Proving isometries of the plane are bijective

Question

So I'm trying to prove that every isometry $I:\mathbb{R}^2 \to \mathbb{R}^2$ is bijective.

I have already proved that I is injective (which is almost immediate) and I also proved $I$ is continuous (because I thought that might be useful) but I'm having trouble with the proof for surjectivity.

Answer 1

Hint: Isometries are continuous. Use this to prove that any isometry of $\mathbb{R}^2$ is, up to a translation, linear. What do you know about the relationship between injective, surjective, and bijective for a linear transformation $T:\mathbb{R}^n \to \mathbb{R}^n$?

Answer 2

There are two ways to prove it- an algebraic and an analytic(=using analysis). I do second:

Let $f\colon\mathbb{R}^2\rightarrow\mathbb{R}^2$ be an isometry.

Case 1. $f(0)=0$. If $B_r$ denotes the closed ball of radius $r$ around $0$, and if we show that $f(B_r)=B_r$ for every $r>0$, then it proves that $f$ is surjective. Now $$x\in B_r\Longrightarrow |x-0|\leq r \Longrightarrow |f(x)-f(0)|\leq r \Longrightarrow |f(x)-0|\leq r \Longrightarrow f(x)\in B_r.$$

Thus $f(B_r)\subseteq B_r$. If possible, let $f(B_r)\neq B_r$. Take $x_0\in B_r\setminus f(B_r)$. Construct a sequence in $f(B_r)$ as follows: $$x_1=f(x_0), x_2=f(x_1), \cdots, x_n=f(x_{n-1}), \cdots .$$

Since $f$ is continuous and $B_r$ is compact, hence $\{x_n\}_{n\geq 1}$ is a sequence in compact set $f(B_r)$, and by Bolzano-Weierstrass theorem, it has a convergent subsequence. But since $x_0\notin f(B_r)$, $$\lambda=d(x_0,f(B_r))=\inf\{d(x_0,y)\colon y\in f(B_r)\} \,\,\,\, \mbox{must be positive},$$ and as $f$ is isometry, we can show that distance between any two terms of sequence $\{x_n\}_{n\geq 1}$ is at least $\lambda$ which implies it has no convergent subsequence. To see this, for example,

$$d(x_1,x_2)=d(f(x_0),f(x_1))=d(x_0,x_1)\geq \lambda.$$

Similarly, you can show $d(x_i,x_j)=d(x_0,x_{j-i})\geq \lambda$ for $i<j$.

Case 2. $f(0)=a\neq 0$. Then compose $f$ with translation by $-a$, i.e. $T_{-a}\colon x\mapsto x-a$. Then $T_{-a}\circ f$ is an isometry fixing $0$, and by Case 1, ..... (can you complete?)