This natural property is not true always, when I mean a bounded set, I mean a set that it's contained in some ball of radius $r$. It's not always true that if this is the case, I can cover fixing a radius $u < r$ cover with balls of radius $u$, my bounded set, using only finite or even numerable balls. An example is the discrete metric, and any no numerable set ($[0,1]$) . What conditions I can impose in my space, to have this nice property? In any bounded set, not only in compact sets (or at least in open balls, that's what I need now)

  • $\begingroup$ The first letter in your post should be capitalized (since it's the first letter in a sentence). "I mean a bounded set" should be "I say a bounded set". "I mean a set that it's" should be "I mean a set that is". The next sentence should be "It's not always true, that if this is the case, I can cover my bounded set using only finitely many or even numerably many balls of a fixed radius u<r." (the word "I" should always be capitalized, and the sentence was badly put together). $\endgroup$ – user57159 Aug 24 '11 at 22:39
  • $\begingroup$ "I can impose in my space" should be "can I impose on my space" ("I can" is for a statement and "can I" is for a question, and "impose in my space" sounds like we already have a specific space and the conditions are on something else). $\endgroup$ – user57159 Aug 24 '11 at 22:39

If Countable Choice, then your property is equivalent to being a Lindelöf space.


I assume we must have $0<u$. $ \; $ (I am not sure if that is part of the definition of balls.)

Let $B_u(x)$ be the open ball of radius $u$ centered at $x$.

Let $\overline{B}_u(x)$ be the closed ball of radius $u$ centered at $x$.

Clearly, $\; B_u(x) \subseteq \overline{B}_u(x) \subseteq B_{\frac12 \cdot (u+r)}(x)$,
so it does not matter whether we cover with open or closed balls.

I also assume you are including finite sets as "numerable", and I include them as "countable".

part 1: Lindelöf spaces have your property

Let $C$ be the set of all open balls of radius $u$.
Since the space is Lindelöf, let $C'$ be a countable subcover of $C$.
$C'$ is a countable cover of every bounded subset, and this works for all $u$, which proves part 1.

part 2: If countable choice, then spaces with your property are Lindelöf

(CC) will mean a use of countable choice.
If the space is finite, then in space is compact, in which case the space is Lindelöf.
Assume the space is infinte. $ \; $ Let $x$ be a point in the space.
By your property, for all non-negative integers $n$,
there is a countable cover of $B_{n+1}(x)$ by balls of radius $\frac1{n+1}$.
(CC) $\;$ For all non-negative integers $n$, let $C_n$ be such a cover.
For all positive integers $n$, since $x_0\in B_n(x)$, $C_n$ is non-empty.

Define $\quad D \; = \; \displaystyle\bigcup_{n\in \{0,1,2,3,...\}} C_n$

(CC) $\;$ By the last property proven here, $D$ is countable.
Since the diameter of the balls in $C_n$ goes to zero as $n$ goes to infinity, and the space is infinite,
the number of balls in $C_n$ must go to infinity as $n$ goes to infinity. $ \; $This shows that $D$ is infinite. $\;\;$ Since $D$ is countably infinite, let $\; f : \{0,1,2,3,...\} \to D \;$ be some bijection.
(CC) $\;$ For all non-negative integers $n$, let $x_n$ be a member of $f(n)$.
That finishes the setup.
Now, let $U$ be a non-empty open subset of the space, and let $y$ be a member of $U$.
Let $r$ be a positive real number such that $\; B_r(y) \subseteq U \;$.
Since the diameter of the balls in $C_n$ goes to zero as $n$ goes to infinity, there is an $n$ such that the diameter of the balls in $C_n$ is less than $r$. $\;\;$ Let $m$ be such an $n$.
Let $B$ be a member of $C_m$ such that $y\in B$. $ \; $ By construction, $\; x_{f^{-1}(B)} \in B \subseteq B_r(y) \subseteq U \;$.
This works for all non-empty open subsets $U$ of the space, so the space is separable.
(CC) $\;$ Therefore, by Theorem 3.12, the space is Lindelof.


  • 1
    $\begingroup$ I agree that this is a particularly badly written question. However, I wouldn't focus on the grammar and capitalization; the site should be accessible to non-native speakers who might not be able to produce perfect grammar. I think the more important flaw is the lack of structure in the exposition of the ideas; ideas are begun, interrupted and then partially repeated, and the implied universal quantifier over $u$ isn't immediately clear. (By the way, the ping doesn't work if you put a space between the '@' and the name.) $\endgroup$ – joriki Aug 24 '11 at 5:15
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    $\begingroup$ I disagree. Your user page gives no indication that you're not a native speaker of Lindelöf's native language, whose name you misspelled, and I don't see why it should. Roughly 95% of all people are not native English speakers. By "ping" I mean the notification you get e.g. when I post a comment under your answer. You can explicitly send one of these using the '@' directly followed by (at least the first three letters of) a user name. $\endgroup$ – joriki Aug 24 '11 at 5:37
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    $\begingroup$ This is not an accent; it's a diacritic that turns the vowel into a merely historically related vowel. In English, which experienced the same vowel shifts but denotes the shifted vowels by different letters, the difference would correspond e.g. to "men" being misspelled as "man". (See en.wikipedia.org/wiki/Trema_(diacritic).) $\endgroup$ – joriki Aug 24 '11 at 6:33
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    $\begingroup$ @Ricky: The specific English errors in Daniel’s questions make it pretty obvious that he’s not a native speaker of English, but he generally manages to make himself understood. $\endgroup$ – Brian M. Scott Aug 24 '11 at 7:21
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    $\begingroup$ @Asaf: Since one has to be online to post a question, one has available at that time that ever ready source called Google. :-) Searching for "Lindelof", "Godel" and "Erdos" will get you plenty of copyable instances of the names. If you ever have to write an umlaut without that vast resource at your fingertips, it's worth knowing that in German they are replaced by "ae", "oe" and "ue", respectively when they're not available, and Wikipedia tells me that this is also done to varying degrees in other languages. $\endgroup$ – joriki Aug 24 '11 at 13:13

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