Exercise: Let $a\in ℝ$. Show that the function $f(x)=x+a$ is an isometry from ℝ to ℝ.
Question: Can this be answered without an explicit distance function d(x,y) for the metric space (ℝ,d)? I need to show d(f(x),f(y)) = d(x,y).
Isometry (metric spaces)
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real-analysis
2 Answers
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I assume $d(\cdot, \cdot)=|\cdot - \cdot|$ because this is the common metric in $\mathbf R$. Then $$d(f(x),f(y))=|x+a-(y+a)|=|x-y|=d(x,y) \qquad \forall x,y\in \mathbf R$$ what shows that $f$ is an isometric.
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If you use a different metric, $f$ might not be an isometry.