Prove that $\mathbb{R}^2$ is not homemorphic to $\mathbb{R}^2 - \{0\}$ using exhaustion by compact sets

Question

I cannot understand example 4.64 from Manetti's topology book, which says that if we consider the following compact exhaustion of $\mathbb{R}^2$ : $\{K_n\}$ , where $K_n = \{(x,y) | x ^ 2 + y ^ 2 \leq n ^ 2\}$ then we can prove that $\mathbb{R}^2$ is not homeomorphic to $Y = \mathbb{R}^2 - \{0\}$. The book proceeds in this way, first of all it is assumed by absurdity that there exists a homeomorphism $f: \mathbb{R}^2 \rightarrow Y$, then the sequence $D_n = f (K_n) $is an exhaustion in compacts of $Y$, therefore there exists an integer $N$ such that the compact $S^1$ is contained in $D_N$. The function $f: Y \rightarrow ]0,+\infty[, f (x, y) = x ^ 2 + y ^ 2$, has on the compact $D_N$ a maximum value $M> 1$ and a minimum value $1> m> 0$. I know there is a maximum and a minimum because we are working with a compact and the function is continuous, but how do I know that the maximum is greater than $1$? and that the minimum is between $1$ and $0$? then the book goes on and says that $\{(x, y) \in Y | x ^ 2 + y ^ 2 < m\}$ is contained in $\{(x, y) \in Y-D_N | x ^ 2 + y ^ 2 <1\}$ and that $\{(x, y) \in Y | x ^ 2 + y ^ 2> M\}$ is contained in $\{(x,y) \in Y-D_N | x ^ 2 + y ^ 2> 1\}$ therefore Y-D_N is the disjoint union of two non-empty openings so $Y-D_N$ is disconnected but $f(R^2 - K_N) = Y-D_N$ but $\mathbb{R}^2 - K_N$ is connected, then I arrive at a contradiction. I did not understand why these two contents are valid $\{(x, y) \in Y | x ^ 2 + y ^ 2 <m\}$ is contained in $\{(x, y) \in Y-D_N | x ^ 2 + y ^ 2 <1\}$ and that $\{(x, y) € Y | x ^ 2 + y ^ 2> M\}$ is contained in $\{(x, y) \in Y-D_N | x ^ 2 + y ^ 2> 1\}$. because, for example, I would say that $A = \{(x, y) \in Y-D_N | x ^ 2 + y ^ 2 <1\}$ is the empty set as $Y-D_N$ corresponds to the complementary of $D_N$ but since $D_N$ is a compact containing $S^1$ or elements such that $x ^ 2 + y ^ 2 = 1$, if we consider the complementary of $D_N$ in $Y$ then surely there are no elements of $\mathbb{R^2}$ in them such that $x ^ 2 + y ^ 2 <1$

Answer 1

We only need to know that compact subsets of $\mathbb R^n$ are closed and bounded to get

Lemma. Let $D \subset Y = \mathbb R^n \setminus \{0\}$ be a compact subset containing the sphere $S^{n-1}$. Then $Y \setminus D$ is not connected.

Proof. Clearly $U = Y \setminus D$ is the union of the two disjoint open sets $U_+ = U \cap \{ x \in Y \mid \lVert x \rVert > 1\}$ and $U_- = U \cap \{ x \in Y \mid \lVert x \rVert < 1\}$.

1. Since $D$ is bounded, we find $M >0$ such that $\lVert x \rVert \le M$ for all $x \in D$. Thus all $x \in Y$ with $\lVert x \rVert > \max(1,M)$ are in $U_+$. That is, $U_+ \ne \emptyset$.

2. Since $D$ is closed in $\mathbb R^n$, its complement $\mathbb R^n \setminus D$ is open. It contains $0$ because $0 \notin D$. Thus there exists $m$ with $1 > m > 0$ such that $V_m = \{ x \in \mathbb R^n \mid \lVert x \rVert < m\} \subset \mathbb R^n \setminus D$. Thus all $x \in V_m \setminus \{0\}$ are in $U_-$. That is, $U_- \ne \emptyset$.

Answer 2

$$f:Y\to \left ]0,\infty \right [,f(x,y)=x^{2}+y^{2}$$ Remark that $f$ is constant on $S^{1}$,so $\max_{x\in S^{1}} f(x)=\min_{x\in S^{1}} f(x)=1$,Since $S^{1}\subset D_{N}$,we have $$1=\max_{x\in S^{1}} f(x)\leq \max_{x\in D_{N}} f(x)$$ $$1=\min_{x\in S^{1}} f(x)\geq \min_{x\in D_{N}} f(x)>0$$ Now $\{(x,y)\in Y \backslash x^{2}+y^{2}<m \}$, is clearly contained in the complement of $D_{N}$ because the minimum of $f$ is $m$,Moreover $m\leq 1$,So : $$\{(x,y)\in Y \backslash x^{2}+y^{2}<m \} \subset \{(x,y)\in Y-D_{N}\backslash x^{2}+y^{2}<1\}$$

Answer 3

I think the argument has a little gap. I do not know which definition of compact exhaustion you are using (see Existence of exhaustion by compact sets), but in the present case we have $K_n \subset \operatorname{int} K_{n+1}$. Thus also $D_n \subset \operatorname{int} D_{n+1}$. Since $S^1$ is compact and $S^1 \subset \bigcup_n \operatorname{int} D_n$, we must have $S^1 \subset \operatorname{int} D_N$ for some $N$. This is stronger than just $S^1 \subset D_N$.

Clearly $f$ attains a maximum $M$ and an minimum $m$ on $D_N$. Since $\operatorname{int} D_N$ is an open neighborhood of $S^1$, $D_N$ contains a point $(x,y)$ with $x^2 + y^2 < 1$ which means $1 > f(x,y) \ge m$, and it contains a point $(x,y)$ with $x^2 + y^2 > 1$ which means $1 < f(x,y) \le M$. Note that $m = 0$ is impossible because $(0,0) \notin D_N$.

Why do we have $\{(x, y) \in Y | x ^ 2 + y ^ 2 < m\} \subset \{(x, y) \in Y \setminus D_N | x ^ 2 + y ^ 2 <1\}$? So let $(x,y) \in Y$ with $x^2 + y^2 < m$. Thus $x^2 + y^2 < 1$ because $m < 1$. Since $m$ is the minimum of $f(x,y) = x^2 + y^2$ on $D_N$, we have $(x,y) \notin D_N$ for $x^2 + y^2 < m$. This shows $(x,y) \in \{(x, y) \in Y \setminus D_N | x ^ 2 + y ^ 2 <1\}$.

The inclusion $\{(x, y) \in Y | x ^ 2 + y ^ 2 > M \} \subset \{(x, y) \in Y \setminus D_N | x ^ 2 + y ^ 2 > 1\}$ is verified similarly.

Answer 4

Alternatively, $\mathbb{R}^2$ is contractible. Hence, $H_1(\mathbb{R}^2)=0$. But, $\mathbb{R}^2\backslash \{0\}$ deformation retracts onto $\mathbb{S}^1$. Hence, $H_1(\mathbb{R}^2\backslash \{0\})=\mathbb{Z}$.