I have some preliminary questions before I attempt this problem in my book.

If $M$ is the metric space of all the real numbers and $x_0 \in M$, prove that there exist exactly two isometries of $M$ that leave $x_0$ fixed.

For completeness, an isometry is a mapping between two metric spaces that preserves distance.

I have two questions.

  1. Am I to assume that the trivial isometry (mapping each point onto itself) "counts" as an isometry in the context of this question?

  2. Are the two isometries the trivial one and the one which will reflect the entire real line over the point $x_0$? Meaning the value of $x_0 + \epsilon$ is swapped with the value of $x_0 - \epsilon$ for $\epsilon > 0$.

I still have to prove that they are the only ones but I want to make sure I am on the right track.

  • $\begingroup$ These are the two isometries you are looking for. Try proving where every other real number must go under isometry with this fixed point. $\endgroup$ Aug 21, 2014 at 15:31

1 Answer 1


An isometry $f$ on the line satisfies $|f(x) - f(y)| = |x - y|$ for any two real numbers $x$ and $y$.

If $f(x_0) = x_0$, you get $|f(x) - x_0| = |x - x_0|$. This leads to only two possible isometries: either $f(x) = x$ or $f(x) = 2x_0 - x$ which are the two you describe.


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