Exercise · ZW-CONVERGENCE-IN-A-METRIC-SPACE-006 · ★★☆ Intermediate

If \(x_n\to x\) and \(x_n\to y\) in a metric space, which argument proves \(x=y\)?

Choose one option.

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  • A. \(d(x,y)\le d(x,x_n)+d(x_n,y)\), an arbitrarily small quantity

Explanation

The triangle inequality makes \(d(x,y)\) smaller than any \(\varepsilon>0\), so \(d(x,y)=0\) and \(x=y\). Completeness is not needed.

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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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