Exercise · ZW-CONVERGENCE-IN-A-METRIC-SPACE-007 · ★★☆ Intermediate
Let \(X=(0,1)\) with the usual distance and \(x_n=\tfrac1{n+1}\). Does it converge in \(X\)?
Choose one option.
A Yes, to \(0\) B Yes, to \(1/2\) C No: its only candidate limit, \(0\), is not in \(X\) D No: it is not bounded
Show answer C. No: its only candidate limit, \(0\), is not in \(X\)
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Explanation The sequence approaches \(0\), but \(0\notin X\). Convergence requires the limit to belong to the 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 \(x\).
\[\forall\varepsilon>0\ \ \exists N:\ n\ge N\Rightarrow d(x_n,x)<\varepsilon\]
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