyle\frac{g(a_{1},a_{2},\ldots,a_{n-1},a_{n})}{a_{n+1}^{r}}$, taking $\sigma^{-1}$ in both members have $\alpha=-\displaystyle\frac{g(a_{1},a_{2},\ldots,a_{n-1},a_{n})}{a_{n+1}^{r}}$ ($\Rightarrow\Leftarrow$)), then $g(a_{1},a_{2},\ldots,a_{n-1},a_{n})=0$; then $a_{1},a_{2},\ldots,a_{n-1},a_{n}=0$.
Case (II). If $k$ has characteristic $p$ with $p$ prime. In that case is $k_{0}$ the separable closure of $k^{a}$ of $k$. If $k\not=k_ {0}$ then repeat what was done in the case (I) taking $\alpha\in{k_{0}\smallsetminus{k}}$. If $k=k_{0}$, then taking $\alpha\in{k^{a}\smallsetminus{k}}$ we have that there is $m\in{\mathbb{N}}$ such that $\alpha^{p^{m}}\in{k}$ (for $\alpha$ is purely inseparable over $k_{0}= k$), then we note that: $F(x_{1},x_{2},\ldots,x_{n},x_{n +1}): = (g(x_{1},x_{2},\ldots,x_{n-1},x_{n})+\alpha{x_{n+1}})^{p^{m}}$ is a polynomial in $k[x_{1},x_{2},\ldots,x_{n},x_{n+1}] $.
$V(f)=\{(0,0,\dots,0,0)\}\subseteq{\mathbb{A}_{k}^{n +1}}$. Indeed: First notice that $f(0,0,\dots,0,0)=(g(0,0,\ldots,0,0)+\alpha{0})^{p^{m}}=0$. On the other hand if $(a_{1},a_{2},\ldots,a_{n},a_{n+1})\in{V(f)}$, then $g(a_{1},a_{2},\ldots,a_{n-1},a_{n})+\alpha{a_{n+1}}=0$, then as $a_{n+1}=0$ (because $\alpha\not\in{k}$) we have $g(a_{1},a_{2},\ldots,a_{n-1},a_{n})=0$, then $a_{1},a_{2},\ldots,a_{n-1},a_{n}=0$.///
Thus, by the above we have that for all $n\in\mathbb{N}$ exists $f\in{k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $V(f)=\{(0,0,\dots,0,0)\}\subseteq{\mathbb{A}_{k}^ {n}}$.
Let $ X $ an affine algebraic set affine space $\mathbb{A}_{k}^{n}$, then there exists an ideal $I$ of $k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]$ such that $X=V(I)$, and $I$ is an ideal finitely generated (for $k [x_{1},x_{2},\dots{x_{n-1},x_{n}}]$ is a NoEther-Ring) have to exist $f_{1},f_{2},\ldots,f_{m-1},f_{m}\in{k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $I=\langle{f_{1},f_{2},\ldots,f_{m-1},f_{m}}\rangle$, then $V(I)= V(\{f_{1},f_{2},\ldots,f_ {m- 1}, f_{m}\})$, then it exists $f\in{k[x_{1},x_{2},\dots{x_{n-1},x_{n}}]}$ such that $V(f)=\{(0,0,\dots,0,0)\}$ must be $X=V(I)=V(f(f_{1},f_{2},\ldots,f_{n-1},f_{n}))$.