# Math 202A Lecture 3: Completions

# Review

- Let \((X, d_X)\) be a metric space. We say that it is
*complete*if every Cauchy sequence converges. - By a
*completion*of \((X, d_x)\) we mean a complete metric space \(\overline{X}\) together with a specific isometry \(d_x \colon X \to \overline{X}\) whose range is dense in \(\overline{X}\).

**Example.** We already looked at \(\mathbb{Q} \hookrightarrow \mathbb{R}\).

**Proposition.** If \((Y_1, j_{y_1})\) and \((Y_2, j_{y_2})\) are both completions of \((X, d_X)\) then there is an isometry \(\varphi \colon Y_1 \to Y_2\) onto, such that \(j_{y_2} = \varphi \circ j_{y_1}\).

The implication of this proposition is that any two completions of a metric space are isomorphic, implying that completion is unique.

\(S\) dense in \(X\) means that for every \(x \in X\), \(B(X, \epsilon) \cap S \neq \emptyset\) for all \(\epsilon > 0\).

# Completions

To build completions, we will essentially construct a new space out of Cauchy sequences. Note that if we have two sequences \(\{ s_n \}, \{ f_n \} \subset X\) converging to the same point \(x \in X\), then \(d_X(s_n, t_n) \to 0\) by the triangle inequality.

In non-complete metric spaces we cannot compare Cauchy sequences based on where they converge, but we can still use this fact to define a notion of equivalence on them:

For any metric space \((Z, d_z)\) and Cauchy sequences \(\{ s_n \}, \{ t_n \} \subset Z\), we say they are *equivalent* if \(d_Z(s_n, t_n) \to 0\) as \(n \to \infty\).

You can check that this equivalence satisfies all the properties of an equivalence relation. Reflexivity and symmetry follow easy, and the triangle inequality gives transitivity.

**Proposition.** Let \((X, d_x)\) and \((Y, d_y\)) be metric spaces, with \((Y, d_y)\) complete. Let \(S\) be a dense subset of \(X\) with the metric from \(X\). Let \(f \colon S \to Y\) be uniformly continuous. Then there exists a continuous extension, \(\bar{f} \colon X \to Y\) of \(f\), where \(\bar{f}|_S = f\). \(\bar{f}\) will also be unique and uniformly continuous.

*Proof.* The full proof is a problem set exercise, so do yourself. We prove existence and continuity.

For \(x \in X\), choose a sequence \(\{ s_n \} \subset S\) converging to \(x\). Because \(f\) is uniformly continuous, \(\{ f(s_n) \}\) is a Cauchy sequence in \(Y\). Since \(Y\) is complete, it to a point \(p \in Y\), define \(\bar{f}(x) = p\).

We must show \(\overline{f}\) is well-defined, that its definition does not depend on our choice of sequence in the previous part. Formally, we must show that if \(\{ t_n \} \subset S\) is another Cauchy sequence converging to \(x\), then \(\{ f_n \}\) also converges to \(p\). Since \(\{ s_n \}, \{ t_n \}\) both converge to \(p\), they are equivalent Cauchy sequences, i.e. \(d_S(s_n, t_n) \to 0\). Since \(f\) is uniformly continuous, \(d(f(s_n), f(t_n)) \to 0\). This implies \(\{ f(t_n) \} \to p\).

Every metric space \((X, d_x)\) has a completion \((\overline{X}, d_{\overline{x}})\).

We will complete the proof in the next lecture.