# Proving this space is not totally bounded.

The real line $\Bbb R$ endowed with the metric $d(x,y)=\min(1,|x-y|)$ is a bounded metric space that isn't totally bounded.

I can easily prove that this is bounded. But to prove this is not totally bounded we have to find $P>0$ such that there is no finite cover of $P$ balls. Can someone help me to prove this?

• I assume "$P$ balls" means balls of radius $P$. – Silvia Ghinassi Nov 23 '15 at 21:46
• Use any $P\le 1$: in that case the $P$-balls in this metric are identical to those in the usual metric. – Brian M. Scott Nov 23 '15 at 21:47
• Just take $P=\frac12$, or anything else smaller than $1$. – 5xum Nov 23 '15 at 21:47
• @BrianM.Scott Don't you mean $<1$? – 5xum Nov 23 '15 at 21:47
• @5xum: No, $1$ also works. If $x$ is in the open $1$-ball about $y$, then $|x-y|<1$. – Brian M. Scott Nov 23 '15 at 21:48

If $P\leq 1$, then $B(x,P)=\{y \in \mathbb R \mid |x-y|<P\}=(x-P,x+P)$ is an open interval centered at $x$ as if $d(x,y)<1$, as the metric $d$ becomes the usual metric on $\mathbb R$.
Since $\mathbb R$ is unbounded with the usual metric (while, as you have proved, it is bounded with this metric) we can see that we can't find finitely many balls of radius $P$ that cover the entire space.
In fact, if we could, that would mean that there exist $x_1,\dots,x_N$ such that $\mathbb R \subset \bigcup_{i=1}^N (x_i-P,x_i+P)$. But the length of the union of such intervals is at most $2NP$, while $\mathbb R$ has infinite length.
• You are right, it is not and $\mathbb R$ is bounded. But if you take $P<1$, then the balls of this metric are the same balls of the usual metric (that is, open intervals) so you can't cover the entire space only with finitely many. – Silvia Ghinassi Nov 23 '15 at 22:13