Exercise · ZW-CONVERGENCE-IN-A-METRIC-SPACE-002 · ★☆☆ Basic

For \(x_n=1/n\) and \(\varepsilon=0.01\) in \(\mathbb{R}\), what is the smallest \(N\) with \(d(x_n,0)<\varepsilon\) for all \(n\ge N\)?

Choose one option.

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  • B. \(101\)

Explanation

We need \(1/n<0.01\), that is \(n>100\). For \(n=100\) it equals \(0.01\), which is not smaller, so \(N=101\).

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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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