Pome

Proposition 17.10.598F175566EB00

Proposition 17.10.59.8F175566EB00 Let $(E,d,\land )$ be a rooftop metric space. Then $(E,d)$ is complete if and only if both of the following hold:

  1. 0E5A5FA43046

    Each increasing Cauchy sequence converges.

  2. 535E2E2720BC

    Each decreasing Cauchy sequence converges.

Proof. The direct implication is trivial.

Conversely, assume that both conditions are true. Let $(x_ j)_{j>0}$ be a Cauchy sequence in $E$. We want to prove that $(x_ j)_ j$ converges. By passing to a subsequence, we may assume that

\[ d\left(x_ j,x_{j+1}\right)\leq 2^{-j}. \]

For $k,j\geq 1$, let

\[ y_ j^ k\coloneqq x_ k\land \cdots \land x_{k+j}. \]

Then $(y_ j^ k)_ j$ is decreasing, and

\[ d\left(y_ j^ k,y_{j+1}^ k\right)\leq d\left(x_{k+j},x_{k+j+1}\right)\leq 2^{-k-j}. \]

So $(y_ j^ k)_ j$ is a decreasing Cauchy sequence. Define

\[ y^ k\coloneqq \lim _{j\to \infty }y_ j^ k. \]

Then

\[ d\left(y^ k,y^{k+1}\right)=\lim _{j\to \infty }d\left(y_{j+1}^ k,y_ j^{k+1}\right)\leq d\left(x_ k,x_{k+1}\right)\leq 2^{-k}. \]

We also need to check that the sequence $(y^ k)_ k$ is increasing. For each $j\geq 1$, we have

\[ y_ j^ k=x_ k\land y_{j-1}^{k+1}\leq y_{j-1}^{k+1}. \]

Equivalently,

\[ y_ j^ k\land y_{j-1}^{k+1}=y_ j^ k. \]

Letting $j\to \infty $ and using the continuity of $\land $ from 17.10.58, we obtain

\[ y^ k\land y^{k+1}=y^ k. \]

Thus $y^ k\leq y^{k+1}$. So $(y^ k)_ k$ is an increasing Cauchy sequence. Let

\[ y\coloneqq \lim _{k\to \infty }y^ k. \]

For every $j\geq 0$, we have

\[ d\left(y_ j^ k,x_ k\right)\leq \sum _{r=k}^{k+j-1}d\left(x_ r,x_{r+1}\right), \]

with the empty sum understood as $0$. Indeed, this is trivial for $j=0$, and the induction step follows from

\[ y_ j^ k=x_ k\land y_{j-1}^{k+1} \]

and

\[ d\left(y_ j^ k,x_ k\right)\leq d\left(y_{j-1}^{k+1},x_ k\right)\leq d\left(y_{j-1}^{k+1},x_{k+1}\right)+d\left(x_{k+1},x_ k\right). \]

Letting $j\to \infty $, we get

\[ d\left(y^ k,x_ k\right)\leq \sum _{r=k}^{\infty }d\left(x_ r,x_{r+1}\right)\leq 2^{1-k}. \]

So $(x_ k)_ k$ converges to $y$. $\square$

Cite this page

Comments

Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.

Read this page’s discussion on GitHubPost a comment on GitHub