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Metric Spaces and Sequences I (Overview) Saturday, August 31, 2024

Metric spaces, are in essence, the realm of analysis (in my experience). Analysis (or as I have heard from those my senior, “applied triangle inequality”) relies on notions of distance and, in particular, arbitrarily precise approximations. (e.g., the $ \varepsilon $-$ \delta $ arguments that you might have seen previously.)

Metric spaces and examples thereof

Consider three different locations A, B, C in our world. Experience tells us that

  1. To get from point A to point A should be $ 0 $ since we don’t have to move at all.
  2. The distance from point A to point B should be the same as the distance from point B to point A.
  3. Walking from A to B to C takes longer than walking from A to C directly.

From this, we get three basic rules that any “sensible” sense of distance should follow: if $ X $ is a space (read: set), a distance function $ d \colon X \times X \to \mathbb{R^{\geq 0}} $ satisfies for all $ x,y,z \in X $:

  1. $ d(x,x) = 0 $,
  2. $ d(x,y) = d(y,x) $, and
  3. $ d(x,y) + d(y,z) \geq d(x,z) $.

Any function satisfying this description is henceforth known as a metric on $ X $.

Note that $ \mathbb{R}^{\geq 0} $ is just the set of nonnegative reals or just $ \left \{ x \in \mathbb{R} : x \geq 0 \right \} $.

Here are many examples of metric spaces:

  1. $ X = \mathbb{R} $ and $ d(x,y) = |x-y| $
  2. $ X = \mathbb{R}^n $ and $ d(x,y) = [\sum_{i=1}^{n} (x_i-y_i)^2]^{1/2} $
  3. $ X = \mathbb{R}^n $ and $ d(x,y) =\sum_{i=1}^{n} |x_i-y_i| $
  4. $ X = \mathbb{R}^n $ and $ d(x,y) = \sup_{i=1}^n |x_i-y_i| $

Notice that we can have different metrics on the same underlying set. Neat! But now we have a few slightly more exotic examples:

  1. $ X = C([0,1], \mathbb{R}) \coloneqq \left \{ f \colon [0,1] \to \mathbb{R} \mid f \text{ continuous} \right \} $ and $ d(f,g) = \sup_{x \in [0,1]} |f(x)-g(x)| $
  2. If $ (X,d) $ is a metric space and $ F \subseteq X $, then $ (F,d) $ is a metric space
  3. Let $ X $ be any set and take $$\begin{equation} d(x,y) = \begin{cases} 1 \quad & x \neq y\\\ 0 \quad & x = y \end{cases} \end{equation}$$

Sequences

We now delve into the one of the most fundamental objects of study in analysis: the sequence.

A sequence is a list of elements from a set (like numbers from $ \mathbb{R} $) indexed by the positive integers $ \mathbb{Z}^+ $.1

To put say this more precisely, given a set $ X $, a sequence is a function $ p \colon \mathbb{Z}^+ \to X $. We also often denote $ p(n) $ by $ p_n $ for simplicity, and we write $ (p_n) $, $ (p_n)_n $, $ (p_n)_{n=1}^\infty $ for the sequence itself.

Also, please note that sequences are therefore infinite in length. ooppp If you need a finite list of things though, just be explicit (by saying “finite sequence” or something like that).

We also have an object called the subsequence, literally a sequence from within a sequence. A subsequence of a sequence $ (p_n) $ is sequence $ (q_m) $ defined by $ q_m = p_{\iota(m)} $ where $ \iota \colon \mathbb{Z}^+ \to \mathbb{Z}^+ $ is a function such that $ \iota(1) < \iota(2) < \iota(3) < \cdots $. We often refer to $ \iota(m) $ as $ n_m $. A shorthand for a subsequence is then $ (p_{n_m})_m $.

From our definition of subsequence, we have an immediate consequence: for all $ m \in \mathbb{Z}^+ $, $ \iota(m) = n_m \geq m $.

An understanding of long term behavior

The power of sequences is not necessarily in simply what occupies your list; it is moreso in its ability to represent the long term behavior of an object.

In particular, these infinitely-long lists can approach or converge to something. There’s obviously very simple cases like the sequence $ p_n = 0 $ for all $ n $, but so much can be studied in this paradigm. However, things can also not converge, like $ p_n = (-1)^n $ or $ p_n = n $.

From this idea of long term behavior, two big ideas fall out: convergence (approaching a single point, or none) and Cauchy-ness (how close the terms get to one another).

We now define both of these a bit more rigorously. Given a metric space $ (X,d) $, we say a sequence $ (p_n) $ in $ X $ converges to $ p \in X $ if for all $ \varepsilon > 0 $ there exists an $ N \in \mathbb{Z}^+ $ such that $ n \geq N $ implies that $ d(p_n,p) < \varepsilon $. In other words $$\begin{equation} (\forall \varepsilon > 0)(\exists N \in \mathbb{Z}^+)[n \geq N \implies d(p_n,p) < \varepsilon ]. \end{equation}$$ We denote this by $\lim_{n\to\infty} p_n = p $ or $ p_n \to p $.

