3 added 182 characters in body

For subsets of the line, it's easy: each open subset is a countable disjoint union of intervals. For subsets of the plane, It's non-trivial, but something sensible can be said. For subsets of $\mathbb R^n$ for $n\ge 3$, I have no idea and I suspect that it's already hopeless in the case of $\mathbb R^3$.

So let $U\subset \mathbb R^2$ be an open subset, and let's assume without loss of generality that it's connected. Then $U$ is completely classified by its space of ends (see comment of Agol below for a reference). That's a compact, separable, zero-dimensional topological space. Those can probably be completely classified, even though I didn't quite manage to do it.

Added later: Actually, there are good reasons why the classification is not feasable (see comments of Clinton Conley below).

The basic operation on such a set $X$ is that of taking its derived subset $$X':=X\setminus \{\mathrm{isolated\ points\ of\ } X\}$$

This operation can be iterated transfinitely, and so we get $X^{(\alpha)}$ for any ordinal $\alpha$ (where $X^{(\alpha)}$ is defined as the intersection of all $X^{(\beta)}$ for $\beta<\alpha$ when $\alpha$ is a limit ordinal). Since $X^{(\alpha)}$ is a descending sequence of subset of $X$, there is a minimal $\gamma$ such that $X^{(\gamma)}=X^{(\gamma+1)}$. This is a countable ordinal and is the first interesting invariant of $X$. There are two options for $X^{(\gamma)}$: either it's empty, or it's a Cantor set: that's the second invariant of $X$. Finally, if $\gamma$ is a successor ordinal, then you can look at the discrete set $X^{(\gamma-1)}\setminus X^{(\gamma)}$. If $X^{(\gamma)}=\emptyset$, then that's a finite non-empty set whose cardinality is an invariant. And if $X^{(\gamma)}$ is a Cantor set, then $X^{(\gamma-1)}\setminus X^{(\gamma)}$ is either finite or infinitely countable.

I recapitulate. The invariants of $X:=\mathrm{Ends}(U)$ are:

• The smallest ordinal $\gamma$ such that $X^{(\gamma)}=X^{(\gamma+1)}$.
• Whether or not $X^{(\gamma)}$ is empty or a Cantor set.
• The cardinality of $X^{(\gamma-1)}\setminus X^{(\gamma)}$.

I thought for a while that those might be a complete set of invariants of $U$, but I was wrong. For example: if $X^{(\gamma)}$ is a Cantor set, then we can also look at the set of accumulation points of $X^{(\gamma-1)}\setminus X^{(\gamma)}$ inside $X^{(\gamma)}$: that's again a compact, separable, zero-dimensional topological space, and so it has its own set of invariants...
We can also look at the minimal ordinal $\beta$ such that the closure of $X^{(\beta)}\setminus X^{(\gamma)}$ intersects $X^{(\gamma)}$ non-trivially...

Note: What I attempted to do was the classification of open of subsets $\mathbb R^n$ ($n=1,2$) up to homeomorphism (that was the question, I think), and not up to ambient homeomorphism. The latter is much more messy, already for $n=1$.

2 added 3 characters in body

For subsets of the line, it's easy: each open subset is a countable disjoint union of intervals. For subsets of the plane, It's non-trivial, but something sensible can be said. For subsets of $\mathbb R^n$ for $n\ge 3$, I have no idea and I suspect that it's already hopeless in the case of $\mathbb R^3$.

So let $U\subset \mathbb R^2$ be an open subset, and let's assume without loss of generality that it's connected. Then $U$ is completely classified its space of ends. That's a compact, separable, zero-dimensional topological space. Those can probably be completely classified, even though I did didn't quite manage to do it.

The basic operation on such a set $X$ is that of taking its derived subset $$X':=X\setminus \{\mathrm{isolated\ points\ of\ } X\}$$

This operation can be iterated transfinitely, and so we get $X^{(\alpha)}$ for any ordinal $\alpha$ (where $X^{(\alpha)}$ is defined as the intersection of all $X^{(\beta)}$ for $\beta<\alpha$ when $\alpha$ is a limit ordinal). Since $X^{(\alpha)}$ is a descending sequence of subset of $X$, there is a minimal $\gamma$ such that $X^{(\gamma)}=X^{(\gamma+1)}$. This is a countable ordinal and is the first interesting invariant of $X$. There are two options for $X^{(\gamma)}$: either it's empty, or it's a Cantor set: that's the second invariant of $X$. Finally, if $\gamma$ is a successor ordinal, then you can look at the discrete set $X^{(\gamma-1)}\setminus X^{(\gamma)}$. If $X^{(\gamma)}=\emptyset$, then that's a finite non-empty set whose cardinality is an invariant. And if $X^{(\gamma)}$ is a Cantor set, then $X^{(\gamma-1)}\setminus X^{(\gamma)}$ is either finite or infinitely countable.

