Metric spaces · Definition
Convergence in a metric space
Before you start you need to know
Saying a sequence «gets close» to a point requires a distance. Metric convergence gives that closeness its exact meaning and is the basis of limits and continuity.
Definition
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 \(x\).
The definition is equivalent to asking that the real sequence \(d(x_n,x)\) tends to \(0\), or that every ball \(B(x,\varepsilon)\) contains all but finitely many terms. In \(\mathbb{R}\) with \(d(x,y)=|x-y|\) one recovers the calculus limit; in Euclidean \(\mathbb{R}^n\), converging is equivalent to each coordinate converging.
The limit, if it exists, is unique: if \(x_n\to x\) and \(x_n\to y\), the triangle inequality gives \(d(x,y)\le d(x,x_n)+d(x_n,y)\), which is smaller than any \(\varepsilon>0\). Also, every subsequence of a convergent sequence converges to the same limit, and a convergent sequence is bounded.
It depends on two things: the metric and the space. With the discrete metric only sequences that are eventually constant converge. And \(1/(n+1)\) converges in \(\mathbb{R}\) but not in \(X=(0,1)\), because its limit \(0\) lies outside \(X\). This phenomenon motivates the next concept, complete spaces.
Think about it
What changes if \(\mathbb{R}^2\) uses the taxi metric instead of the Euclidean one?
Nothing about which sequences converge or their limits: \(d_\infty\le d_2\le d_1\le 2d_\infty\), so \(d_1(x_n,x)\to0\) iff \(d_2(x_n,x)\to0\). The value of \(N\) for a given \(\varepsilon\) does change.
Does \(x_n=(-1)^n/n\) converge in \(\mathbb{R}\)?
Yes, to \(0\), since \(d(x_n,0)=1/n\to0\); the sign does not affect the distance.
Exercises 10
- In usual \(\mathbb{R}\), what is the limit of \(x_n=1/n\)?
- 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 \(n\ge N\)?
- In Euclidean \(\mathbb{R}^2\), to which point does \(x_n=\left(\tfrac1n,\tfrac{n+1}{n}\right)\) converge?
- In \(\mathbb{R}\) with the discrete metric, which of these sequences converges?
- In usual \(\mathbb{R}\), what happens with \(x_n=(-1)^n\)?
- If \(x_n\to x\) and \(x_n\to y\) in a metric space, which argument proves \(x=y\)?
- Let \(X=(0,1)\) with the usual distance and \(x_n=\tfrac1{n+1}\). Does it converge in \(X\)?
- 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 \(n\ge N\)?
- Which statement is equivalent to \(x_n\to x\)?
- In usual \(\mathbb{R}\), let \(x_n=1+\tfrac1n\) and \(y_n=3-\tfrac1n\). What number does \(d(x_n,y_n)\) approach?
What comes next
Related
References
- James R. Munkres, Topology, 2nd edition, Pearson, 2000. §21
- Bert Mendelson, Introduction to Topology, 3rd edition, Dover Publications, 1990. cap. 2