Is [0,1] in R totally bounded? Although the concept of boundedness is quite understandable, I am struggling with the concept of total boundedness. 
If I understand properly, then the metric space $[0,1] \subset \mathbb{R}$ with the ordinary distance $d(x,y)=|x-y|$ is not totally bounded. This because I cannot find a finite union of balls of radius $\varepsilon$, with $\varepsilon$ that can be arbitrary small, that "covers" $[0,1]$. 
For example, for $\varepsilon = 0.5$ I can find a finite number of balls to cover $ [0,1]$ (for instance, two balls are enough), for $\varepsilon = 0.2$ I need e.g. five balls, etc. But for arbitrary small $\varepsilon$ I would need an infinite number of balls, and therefore the defined metric space is not totally bounded. Is that reasoning correct? 
If so, I I have the feeling that a necessary condition to have totally boundedness is that the set in the metric space must be finite (the condition  $\forall \varepsilon>0$ bothers me a lot :) )
 A: The condition is that for every $\epsilon > 0$ there is some finite $N(\epsilon)$ (a natural number) so that $N(\epsilon)$ many balls of radius $\epsilon$ cover $[0,1]$. There is no other condition except that $N(\epsilon)$ is finite for every $\epsilon > 0$, and this number will often grow with $\epsilon$ getting smaller, but that's OK. The finite number depends on $\epsilon$ and that's fine.
A: The condition “for all $\varepsilon>0$” should not bother you and it won't once you grasp its meaning.
The statement that “$[0,1]$ is totally bounded” can be seen as a challenge you are sure to win. The game is that you ask the challenger to select a positive number $\varepsilon$ and you win if you are able to show a finite number of balls of radius $\varepsilon$ that cover $[0,1]$.
In the rules of the game there is no bound on the number of balls you have to list: you're free to list $3$, $1000$ or more after the choice of $\varepsilon$ has been made.
You are sure to win the challenge because, if a challenger selects the number $\varepsilon$, you can take an integer $n>1/\varepsilon$ and the balls of radius $\varepsilon$ centered at $1/n, 2/n, 3/n, \dots, (n-1)/n$ will cover $[0,1]$.
Maybe you can do better, that is, with less balls, but it's not in the rules of the game that you have to provide the minimum number of balls.
Since you know a general procedure for listing the needed balls, you are sure to win whichever value for $\varepsilon$ the challenger actually chooses.
