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Metric spaces

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Overview

A metric space is a collection of points that additionally has a way to define the distance 𝑑(𝑖,𝑗) between every pair of points, which must satisfy the following conditions:

  1. 𝑑(𝑖,𝑖)=0for all 𝑖
  2. 𝑑(𝑖,𝑗)>0for all𝑖≠𝑗
  3. 𝑑(𝑖,𝑗)=𝑑(𝑗,𝑖)for all𝑖≠𝑗
  4. 𝑑(𝑖,𝑗)≤𝑑(𝑖,𝑘)+𝑑(𝑘,𝑗)for all𝑖,𝑗,𝑘 Some examples of metric spaces include:
  • ℝ𝑛 with Euclidean distance
  • ℝ𝑛 with Manhattan distance
  • a weighted graph with shortest-path distance

A fun visual for the last example, courtesy of this math exchange answer:

notes from napkin

Convergence

2.2.4: The convergent sequences in a discrete metric space (where different points have distance 1 and all others have distance 0) are precisely those which eventually become constant.

Continuity

2.3.4: Given 𝜀−𝛿 continuity and a sequence 𝑥𝑖 converging to 𝑝, our goal is to find, for all 𝜀>0, some 𝑁 such that ∀𝑛≥𝑁,𝑑𝑁(𝑓(𝑥𝑛),𝑓(𝑝))<𝜀. By 𝜀−𝛿 continuity, there must be some 𝛿 such that 𝑑𝑀(𝑥𝑛,𝑝)<𝛿⇒𝑑𝑁(𝑓(𝑥𝑛),𝑓𝑝)<𝜀; therefore, it suffices to find some 𝑁 such that ∀𝑛≥𝑁,𝑑𝑀(𝑥𝑛,𝑝)<𝛿. Such an 𝑁 must exist by the convergence of 𝑥𝑖. ∎

To reiterate the converse proof shown above, if 𝜀−𝛿 continuity does not hold, we can explicitly construct a sequence 𝑥𝑖 such that 𝑥𝑖 converges to 𝑝, but 𝑓(𝑥𝑖) does not converge to 𝑓(𝑝).

𝑓 is always continuous. As noted before, any convergent sequence 𝑥𝑖 in 𝐷 must become some constant 𝑝 after some 𝑁. Therefore, 𝑓(𝑥𝑖) always converges to 𝑓(𝑝). Intuitively, this is because no two points in 𝐷 are “next” to each other, and therefore they are all independent. ∎

Homeomorphisms

Why must we require the inverse also to be continuous?

Open sets

(0,1) is open in ℝ (but not in ℝ2), while [0,1] is open in neither.

Question 2.6.7. What are the open sets of the discrete space?

All subsets are open, since we can always select 𝑟=0.01. ∎

For a), just set 𝑟 for each point equal to the minimum viable 𝑟 among all the intersected sets. The same choice of 𝑟 suffices for 𝑏.

An infinite collection of open sets in ℝ whose intersection is {0} are the sets 𝑆𝑛=(−1𝑛,1𝑛). ∎

Closed sets

First, we show that if 𝑆 is closed, 𝑀\𝑆 must be open. Assume for contradiction that there exists some 𝑝∉𝑆 such that for every 𝜀>0, there exists some 𝑝′∈𝑆 such that 𝑑(𝑝,𝑝′)<𝜀. Then, we can write a sequence 𝑥𝑛 containing these values of 𝑝′ for 𝜀=1𝑛, which thus converges to 𝑝. This means 𝑝∈lim𝑆, a contradiction.

A similar contradiction argument works in the reverse direction. Suppose 𝑆 is open, and there exists a sequence of elements 𝑥𝑖∉𝑆 that converges to an element 𝑝 in 𝑆. Then 𝑝 has no 𝜀-neighborhood in 𝑆 for any 𝜀>0, contradicting the openness of 𝑆. ∎

Completeness

If the sequence converges, can we select 𝑚,𝑛≥𝑁 such that for all 𝑛≥𝑁,𝑑(𝑥𝑛,𝑝)<𝜀2. By the triangle inequality, this means 𝑑(𝑥𝑚,𝑥𝑛)≤𝑑(𝑥𝑚,𝑝)+𝑑(𝑥𝑛,𝑝)<𝜀, as desired. ∎

We first prove that if 𝑆 is complete, it is closed in 𝑀. This is because any convergent sequence in 𝑆 is a Cauchy sequence, which by completeness must converge to another element in 𝑆.

Moreover, because 𝑀 is a complete metric space, any Cauchy sequence in 𝑆 must converge to some element in 𝑀. Therefore, if 𝑆 is closed in 𝑀, all such convergent points must be contained in 𝑆, making it closed. ∎

Sketch. Take any starting point 𝑥0; we will show that iterating 𝑇 on 𝑥0 will eventually converge to a fixed point. This is because the sequence 𝑇𝑛(𝑥0) is Cauchy, and 𝑀 is a complete metric space. ∎

Graph View

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