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.
A \(d(x,y)\le d(x,x_n)+d(x_n,y)\), an arbitrarily small quantity B The symmetry of \(d\) C That \((x_n)\) is bounded D The completeness of \(X\)
Show answer A. \(d(x,y)\le d(x,x_n)+d(x_n,y)\), an arbitrarily small quantity
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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.
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 \(x\).
\[\forall\varepsilon>0\ \ \exists N:\ n\ge N\Rightarrow d(x_n,x)<\varepsilon\]
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