In ℝ^2 with d_∞(p,q)=max(|p_1-q_1|,|p_2-q_2|), let x_n=(tfrac1n,tfrac1n^2)→(0,0). What is the smallest N with d_∞(x_n,(0,0))<0.1 for all n≥ N?

Exercise · ZW-CONVERGENCE-IN-A-METRIC-SPACE-008 · ★★☆ Intermediate

In \(\mathbb{R}^2\) with \(d_\infty(p,q)=\max(|p_1-q_1|,|p_2-q_2|)\), let \(x_n=\left(\tfrac1n,\tfrac1{n^2}\right)\to(0,0)\). What is the smallest \(N\) with \(d_\infty(x_n,(0,0))<0.1\) for all \(n\ge N\)?

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A. \(11\)

Explanation

\(d_\infty(x_n,0)=\max(1/n,1/n^2)=1/n\). We need \(1/n<0.1\), that is \(n>10\); for \(n=10\) it equals \(0.1\). Hence \(N=11\).

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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 \(x\).

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

In \(\mathbb{R}^2\) with \(d_\infty(p,q)=\max(|p_1-q_1|,|p_2-q_2|)\), let \(x_n=\left(\tfrac1n,\tfrac1{n^2}\right)\to(0,0)\). What is the smallest \(N\) with \(d_\infty(x_n,(0,0))<0.1\) for all \(n\ge N\)?

Explanation

Review the concept: Convergence in a metric space

More practice problems: convergence in a metric space