# Review

1. Let $$(X, d_X)$$ be a metric space. We say that it is complete if every Cauchy sequence converges.
2. 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.