I recapitulate. The invariants of $X:=\mathrm{Ends}(U)$ are:

• The smallest ordinal $\gamma$ such that $X^{(\gamma)}=X^{(\gamma+1)}$.
• Whether or not $X^{(\gamma)}$ is empty or a Cantor set.
• The cardinality of $X^{(\gamma-1)}\setminus X^{(\gamma)}$.

I thought for a while that those might be a complete set of invariants of $U$, but I was wrong. For example: if $X^{(\gamma)}$ is a Cantor set, then we can also look at the set of accumulation points of $X^{(\gamma-1)}\setminus X^{(\gamma)}$ inside $X^{(\gamma)}$: that's again a compact, separable, zero-dimensional topological space, and so it has its own set of invariants...
We can also look at the minimal ordinal $\beta$ such that the closure of $X^{(\beta)}\setminus X^{(\gamma)}$ intersects $X^{(\gamma)}$ non-trivially...

Note: What I attempted to do was the classification of open of subsets $\mathbb R^n$ ($n=1,2$) up to homeomorphism (that was the question, I think), and not up to ambient homeomorphism. The latter is much more messy, already for $n=1$.

1

For subsets of the line, it's easy: each open subset is a countable disjoint union of intervals. For subsets of the plane, It's non-trivial, but something sensible can be said. For subsets of $\mathbb R^n$ for $n\ge 3$, I have no idea and I suspect that it's already hopeless in the case of $\mathbb R^3$.

So let $U\subset \mathbb R^2$ be an open subset, and let's assume without loss of generality that it's connected. Then $U$ is completely classified its space of ends. That's a compact, separable, zero-dimensional topological space. Those can probably be completely classified, even though I did quite manage to do it.

The basic operation on such a set $X$ is that of taking its derived subset $$X':=X\setminus \{\mathrm{isolated\ points\ of\ } X\}$$

This operation can be iterated transfinitely, and so we get $X^{(\alpha)}$ for any ordinal $\alpha$ (where $X^{(\alpha)}$ is defined as the intersection of all $X^{(\beta)}$ for $\beta<\alpha$ when $\alpha$ is a limit ordinal). Since $X^{(\alpha)}$ is a descending sequence of subset of $X$, there is a minimal $\gamma$ such that $X^{(\gamma)}=X^{(\gamma+1)}$. This is a countable ordinal and is the first interesting invariant of $X$. There are two options for $X^{(\gamma)}$: either it's empty, or it's a Cantor set: that's the second invariant of $X$. Finally, if $\gamma$ is a successor ordinal, then you can look at the discrete set $X^{(\gamma-1)}\setminus X^{(\gamma)}$. If $X^{(\gamma)}=\emptyset$, then that's a finite non-empty set whose cardinality is an invariant. And if $X^{(\gamma)}$ is a Cantor set, then $X^{(\gamma-1)}\setminus X^{(\gamma)}$ is either finite or infinitely countable.

I recapitulate. The invariants of $X:=\mathrm{Ends}(U)$ are:

• The smallest ordinal $\gamma$ such that $X^{(\gamma)}=X^{(\gamma+1)}$.
• Whether or not $X^{(\gamma)}$ is empty or a Cantor set.
• The cardinality of $X^{(\gamma-1)}\setminus X^{(\gamma)}$.

I thought for a while that those might be a complete set of invariants of $U$, but I was wrong. For example: if $X^{(\gamma)}$ is a Cantor set, then we can also look at the set of accumulation points of $X^{(\gamma-1)}\setminus X^{(\gamma)}$ inside $X^{(\gamma)}$: that's again a compact, separable, zero-dimensional topological space, and so it has its own set of invariants...
We can also look at the minimal ordinal $\beta$ such that the closure of $X^{(\beta)}\setminus X^{(\gamma)}$ intersects $X^{(\gamma)}$ non-trivially...

Note: What I attempted to do was the classification of open of subsets $\mathbb R^n$ ($n=1,2$) up to homeomorphism (that was the question, I think), and not up to ambient homeomorphism. The latter is much more messy, already for $n=1$.