Metric completion

The completion $―― 𝑋$ of a metric space $𝑋$ may be thought of as the smallest possible metric space containing $𝑋$ but with all limits added, #m/def/anal i.e. a complete metric space. This notion is made rigorous by the universal property, which ensures uniqueness up to unique isomorphism.

Universal property

The metric completion is characterized up to unique isomorphism in $𝖨 𝗌 𝗈 𝖬 𝖾 𝗍$ by the following universal property:

$―― 𝑋$ is complete. If $𝑌$ is a complete metric space and $𝑓 \in 𝖨 𝗌 𝗈 𝖬 𝖾 𝗍 (𝑋, 𝑌)$ such that $𝑓 (𝑋)$ is dense in $𝑌$ , then there exists a unique isometry $¯ 𝑓 \in 𝖨 𝗌 𝗈 𝖬 𝖾 𝗍 (―― 𝑋, 𝑌)$ such that $𝑓 = ¯ 𝑓 𝜄$ , i.e. the following diagram commutes

This of course forms a Free-forgetful adjunction into the full subcategory of complete metric spaces and isometries.

Construction

Let $˜ 𝑋$ denote the set of all Cauchy sequences on $𝑋$ , For any sequences $𝑥 •, 𝑦 • \in ˜ 𝑋$ , let

𝑥 • \sim 𝑦 • ⟺ l i m 𝑛 \to \infty 𝑑 (𝑥 𝑛, 𝑦 𝑛) = 0

which defines an equivalence relation. The completion $―― 𝑋$ is the quotient $˜ 𝑋 / \sim$ with a metric $𝑑$ given by

𝑑 ([𝑥 •], [𝑦 •]) = l i m 𝑛 \to \infty 𝑑 (𝑥 𝑛, 𝑦 𝑛)

and $𝜄 (𝑥) = [(𝑥, 𝑥, 𝑥, \dots)]$ .

Validity of construction

First we will show that $\sim$ indeed defines an equivalence relation. reflexivity and symmetry are clear, and transitivity follows from triangle inequality: If $𝑥 • \sim 𝑦 •$ and $𝑦 • \sim 𝑧 •$ , then $l i m 𝑛 \to \infty 𝑑 (𝑥 𝑛, 𝑧 𝑛) \leq l i m 𝑛 \to \infty (𝑑 (𝑥 𝑛, 𝑦 𝑛) + 𝑑 (𝑦 𝑛, 𝑧 𝑛)) \leq 0$ so $𝑥 • \sim 𝑧 •$ .

Next we show that the metric on $―― 𝑋$ is well-defined. Let $𝑥 • \sim 𝜉 •$ and $𝑦 • \sim 𝜂 •$ . Then by the triangle inequality

𝑑 ([𝑥 •], [𝑦 •]) = l i m 𝑛 \to \infty 𝑑 (𝑥 𝑛, 𝑦 𝑛) \leq l i m 𝑛 \to \infty (𝑑 (𝜉 𝑛, 𝜂 𝑛) + 𝑑 (𝑥 𝑛, 𝜉 𝑛) + 𝑑 (𝑦 𝑛, 𝜂 𝑛)) = l i m 𝑛 \to \infty 𝑑 (𝜉 𝑛, 𝜂 𝑛) = 𝑑 ([𝜉 •, 𝜂 •])

but by symmetry the reverse inequality holds too, so $𝑑 ([𝑥 •], [𝑦 •]) = 𝑑 ([𝜉 •, 𝜂 •])$ . symmetry and positive definite follow immediately. For positive definite note

𝑑 ([𝑥 •], [𝑦 •]) = l i m 𝑛 \to \infty 𝑑 (𝑥 𝑛, 𝑦 𝑛) \leq l i m 𝑛 \to \infty (𝑑 (𝑥 𝑛, 𝑧 𝑛) + 𝑑 (𝑧 𝑛, 𝑦 𝑛)) = 𝑑 ([𝑥 •], [𝑧 •]) + 𝑑 ([𝑧 •], [𝑦 •])

as required.

