Homeomorphism between $\mathbb{R^2}$ without n dots to $\mathbb{R^2}$ without n other dots I need to prove that if we take off from $\mathbb{R^2}$ n dots,it doesnt matter which dots we take off from this space,meaning:
Let $n$ be a natural  number,and let $K,K'$ be subsets of $\mathbb{R^2}$ such that that $K=${$p_1$,...,$p_n$},and $K'=${$p'_1$,...,$p'_n$}.
Prove that $\mathbb{R^2}$\K is homeomorphic to $\mathbb{R^2}$\K'.
So I thought about the following functions:
f: $\mathbb{R^2}$\K->$\mathbb{R^2}$\K'
Defined as:
f(p)=p if p isn't in $K'$.,and $p_i$ if it is in $K'$.
(meaning that if p is in $K'$,with index i,so its image will be the $i$-th in $K$ set).
g will be the same just the opposite.
So the composition of f and g is the ID function,But I'm not sure how to prove that these functions are continuous(if at all).
 A: The case-wise definition should have immediately raised suspicion: $f$ is not continouous at points of $K'\setminus K$. As a matter of fact. it may not even be a map $\Bbb R^2\setminus K\to\Bbb R\setminus K'$.
Lemma 1. Let $\Bbb H=\Bbb R\times(0,\infty)$,  $\overline{\Bbb H}=\Bbb R\times [0,\infty)$, $a,b\in\Bbb H$. Then there exists a homeomorphism $\overline {\Bbb H}\setminus\{a\}\to \overline{\Bbb H}\setminus\{b\}$ that is the identity on $\Bbb R\times \{0\}$.
Proof. Try $(x,y)\mapsto (x+\alpha y,\beta y)$ for suitable $\alpha,\beta\in\Bbb R$. $\square$
Lemma 2. Let $\Bbb I=\Bbb R\times(0,1)$,  $\overline{\Bbb I}=\Bbb R\times [0,1]$, $a,b\in\Bbb I$. Then there exists a homeomorphism $\overline {\Bbb I}\setminus\{a\}\to \overline{\Bbb I}\setminus\{b\}$ that is the identity on $\Bbb R\times \{0,1\}$.
Proof. Try $(x,y)\mapsto (x+\alpha y(1-y),y+\beta y(1-y))$ for suitable $\alpha,\beta\in\Bbb R$. $\square$
Now for the problem at hand, we may assume wlog. (namely, by applying a rotation if necessary) that the points in $K$ have pairwise different $y$-coordinates.
Then we can split $\Bbb R^2\setminus K$ by horizontal lines into parts that are $\approx \Bbb H\setminus \{a\}$ or $\approx \Bbb I\setminus\{a\}$ via homeomorphisms of the form $(x,y)\mapsto (x,\alpha +\beta y)$. We can do the same for $\Bbb R^2\setminus K'$. Then for these parts in top to bottom order apply lemma 1 and 2. Verify that these homeomorphisms glue nicely to a homeomorphism $\Bbb  R^2\setminus K\to \Bbb r^2\setminus K'$.
