By definition (in the book that I am following), an affine variety is zero set of a collection of finitely many polynomials.
Question is to prove that $\mathbb{A}^1\setminus\{0\}$ is an affine variety.
For that I need to prove that $\mathbb{A}^1\setminus\{0\}$ is zero set of (finitely many) polynomial(s).
I was having this misconception that I have to prove that $\mathbb{A}^1\setminus\{0\}$ is zero set of polynomial in one variable as we are in $\mathbb{A}^1$.
Then I read the definition again and it says:
$X\subset k^n$ is an affine variety if it is zero set of polynomials in $n$ variables.
As $\mathbb{A}^1\setminus\{0\}$ and $\{(x,y)\in k^2:xy=1\}$ are homeomorphic and $\{(x,y)\in k^2:xy=1\}$ being zero set of polynomial $f(x,y)=xy-1$, is an affine variety and so $\mathbb{A}^1\setminus \{0\}$ should also be an affine variety though not as subvariety of $\mathbb{A}^1$.
Is this justification sufficient enough to say that $\mathbb{A}^1\setminus\{0\}$ is an affine variety or am I missing something?
Let $X=\mathbb{A}^1\setminus\{0\}$, it is an open subset of $\mathbb{A}^1$ (w.r.t. Zariski topology); by defintion, only the finite subsets (a.k.a. Zariski closed subsets) of $\mathbb{A}^1$ are affine (sub)varieties of $\mathbb{A}^1$: $X$ is not an affine subvariety of $\mathbb{A}^1$!
As the OP-user affirms: $X$ is homeomorphic to hyperbola $V(xy-1)=\{(x,y)\in\mathbb{A}^2\mid xy-1=0\}$ (w.r.t. Zariski topologies); therefore one can pull-back from $V(xy-1)$, via a fixed homeomorphism, an affine algebraic variety structure on $X$, and $X$ with a such structure is an abstract affine algebraic variety.
The same trick works, for example, with the graph $G$ of the absolute function $|\cdot|$: as subset of $\mathbb{R}^2$, it is homeomorphic to $\mathbb{R}$ (w.r.t. natural topologies); therefore $G$ can be structured as an analityc manifold.