Exercise · ZW-CONVERGENCE-IN-A-METRIC-SPACE-010 · ★★★ Advanced

In usual \(\mathbb{R}\), let \(x_n=1+\tfrac1n\) and \(y_n=3-\tfrac1n\). What number does \(d(x_n,y_n)\) approach?

Choose one option.

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  • C. \(2\)

Explanation

\(x_n\to1\) and \(y_n\to3\), and distance is continuous: \(d(x_n,y_n)=|2-2/n|\to2=d(1,3)\).

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Review the concept: Convergence in a metric space

A sequence \((x_n)\) in a metric space \((X,d)\) converges to \(x\in X\) if, for every \(\varepsilon>0\), there is an index \(N\) from which \(x_n\) lies within distance \(\varepsilon\) of \(x\).

\[\forall\varepsilon>0\ \ \exists N:\ n\ge N\Rightarrow d(x_n,x)<\varepsilon\]

More practice problems: convergence in a metric space

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