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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 :) )

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    $\begingroup$ You are not quite applying the definition correctly. The number of balls needed is allowed to grow when $\epsilon$ gets smaller. It just needs to be finite for any choice of $\epsilon$ (which it does in your example). $\endgroup$
    – Tobias Kildetoft
    Oct 29, 2015 at 9:10
  • $\begingroup$ Ok, but for $\varepsilon \to 0$ then you need infinite balls, no? I am getting confused... $\endgroup$
    – Barzi2001
    Oct 29, 2015 at 9:17
  • $\begingroup$ No, you never need an infinite number of balls. You need to understand the difference between a sequence of natural numbers diverging to infinity and actually needing an infinite number of things. $\endgroup$
    – Tobias Kildetoft
    Oct 29, 2015 at 9:20
  • $\begingroup$ Ohi, that's hard. Is that every element of the diverging sequence you mentioned represent the number of things (balls) that I need? $\endgroup$
    – Barzi2001
    Oct 29, 2015 at 9:28
  • $\begingroup$ Yes, precisely. $\endgroup$
    – Tobias Kildetoft
    Oct 29, 2015 at 9:29

2 Answers 2

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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.

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  • $\begingroup$ Ok, so I don't mind if it holds $\lim_{\varepsilon \to 0} N(\varepsilon) = +\infty$? It is enough that $N(\varepsilon)$ is finite for all $\varepsilon >0$, right? $\endgroup$
    – Barzi2001
    Oct 29, 2015 at 9:24
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    $\begingroup$ @UbaldoTiberi Indeed. Any finite positive $\epsilon$ has a corresponding finite number $N(\epsilon)$. Nothing else. $\endgroup$
    – Henno Brandsma
    Oct 29, 2015 at 9:25
  • $\begingroup$ Ok! That was what confused me. :) If it it was stated "Any finite positive $\varepsilon$" from the beginning, then I would have been OK, but given that the definition should hold for $\forall \varepsilon >0$ I understand that $N(\varepsilon)$ finite shall hold also for "infintesimally small" epsilon. To indicate finite (but not infintesimally small) I would write $\forall \varepsilon > a, a>0$ instead. :) $\endgroup$
    – Barzi2001
    Oct 29, 2015 at 9:32
  • $\begingroup$ @UbaldoTiberi there are no infinitesimally small $\epsilon$. There are only finite $\epsilon$'s that we can choose as small as we like. We use these $\epsilon$-$\delta$ definitions to avoid "infinitesimal numbers". (except in non-standard analysis, but that's a different approach). $\endgroup$
    – Henno Brandsma
    Oct 29, 2015 at 9:35
  • $\begingroup$ Ok, so I have one thing in mind. If we see $N(\varepsilon)$ as a piecewise constant function $N:\mathbb{R^+}\to \mathbb{R^+}$ that associates to each $\varepsilon$ the number of balls $N(\varepsilon)$ needed to cover the set considered in the metric space, can we conclude that every metric space can be rendered totally bounded provided that choice of the distance function lead to a bounded $N(\varepsilon)$ (or, at most bounded plus $lim_{\varepsilon \to 0} N(\varepsilon) +\infty$)? $\endgroup$
    – Barzi2001
    Oct 29, 2015 at 9:47
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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.

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