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If a metric space is separable, then any open set is a countable union of balls. Is the converse statement true?

UPDATE1. It is a duplicate of the question here https://math.stackexchange.com/questions/94280/if-every-open-set-is-a-countable-union-of-balls-is-the-space-separable/94301#94301

UPDATE2. Let me summarize here the positive answer following Joel David Hamkins and Ashutosh. It is a matter of taste, but I omit using ordinals and use Zorn lemma instead, which may be more usual for most mathematicians (at least, it is for me).

Lemma 1. If $(X,d)$ is non-separable metric space, then for some $r>0$ there exists an uncountable subset $X_1\subset X$ such that $d(x,y)>r$ for any two points $x\ne y$ in $X_1$.

Proof. For each $r=1/n$ consider the maximal (by inclusion) subset with such property. If it is countable, then $X$ has a countable $1/n$-net for each $n$, hence it is separable.

Now consider two cases. Define $X_2\subset X_1$ as a set of points $x\in X_1$ for which there exist a point $y_x\in X$ such that $0<d(x,y_x)<r/10$. Consider two cases.

1) $X_2$ is uncountable. Consider the union of open balls $U=\cup_{x\in X_2} B(x,d(x,y_x))$. Consider any open ball $B(z,a)$ containing in $U$. We have $z\in U$, so $d(z,x)<d(x,y_x)$ for some $x$, but $r/5>2d(x,y_x)\geq d(x,z)+d(x,y_x)\geq d(z,y_x)>a$ since $y_x\notin B(z,a)$. It implies that $B(z,a)$ is contained in a unique ball $B(x,d(x,y_x))$, hence we need uncountably many such balls to cover whole $U$.

2) $X_3=X\setminus X_2$ is uncountable. For any $x\in X_3$ define $R(x)>0$ as a radius of maximal at most countable open ball centered in $x$. Clearly $R(x)\geq r/10$ for any $x\in X_3$. For any $x\in X_3$ define a star centered in $x$ as a union $D=x\cup C$, where $C=\{z_1,z_2,\dots\}\subset X_3$ is a countable sequence of points with $d(x,z_i)\rightarrow R(x)+0$. Choose a maximal disjoint subfamily of stars. Clearly it is uncountable, else we may easily increase it. Denote by $U$ the set of centers of chosen stars. It is open (as any subset of $X_3$), assume that it is a countable union of balls $U=\cup_{i=1}^{\infty} B(x_i,r_i)$, $x_i\in U$. We have $r_i> R(x_i)$ for some $i$, else $U$ is at most countable. But then $B(x_i,r_i)$ contains infinitely many points of the star $D$ centered in $x_i$, while by our construction $U\cap D=\{x_i\}$. A contradiction.

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    $\begingroup$ A surprising cute question. Perhaps it's more than just cute. $\endgroup$ – Włodzimierz Holsztyński Sep 18 '14 at 19:19
  • $\begingroup$ One could also consider a closely related question, namely a metric space $\ (Y\ d),\ $ and $\ X\subseteq Y$, and traces of balls $\ B(y\ r)\cap X$ for $\ y\in Y\ $ (instead of simply $\ X\ $ and balls in $\ X\ $). $\endgroup$ – Włodzimierz Holsztyński Sep 18 '14 at 19:29
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    $\begingroup$ If we allow exterior centers, the answer becomes obviously no: take arbitrary X, all distances are equal to 1, then for any subset Z of X choose a new point y(Z) with distances 1 to Z and 2 to all other points. $\endgroup$ – Fedor Petrov Sep 18 '14 at 19:41
  • $\begingroup$ Are there any clear connections between this question and the Baire Category theorem? $\endgroup$ – The Masked Avenger Sep 19 '14 at 0:39
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    $\begingroup$ Andres, comments below suggest that my initial tag set-theory was quite reasonable) $\endgroup$ – Fedor Petrov Sep 19 '14 at 5:28
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Towards a contradiction, let us assume that we have a metric space $X = \{x_i : i < \omega_1\}$ in which any two points are at least unit distance apart and every subset of $X$ is the union of a countable family of open balls. Let $r_i$ be the supremum of all $r > 0$ such that $B(x_i, r)$ is countable. Construct $\{C_i : i < \omega_1\}$ such that

(1) Each $C_i$ is countable and the infimum of $\{d(x_i, y): y \in C_i\}$ is $r_i$

(2) If $i < j$, then $x_i \notin C_j$

Now let $I \in [\omega_1]^{\omega_1}$ be such that whenever $i < j$, $x_j \notin C_i$. Suppose $Y = \{x_i : i \in I\}$ can be covered by a family $F$ of countably many balls. Let $i \in I$ be least such that $B(x_i, r) \in F$ for some $r > r_i$. Pick $y \in C_i \cap B(x_i, r)$. So $y \in \bigcup F = Y$ so that $y = x_j$ for some $j \in I$ which is impossible.

