Let A and B be two finite subsets of $\mathbb R$. Describe a necessary and sufficient condition for the spaces $\mathbb R\setminus A$ and $\mathbb R\setminus B$ to be homeomorphic.
I think $|A|=|B|$. Am I right?
Let A and B be two finite subsets of $\mathbb R$. Describe a necessary and sufficient condition for the spaces $\mathbb R\setminus A$ and $\mathbb R\setminus B$ to be homeomorphic.
I think $|A|=|B|$. Am I right?
I’m going to assume that you know that all open intervals in $\Bbb R$, including the open rays of the form $(x,\to)$ and $(\leftarrow,x)$, are homeomorphic to one another; if you don’t, you should try to prove it. In what follows I’ll use the term open interval to include the open rays.
If $A$ is finite, then $\Bbb R\setminus A$ is the union of $|A|+1$ open intervals, so it’s homeomorphic to the disjoint union of $|A|+1$ copies of $(0,1)$. Clearly, then, $\Bbb R\setminus A$ is homeomorphic to $\Bbb R\setminus B$ whenever $|A|=|B|$: both are homeomorphic to the disjoint union of $|A|+1$ copies of $(0,1)$.
Suppose that $|A|=m$ and $|B|=n$, where $m
Try proving by induction on $n$ that if $A$ is a finite subset of $\mathbb{R}$ with $|A| = n$, then $\mathbb{R} \setminus A$ is homeomorphic to a disjoint union of $n+1$ copies of $\mathbb{R}$. This suffices.
Assume $A=\{a_1, \ldots, a_n\}$ with $a_1<\ldots
If $n\ne m$, then $\mathbb R\setminus A$ and $\mathbb R\setminus B$ cannot be homeomorphic because we can recover $n$ from $A$ as a topological invariant: It is one less than the number of connected components, i.e. the maximal number of disjoint (relatively) open sets that can cover the whole space. To see this you need to know that each interval is connected, which is not hard.