Let $\left (a,b\right)\subset\mathbb {R} $ be endowed with the absolute value metric. In other words to show that $\left (a,b\right)$ is totally bounded in $\mathbb {R}$ I used the following facts:
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In $\mathbb {R}$ endowed with the Euclidian Metric a set $K$ is compact iff it is closed and bounded.
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Every compact set $K$ in a metric space $\left (X,d\right)$ is totally bounded.
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Every non empty subset $B$ of a totally bounded set $A$ is totally bounded.
I think that with this aproach the problem is solved. Nevertheless I would be tremendously thankful if someone could provide me a hint to prove it using the definition. This is:
$\forall\epsilon>0$ there exist a finite collection of points $x_{1},…,x_{n} \in \left (a,b\right)$ such that
$\left (a,b\right)\subseteq
\bigcup_{i=1}^{n} B\left (x_{i},\epsilon\right) $
Thank you in advanced for any help provided.
Best Answer
By the Archimedean property of $\mathbb R$, for all $\varepsilon>0$, there exists $n\in\mathbb N$ such that $n\varepsilon>(b-a)$, and we may take $n$ to be the least such integer. Consider the points $x_k=a+k\varepsilon$ for $k=1,2, \ldots,n-1$, and the intervals $B(x_k,\varepsilon)$ will cover $(a,b)$.
(The above doesn't quite work in case $\varepsilon>b-a$, in which case instead you can take $x_1=\frac{a+b}{2}$ and cover $(a,b)$ with $B(x_1,\varepsilon)$.)