Is $\mathbb{A}^1$ isomorphic to one of its quasi-affine subsets? It's well known that $\mathbb{A}^1$ is not isomorphic to any proper open subset of itself. Just out of curiosity, is $\mathbb{A}^1$ isomorphic to any proper quasi-affine subset of itself, or is this also impossible? 
To clarify my definition, I say a set is quasi-affine if it locally closed in the Zariski topology, i.e., $Z$ is quasi-affine if $Z=U\setminus W$, where $U$ and $W$ are both Zariski-closed. Here, everything is over some algebraically closed field.
Edit: Is this the correct idea? The closed sets of $\mathbb{A}^1$ are $\mathbb{A}^1$ itself, and the finite sets. But $\mathbb{A}^1$ cannot be isomorphic to a difference of finite sets for cardinality reasons, nor can it be isomorphic to a difference of $\mathbb{A}^1$ and a nonempty finite set, since the latter is disconnected?
 A: It is important to note that if $X$ is an irreducible variety, and $U\supset W$ are closed subsets, then either $X=U$ or $\dim U < \dim X$.  So the only locally closed subsets of $X$ that could possibly be isomorphic to $X$ are the open subsets.
In fact, I believe that no noetherian topological space is homeomorphic to a non-open, locally-closed subset of itself, but I haven't thought about this in too much detail.

Note that $\mathbb{A}^1$ is homeomorphic to $\mathbb{A}^1 \setminus \{0\}$ as topological spaces, as both are cofinite topologies on sets of the same cardinality.
But these are not isomorphic as varieties; in fact, I claim that the number of missing points is an isomorphism invariant: if $\mathbb{A}^1 \setminus \{p_1,\cdots , p_m\} \cong \mathbb{A}^1 \setminus \{p_1,\cdots , p_n\}$ for $m,n\geq 0$, then $m=n$.
This is easy to see using the uniqueness of completion for curves, but here is a "low-tech" proof: Let $X=\mathbb{A}^1 \setminus \{p_1,\cdots , p_n\}$.  The coordinate ring $R$ of $X$ is $k[T]$, with $T-p_1, \cdots , T-p_n$ inverted.
We can compute that $R^\times / k^\times \cong \mathbb{Z}^n$ as abelian groups.  And $\mathbb{Z}^m \cong \mathbb{Z}^n \implies m=n$ (exercise).
