Question: A scheme $(X, \mathcal{O}_X)$ is a pair where $X$ is a topological space and $\mathcal{O}_X$ is a sheaf of rings on $X$, with the property that there is an open cover $U_i$ of $X$ with $U_i \cong Spec(A_i)$ an affine scheme. It is easy to construct an example of a topological space $X_t$ and a strict open subspace $U_t \subsetneq X_t$ with an abstract homeomorhism $\phi_t: U_t \cong X_t$. Does a similar property hold for a scheme of finite type over a field?
Example: Let $k$ be an algebraically closed field and let $S:=\mathbb{A}^2_k$ be the affine plane and let $U \subsetneq S$ be a strict open subscheme. Can you give an elementary example of such $U$ where there is an abstract isomorphism $\phi: U \cong S$ of schemes? What about a scheme $X$ of finite type over a field $k$?
As a consequence: If $U_t$ is the topological space of $U$ it follows there is a strict subset $U_t \subsetneq S_t$ with an abstract homeomorphism $\phi_t:U_t \cong S_t$.
Note: Two schemes (or ringed spaces) $(X, \mathcal{O}_X), (Y, \mathcal{O}_Y)$ are "isomorphic" iff there are maps $f:(X, \mathcal{O}_X) \rightarrow (Y, \mathcal{O}_Y), g:(Y, \mathcal{O}_Y)\rightarrow (X, \mathcal{O}_X)$ with $f\circ g =Id,g \circ f=Id$. In particular the maps $f,g$ induce homeomorphisms of topological spaces $X \cong Y$. If a strict open subscheme $U \subsetneq V$ is abstractly isomorphic to $V$ you end up with a topological space $U_t \subsetneq V_t$ realizable as a strict subspace of $V_t$, but where there is an abstract isomorphism $U_t \cong V_t$. This stuation is "paradoxical" - the topological space $V_t$ would then be homeomorphic to a strict subspace of itself.
Example: For the affine line $\mathbb{A}^1_k$ with $k$ an algebraically closed field, it follows there is no abstract isomorphism $\phi: U \cong \mathbb{A}^1_k$ for any strict open subscheme $U \subsetneq \mathbb{A}^1_k$:
Let $k$ be an algebraically closed field and let $A:=k[x]$ be the polynomial ring in one variable $x$ over $k$. Let
$$f(x):=(x-u_1)^{l_1}\cdots (x-u_m)^{l_m}$$
with $u_i \neq u_j\in k$. Let $X:=Spec(A)$ and $U:=D(f) \subsetneq X$ be a strict open subscheme of $X$ and assume $X \cong D(f)$ is an abstract isomorphism of schemes over $k$, induced by a map of $k$-algebras $\tilde{\phi}: A \cong A_f$. The map $\tilde{\phi}$ induce a map of $k$-algebras
$$\phi:k[x] \rightarrow k[x]$$
with $\phi(x)=a(x)$ and $\phi(f(x)):=f(a(x))\in k^*$ a unit. It follows $\phi(a(x)):=a_0\in k^*$ is a unit with $a_0\neq u_i$ for all $i$. Hence the map $\phi$ factors as follows
$$ k[x] \rightarrow k[x]/(x-a_0) \rightarrow k \subseteq k[x].$$
Hence the induced map $k[x]\rightarrow A_f \rightarrow A$ is the "constant map": The induced map of affine schemes
$$ D(f) \rightarrow X$$
identifies $D(f) \cong \{(x-a_0)\} \subseteq X$ as a topological space with one point which contradicts the assumption that $D(f) \subseteq X$ is an open subscheme.
Hence for the affine line there are no such strict open subschemes. Can you construct such a strict open subscheme $U \subsetneq \mathbb{A}^2_k$ for the affine plane? What about a scheme $X$ of finite type over a field $k$?
Note: I post this question to set theorists and algebraic geometers - I want response from both groups of people.