Complete Spaces

1. A space is complete if every Cauchy sequence converges; any incomplete normed linear space has a completion.
2. The completion of a space is unique in the sense of isometric isomorphism.
   • In categorical terms, the completion is unique up to a unique isometric isomorphism: for any two completions of a given normed space, there exists a unique isometric isomorphism between them that restricts to the identity on the original space.
3. This uniqueness relies on density.

The Importance of Density

\[X:=\left\{(x_1, x_2,...)\in \mathbb R^\infty,\ \sup_{n\in \mathbb N} |x_n|<\infty\right\}\] \[X_1:=\left\{(x_1, x_2,...)\in \mathbb R^\infty,\ x_1=0\ \text{or}\ 1,\ \sup_{n\in \mathbb N} |x_n|<\infty\right\}\] \[X_2:=\left\{(0,x_1, x_2,...)\in \mathbb R^\infty,\ \sup_{n\in \mathbb N} |x_n|<\infty\right\}\]

Clearly, $(X, d)$ is a complete space, and it is its own completion. It is easy to prove that $(X_1, d)$ is also a completion of $(X, d)$. Now define

\[X_3:=\left\{(1,x_1, x_2,...)\in \mathbb R^\infty,\ \sup_{n\in \mathbb N} |x_n|<\infty\right\}\]

We aim to prove that $(X_1, d)$ and $(X, d)$ are not homeomorphic. In fact, $(X_1, d)$ is not connected¹. Let us prove this. The key is to show that both $X_2$ and $X_3$ are open sets. Take $x\in X_3$, then $B(x,\tfrac 1 2)\cap X_1\subset X_3$ — this shows that every point in $X_3$ has an open neighborhood. Similarly, $X_2$ is also open. Since $X_2\cup X_3= X_1$, the latter is not connected.

From this proof, we can see the importance of density. In fact, the definition of “minimal” gives a partial order of metric spaces under isometries. However, this order is not total, so it is possible to have two completely unrelated chains, i.e., two different minimal elements.

Isometric isomorphism alone does not guarantee homeomorphism.