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  • $\begingroup$ Nice conclusion. $\endgroup$ – Bill Johnson Sep 19 '14 at 9:29
  • $\begingroup$ Great! This seems to get rid of the need for any CH assumption in the final case of my argument. $\endgroup$ – Joel David Hamkins Sep 19 '14 at 9:35
  • $\begingroup$ I don’t understand how this answers the question. If $X$ is a counterexample to the OP, it does contain an uncountable subspace $Y$ whose points are at least a fixed distance apart. Now, any open set $U$ is a countable union of balls in $X$, however their centres may be outside $Y$, hence it doesn’t follow that $U\cap Y$ is a countable union of balls in $Y$. $\endgroup$ – Emil Jeřábek Sep 19 '14 at 11:12
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    $\begingroup$ Oh, I see. So this is not meant as a standalone answer, but as a replacement of the last part of your (Joel’s) proof? @Ashutosh, I think you should make this clear in the answer. $\endgroup$ – Emil Jeřábek Sep 19 '14 at 11:40
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    $\begingroup$ What does the notation $I \in [\omega_1]^{\omega_1}$ mean? That's the one place where this topologist gets lost. $\endgroup$ – Lee Mosher Sep 19 '14 at 15:59
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The answer is yes.

My original argument made use of the continuum hypothesis, or actually just the assumption that $2^\omega<2^{\omega_1}$), but this assumption has now been omitted by the argument of Ashutosh, which handles the case where I had used that assumption. So no extra hypothesis is needed.

Theorem. If every open set in a metric space is a countable union of balls, then the space is separable.

Proof. Suppose that metric space $X$ is not separable. Let us first build an $\omega_1$-sequence of points $\langle x_\alpha\mid\alpha<\omega_1\rangle$, such that no $x_\alpha$ is in the closure of the previous points. This is easy from non-separability. Start with any $x_0$, and at any countable stage $\alpha$, you've got countably many points $x_\beta$ for $\beta<\alpha$, and by assumption this is not dense, so there is some point $x_\alpha$ not in the closure of the previous points.

Now, we can place a ball $B_{r_\alpha}(x_\alpha)$ of rational radius $r_\alpha>0$ around $x_\alpha$ not containing any $x_\beta$ for $\beta<\alpha$. (This ball may contain later points, but never mind about that just yet.) Since we've got uncountably many $\alpha$, we must have chosen the same radius $r_\alpha$ uncountably often. So let's simply throw out all the other points, keeping only the points where we had used that fixed radius $r$, and without loss of generality we reduce to the case where every $x_\alpha$ has distance at least $r$ to all earlier $x_\beta$. It follows that the points $x_\alpha$ are all also at least distance $r$ from the later points as well.

I claim that this violates the assumption that every open set is the union of countably many balls. Consider first the case where uncountably many of the $x_\alpha$ are not isolated. We may consider an open set $U$ consisting of tiny balls around each $x_\alpha$. Specifically, pick $p_\alpha$ within $\frac r4$ of $x_\alpha$, and let $U$ be the union of all $B_{d(x_\alpha,p_\alpha)}(x_\alpha)$. This is an open set and $p_\alpha\notin U$. If a ball $B$ is centered within $\frac r4$ of some $x_\gamma$ and contains some $x_\alpha$ and $x_\beta$, then the radius of $B$ must exceed $\frac r2$, in which case it will contain $p_\gamma$ by an instance of the triangle inequality. So if $U$ is a union of open balls, then each ball contains at most one $x_\alpha$, and so $U$ is not a countable union of open balls.

So we have reduced to the case where we have uncountably many points $x_\alpha$ that are isolated points in $X$, so that $B_r(x_\alpha)$ contains only $x_\alpha$ for a fixed rational $r>0$. In particular, every subset of $\{x_\alpha\mid\alpha<\omega_1\}$ is open in $X$. This case is exactly handled by the construction in Ashutosh's argument, which explains how to build such a set that cannot be a countable union of balls.

(My original argument handled this case with the assumption that $2^\omega<2^{\omega_1}$, a consequence of the continuum hypothesis, by pointing out that there are $2^{\omega_1}$ many open subsets of $\{x_\alpha\mid\alpha<\omega_1\}$, but only $\omega_1^\omega=2^\omega=\frak{c}$ many choices of countably many balls of rational radius. So there are just too many open sets for them all to be realized as the union of countably many balls.)

