# Every metric space has a $(1, 1)$-net

I'm trying to show that every metric space $X$ has a $(1, 1)$-net but struggling - surely $(1, 1)$ is just arbitrary and I've run out of obvious subgroups of $X$ to play with. Any help plz!

Here a subset $N$ of a metric space $X$ is called an $(\epsilon, \delta)$-net is for every $x \in X$, there is some $n \in N$ such that $d(x, n) \leq \epsilon$ and for every $n_1, n_2 \in N$, $d(n_1, n_2) \geq \delta$.

• Can you add in the definition of a (1,1)-net? – SomeEE Mar 1 '14 at 0:08
• sorry, have done! – jemima Mar 1 '14 at 17:38
• What is $N$ here? – Andrés E. Caicedo Mar 1 '14 at 17:40
• I think what you mean is that a metric space has an $(\epsilon,\delta)$-net iff there is a set $N\subseteq X$ with the property that you describe. It is the set $N$ that one calls the net, not the set $X$ itself. – Andrés E. Caicedo Mar 1 '14 at 17:41
• ugh yupp, sorry! – jemima Mar 1 '14 at 18:37

Pick a maximal subset $N$ of $X$ with the property that any two points in $N$ have distance at least $1$ from each other. Maximal means here that if $N\subsetneq N'\subseteq X$, then $N'$ contains points $n,n'$ with $n\ne n'$ and $d(n,n')<1$. That $N$ exists is an immediate consequence of Zorn's lemma.
The point is that $N$ is a $(1,1)$-net, because if $x\in X$ and $d(x,n)>1$ for all $n\in N$, then $x\notin N$ and $N\cup\{x\}$ contradicts the maximality of $N$.