Does metric space have to contain zero element?

Question

Let $(X,d)$ be a metric space and $\{x_n\}$ be a Cauchy sequence. I want to show, by only using the metric $d$, that the sequence cannot be "unbounded" (in some sense). I would like to show this as follows:

$$d(x_n, x_N)\le d(x_n, x_{N+1})+d(x_{N+1}, x_N)\le d(x_n, x_{N+2})+d(x_{N+2}, x_{N+1})+\dots + d(x_{N+K}, x_{N+K-1})+d(x_N, x_{N+K-1})<(N+K+2)\varepsilon$$

Initially I was trying to show this by analogy with a normed space, but using things like $d(x_n, 0)$.

Hence my question: does there need to be the zero element in a metric space?

Answer 1

There is a precise sense of “bounded” in a metric space. A subset $A$ (in your case the set of terms of the sequence) is bounded if $$ \sup\{d(x,y):x\in A\} $$ is finite or, which is the same, there exists $K$ such that $$ d(x,y)\le K,\quad\text{for all $x,y\in A$} $$ You need no “zero”, which doesn't necessarily exist.*

Since the sequence is Cauchy, there exists $N$ such that, for $m,n>N$, $d(x_m,x_n)<1$.

Let $K'=\max\{d(x_{N+1},x_n):0\le n\le N\}$ and $K=2(K'+1)$. Then…

^{* Usually metric spaces are taken to be non empty, but there is no real reason to make such an assumption.}

Answer 2

A metric space does not have to have a designated zero element (what would that even mean, since you can't add elements together?).

However, you can try to pick any arbitrary point from the space and use that to function as a zero element, and see if that works for your needs.

Alternatively, you can choose a more carefully selected point as your zero point.

Answer 3

Not necessarily.

A metric space is a pair $(X,d)$ where $X$ is a set and $d:X^2\to \mathbb R$ is a function that satisfies some conditions.

That's all.

A zero only exists in some context: there must be some operation on $X$ and the zero then serves as some neutral element.