Now we need to show that the given construction is indeed complete. We make the following observations

Let $―― 𝑥 \in ―― 𝑋$ . If $𝑥 ∙ \in ―― 𝑥$ , then so too is every subsequence $𝑥 𝑛 ∙ \in ―― 𝑥$ .
Let $𝑥 ∙ \in ―― 𝑥 \in ―― 𝑋$ . Since $𝑥 ∙$ is Cauchy, for every $𝜖 > 0$ there exists a subsequence $𝑥 𝑛 ∙$ such that $𝑑 (𝑥 𝑛 𝑘, 𝑥 𝑛 ℓ) < 𝜖$ for all $𝑘, ℓ \in ℕ$ .

Let $―― 𝑥 ∙$ be a Cauchy sequence in $―― 𝑋$ . For each $𝑛 \in ℕ$ , we fix a representative $𝑥 𝑛, ∙ \in ―― 𝑥 𝑛$ in $𝑋$ such that for all $𝑘, ℓ \in ℕ$ we have $𝑑 (𝑥 𝑛, 𝑘, 𝑥 𝑛, ℓ) < 1 𝑛$ . Since $―― 𝑥 ∙$ is Cauchy, for every $𝑗 \in ℕ$ there exists an $𝑁 𝑗 \in ℕ$ such that for all $𝑚, 𝑛, 𝑘 \geq 𝑁 𝑗$ we have $𝑑 (𝑥 𝑛, 𝑘, 𝑥 𝑚, 𝑘) < 1 𝑗$ (using that sufficiently large $𝑚, 𝑛$ have $l i m 𝑘 \to \infty 𝑑 (𝑥 𝑚, 𝑘, 𝑦 𝑛, 𝑘) < 1 𝑗$ ). Now let $𝜆 𝑗 = 𝑥 𝑁 𝑗, 𝑁 𝑗$ which is Cauchy in $𝑋$ , since for any $ℓ \in ℕ$ if $𝑖, 𝑗 \geq ℓ$ and thus $𝑁 𝑖, 𝑁 𝑗 \geq 𝑁 ℓ$ we have

𝑑 (𝜆 𝑖, 𝜆 𝑗) = 𝑑 (𝑥 𝑁 𝑖, 𝑁 𝑖, 𝑥 𝑁 𝑗, 𝑁 𝑗) \leq 𝑑 (𝑥 𝑁 𝑖, 𝑁 𝑖, 𝑥 𝑁 𝑖, 𝑁 𝑗) + 𝑑 (𝑥 𝑁 𝑖, 𝑁 𝑗, 𝑥 𝑁 𝑗, 𝑁 𝑗) \leq 1 𝑖 + 1 ℓ \leq 2 ℓ

We claim $―― 𝑥 = [𝜆 ∙]$ is the limit of $―― 𝑥 ∙$ .

Let $𝑗 \in ℕ$ and $𝑛 \geq m a x {𝑁 𝑗, 𝑗}$ . Then

𝑑 (―― 𝑥 𝑛, ―― 𝑥) = l i m 𝑘 \to \infty 𝑑 (𝑥 𝑛, 𝑘, 𝜆 𝑘) = l i m 𝑘 \to \infty 𝑑 (𝑥 𝑛, 𝑘, 𝑥 𝑁 𝑘, 𝑁 𝑘) \leq l i m 𝑘 \to \infty (𝑑 (𝑥 𝑛, 𝑁 𝑘, 𝑥 𝑁 𝑘, 𝑁 𝑘) + 𝑑 (𝑥 𝑛, 𝑘, 𝑥 𝑛, 𝑁 𝑘)) \leq 1 𝑗 + 1 𝑛 \leq 2 𝑗

so $―― 𝑥 ∙ \to ―― 𝑥$ .

It remains to show that $―― 𝑋$ satisfies the universal property. Let $𝑌$ be a complete metric space and $𝑓 \in 𝖨 𝗌 𝗈 𝖬 𝖾 𝗍 (𝑋, 𝑌)$ .

#state/tidy | #lang/en | #SemBr

Metric completion

Universal property

Construction

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