To an algebraist the simplest geometric spaces are affine varieties, determined by their ring of functions. Unfortunately, students in algebraic geometry discover early that subvarieties are (a) interesting to study since functions and sheaves restrict to them and (b) indistinguishable by their function rings alone.
Using (b) to detect non-affineness, we can show rigorously that the quasiaffine variety $\Bbb{A}^2 \setminus \{(0,0)\}$ is not affine.

$\underline{\text{Identifying } \mathcal{O}(\mathbb{A}^2 \setminus \{(0,0)\})}$: We have an injection $k[x,y] \to \mathcal{O}(\mathbb{A}^2 \setminus \{(0,0)\})$ by restriction of regular functions on $\mathbb{A}^2$: if $f, g$ are regular functions on $\mathbb{A}^2$, then $f-g : \mathbb{A}^2 \to k$ is also regular and in particular continuous. If $f, g$ are identical except possibly at $(0,0)$, then since $f-g$ is nonzero at $(0,0)$ if it is nonzero in an open neighbourhood of $(0,0)$ we see that $f-g$ must be identically $0$. Hence the restriction map is injective. We can also determine the regular functions sheaf-theoretically by using the open cover $D(x)\cup D(y)$ for $\mathbb{A}^2 \setminus \{(0,0)\}$. These sets contain the same points they would contain in $\mathbb{A}^2$, and we know that $\mathcal{O}(D(x)) = k[x,y]_{x} = k[x^{\pm 1},y]$, $\mathcal{O}(D(y)) = k[x, y^{\pm 1}]$. A regular function on $\mathbb{A}^2 \setminus \{(0,0)\}$ is then given by a pair of $f \in k[x^{\pm1}, y], g \in k[x, y^{\pm 1}]$ so that $f$ and $g$ have the same image in $\mathcal{O}(D(xy)) = k[x, y]_{xy} = k[x^{\pm 1}, y^{\pm 1}]$. The image of $f$ is $\frac{f}{1}$ and that of $g$ is $\frac{g}{1}$, so this means that there is some $h$ in $(xy)$ with $h(f-g) = 0 \implies f =g$ as $k[x, y]$ is a domain. Clearly we have $\mathcal{O}(D(x)) \cap \mathcal{O}(D(y)) = k[x,y]$ (happening in the function field $k(x, y)$), so we see that $\mathcal{O}(\mathbb{A}^2 \setminus \{(0,0)\}) = k[x,y]$.

Now when $X$ and $Y$ are varieties and $Y$ is affine there is an isomorphism $\text{Hom}(X, Y) \cong \text{Hom}(A(Y), \mathcal{O}(X))$. This means that for any affine $Y$ with $A(Y) \cong k[x,y]$ one has $Y \cong \mathbb{A}^2$, so to determine the difference between $\Bbb{A}^2$ and $\Bbb{A}^2 \setminus \{(0,0)\}$ using maps of rings we will need to look at maps from $\Bbb{A}^2 \setminus \{(0,0)\}$ to $\Bbb{A}^2$. Alternatively we could compare their topologies, but this turns out to be surprisingly delicate: a result Wie78 of Wiegand shows that all nonempty open subsets of $\mathbb{A}^2$ are homemorphic in positive characteristic.

$\underline{\mathbb{A}^2 \setminus \{(0,0)\} \text{ is not isomorphic to } \mathbb{A}^2}$: Since a morphism of varieties induces a map of the rings of regular functions, an isomorphism $f: \mathbb{A}^2 \xrightarrow{\sim} \mathbb{A}^2 \setminus \{(0,0)\}$ would correspond to an automorphism of $k[x,y]$. Let $f_{*}: k[x,y] \to k[x,y], g \mapsto g \circ f$ be the associated isomorphism of rings (thinking of $k[x,y]$ as regular functions on each space), and notice that $f$ induces an isomorphism $k \cong k[x,y]/(x,y) \to k[x, y]/(x \circ f, y \circ f)$. It follows that $(x\circ f, y \circ f)$ is a maximal ideal, so $(x\circ f, y\circ f) = (x-a, y-b)$ for some $a, b \in k$.

Now if $f : X \to Y$ is a morphism of varieties and $g \in \mathcal{O}(Y)$, then $V(g)$ is closed in $Y$ and $V(g\circ f)$ is closed in $X$, $f(V(g\circ f)) \subseteq V(g)$ . If $f$ is an isomorphism, then applying this reasoning in the other direction gives $f(V(g\circ f)) = V(g)$. In our situation, this implies that $f(V(x-a, y - b)) = V(x, y) = \varnothing$, which is a contradiction as $V(x-a, y-b)$ is a point of $\mathbb{A}^2$ for all $a, b \in k$. We see that $\mathbb{A}^2 \setminus \{(0,0)\}$ cannot be isomorphic to $\mathbb{A}^2$.

Note that our argument identifying the regular functions on $\mathbb{A}^2 \setminus \{(0,0)\}$ will work similarly for $\mathbb{A}^n \setminus \{(0,0)\}$ with $n \geq 3$. In the $n = 1$ case, $\mathbb{A}^1 \setminus \{0\}$ is a variety, and in this setting we notice that $\mathbb{A}^1 \setminus \{0\} = D(x)$ with functions $k[x]_x$.

Further reading:

• BFH19 for more study of the topology of $\mathbb{A}^2$ via subvarieties
• MSEAffineJacobian for relating this result to algebraic Hartog's theorem