2
$\begingroup$

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:

  • In $\mathbb {R}$ endowed with the Euclidian Metric a set $K$ is compact iff it is closed and bounded.

  • Every compact set $K$ in a metric space $\left (X,d\right)$ is totally bounded.

  • 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.

$\endgroup$
3
$\begingroup$

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 $B(x_1,\varepsilon)$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.