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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.

nullUser
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elfgot5
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2 Answers2

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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 $T:\mathbb{R}^n \to \mathbb{R}^n$?

nullUser
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  • as far as I know, a linear transformation does not have to be bijective – elfgot5 Feb 08 '15 at 06:30
  • You know more than that it is linear. You know that it is injective and linear. What do you know about the relationship between injective, surjective, and bijective for linear transformations from $\mathbb{R}^n$ to itself? Hint: Rank-nullity theorem. – nullUser Feb 08 '15 at 06:32
  • The isometry is affine, not necessarily linear. – copper.hat Feb 08 '15 at 07:46
  • @nullUser Many isometries are not linear at all. For example, a translation is linear iff it is trivial (translation by the zero vector) – Timbuc Feb 08 '15 at 08:08
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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?)

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