Exercise · ZW-COMPLETE-METRIC-SPACE-009 · ★★★ Advanced
On \(\mathbb{R}\) define \(d(x,y)=|\arctan x-\arctan y|\) (a genuine distance). Is \((\mathbb{R},d)\) complete?
Choose one option.
A Yes: it agrees with the usual distance. B No: \(x_n=n\) is Cauchy, since \(\arctan n\to\pi/2\), but it does not converge in \(\mathbb{R}\). C No: \(d\) fails the triangle inequality. D Yes: \(\arctan\) is a bounded function. Show answer B. No: \(x_n=n\) is Cauchy, since \(\arctan n\to\pi/2\), but it does not converge in \(\mathbb{R}\).← Back to the question
Explanation Since \(\arctan n\to\pi/2\), \(d(x_m,x_n)\to0\). A limit \(x\) would need \(\arctan x=\pi/2\), impossible. Although \(d\) and the usual distance give the same open sets , completeness is not preserved.
A metric space \(X\) is complete if every Cauchy sequence \((x_n)\) in \(X\) converges to a point \(x\) of \(X\). A sequence is Cauchy if its terms eventually get arbitrarily close to one another.
\[(x_n)\ \text{Cauchy}\ \Longrightarrow\ \exists\,x\in X:\ x_n\to x\]
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