In other other words, our sequence converges to $ p $ if we can get arbitrarily close to $ p $. Now we say our sequence is Cauchy if for all $ \varepsilon > 0 $ there exists $ N \in \mathbb{Z}^+ $ such that whenever $ n,m \geq N $, $ d(p_n,p_m) < \varepsilon $. In quite similar words as before, $$\begin{equation} (\forall \varepsilon > 0)(\exists N \in \mathbb{Z}^+)[n,m \geq N \implies d(p_n,p_m) < \varepsilon]. \end{equation}$$ In other other quite similar words as before, our sequence is Cauchy if the terms of the sequence get arbitrarily close together.

This may seem almost redundant: if they get close together, don’t they have to get to the same point? Yes! But also no. Indeed, they do approach a point, BUT that point may not be in your space…. This property of having the limit of every Cauchy sequence in your space is called completeness and we will talk about it later.

Boundedness

We first have to discuss sets being bounded. A set is bounded if it can be entirely contained in a ball of finite radius around some point of the space. Now, a sequence $ (p_n) $ is bounded if the set $ \left \{ p_n : n \in \mathbb{Z}^+ \right \} $ is bounded. Symbolically, $ (p_n) $ is bounded if and only if: $$\begin{equation} (\exists q \in X)(\exists r > 0)\left [ \left \{ p_n : n \in \mathbb{Z}^+ \right \} \subseteq \left \{ p \in X : d(p,q) < r \right \} \right ] \end{equation}$$

Some results on sequences

  1. If a sequence converges, its limit is unique.
  2. Every convergent sequence is bounded.
  3. Every Cauchy sequence is bounded.
  4. Every subsequence of a convergent sequence converges to the parent sequence’s limit.
  5. Any convergent sequence is a Cauchy sequence.
  6. Given a Cauchy sequence with a convergent subsequence, all subsequences share the same limit.

For the following proofs, let $ (X,d) $ be a metric space with a sequence $ (p_n) $ in $ X $.

Statement #1.

Solution

Suppose $ (p_n) $ converges to $ p,p^{\prime} \in X $. Then for all $ \varepsilon > 0 $, there exists $ N $ such that $ n \geq N $ such that $ d(p_n,p) < \varepsilon/2 $ and $ d(p_n,p^{\prime}) < \varepsilon/2 $. Then $$\begin{equation} d(p,p^{\prime}) \leq d(p,p_N) \end{equation}$$


Statement #2.

Solution

Suppose $ p_n \to p \in X $. Then there exists some $ N $ such that $ d(p_n,p) < 1 $ whenever $ n \geq N $. Then taking $$\begin{equation} r = \max \left \{ d(p_i,p) : 1 \leq i \leq N \right \} \cup \left \{ 1 \right \} \end{equation}$$ we get that $ d(p_i,p) < r $ for all $ i \in \mathbb{Z}^+ $.


Statement #3.

Solution

Suppose $ (p_n) $ is Cauchy. Then there exists some $ N $ such that $ n,m \geq N $ implies that $$\begin{equation} d(p_n,p_m) < 1. \end{equation}$$ Then we can apply the same argument as with the previous statement, but center our bounding set around $ p_N $ with radius $$\begin{equation} r = \max \left \{ d(p_i,p_N) : 1 \leq i \leq N \right \} \cup \left \{ 1 \right \} \end{equation}$$


Statement #4.

Solution

Suppose $ (p_{n_m}) $ is a subsequence of $ (p_n) $ and that $ p_n \to p \in X $. Let $ \varepsilon > 0 $. Then there exists an $ N $ such that $ n \geq N $ implies $ d(p_n,p) < \varepsilon $. Now, taking $ m \geq N $, we get that $$\begin{equation} d(p_{n_m},p) < \varepsilon \end{equation}$$ as $ n_m \geq m \geq N $.


Statement #5.

Solution

Suppose $ p_n \to p \in X $. Let $ \varepsilon > 0 $. Then there exists some $ N $ such that $ d(p_n,p) < \varepsilon/2 $ whenever $ n \geq N $. Then, taking $ n,m \geq N $, we get that $$\begin{equation} d(p_n,p_m) \leq d(p_n,p) + d(p,p_m) < \frac{ \varepsilon }{ 2 } + \frac{ \varepsilon }{ 2 } = \varepsilon. \end{equation}$$ Thus $ (p_n) $ is Cauchy.


Statement #6.

Solution

Suppose $ (p_n) $ is a Cauchy sequence and suppose $ (p_{n_m}) $ is a subsequence of $ (p_n) $ which converges to $ p \in X $. Now suppose $ (p_{n_i}) $ is another subsequence. Then, for any $ \varepsilon > 0 $, there is some $ M $ large enough so that $ j,k,m \geq M $ implies that $ d(p_{n_m},p) < \varepsilon/2 $ and $ d(p_j,p_k) < \varepsilon/2 $. Then taking $ i \geq M $, $$\begin{equation} d(p_{n_i},p) \leq d(p_{n_i}, p_{n_M}) + d(p_{n_M}, p) < \frac{ \varepsilon }{ 2 } + \frac{ \varepsilon }{ 2 } = \varepsilon. \end{equation}$$


  1. Most people index by either $ 0 $ or $ 1 $, depending on what’s convenient to them at that moment. ↩︎