So we've got a solution in ZFC, without any extra assumption. QED

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    $\begingroup$ It is not very much important, but in order to construct uncountable set with pairwise distances at least $r$ you may just take the maximal (by inclusion) set with this property for small enough $r$: if for any $r=1/n$ such a maximal set is countable, the space is clearly separable. $\endgroup$ – Fedor Petrov Sep 18 '14 at 20:49
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    $\begingroup$ You are right! My argument is more algorithmic, like the proof of Zorn, rather than using Zorn itself. $\endgroup$ – Joel David Hamkins Sep 18 '14 at 20:51
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    $\begingroup$ I did a quick MathSciNet search to see if this result was known. The closest hit I got was a paper by Wacław Sierpiński, "Sur l'inversion du théorème de Bolzano-Weierstrass généralisé," Fund. Math. 34 (1947), 155–156, which shows that if the continuum hypothesis is true, then every nonseparable metric space contains a sequence of subsets with no convergent subsequence. While this result doesn't seem directly relevant to the question at hand, it is sort of interesting that Sierpiński found it necessary to assume CH. So maybe CH is needed here as well, after all. $\endgroup$ – Timothy Chow Sep 18 '14 at 22:07
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    $\begingroup$ Joel, it seems you had answered this same question a few years ago math.stackexchange.com/a/94301. Perhaps you should remove the CH hypothesis over there too. $\endgroup$ – Ramiro de la Vega Sep 19 '14 at 12:30
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    $\begingroup$ Joel, I'm probably being thick, but why must a ball $B_s(x)$ that contains two $x_\alpha$, $x_\beta$ also exclude $p_\alpha$? $\endgroup$ – Todd Trimble Sep 28 '14 at 14:14
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This is really a followup to my comment to Joel David Hamkins's answer but it is too long for a comment. Joel asked if Sierpiński's argument uses the full strength of CH. It looks to me that it does, but here are the two key paragraphs from his paper so that you can judge for yourself. He doesn't say what $\Omega$ is but I'm assuming it's the first uncountable ordinal.

L'espace métrique $M$ étant non séparable, il existe, comme on sait, un nombre positif $d$ et une suite transfinie $\{p_\xi\}_{\xi<\Omega}$ du type $\Omega$ formée de points de $M$ tels que $\varrho(p_\xi,p_\eta)\ge d$ pour $\xi<\eta<\Omega$, $\varrho(p,q)$ désignant la distance des points $p$ et $q$ de $M$.

Or, la famille de toutes les suites infinies de nombres naturels étant de puissance du continu, il résulte de l'hypothèse $2^{\aleph_0}=\aleph_1$ qu'il existe une suite transfinie du type $\Omega$, $\{s^\xi\}_{\xi<\Omega}$, formée de toutes les suites infinies de nombres naturels. Soit (pour $\xi<\Omega$) $s^\xi$ la suite infinie de nombres naturels $n_1^\xi,n_2^\xi,n_3^\xi,\ldots$; $k$ étant un nombre naturel donné, désignons par $E_k$ l'ensemble formé de tous les points $p_\xi$, tels que $k\in \{n_1^\xi,n_2^\xi,\ldots\}$. Je dis que la suite infinie d'ensembles $E_1,E_2,E_3,\ldots$ ne contient aucune sous-suite convergente. Soit, en effet, $E_{k_1}, E_{k_2},\ldots$ où $k_1<k_2<\cdots$ une sous-suite quelconque de la suite $E_1,E_2,\ldots$ D'après la définition de la suite transfinie $\{s^\xi\}_{\xi<\Omega}$ il existe un nombre ordinal $\alpha<\Omega$, tel que $s^\alpha$ est la suite infinie $k_2, k_4, k_6, \ldots$, donc que $n_i^\alpha=k_{2i}$ pour $i=1,2,\ldots$ Vu que $k_1<k_2<k_3<\cdots$, on a donc $k_{2i}\in\{n_1^\alpha,n_2^\alpha,\ldots\}$ et $k_{2i-1} \notin \{n_1^\alpha,n_2^\alpha,\ldots\}$, donc $p_\alpha\in E_{k_{2i}}$ et $p_\alpha \notin E_{k_{2i-1}}$ pour $i=1,2,\ldots$ La sphère ouverte au centre $p_\alpha$ et au rayon $d$ (qui se réduit évidemment a un seul point $p_\alpha$) contient donc un point commun avec chacun des ensembles $E_{k_2},E_{k_4},E_{k_6},\ldots$ et ne contient aucun point des ensembles $E_{k_1},E_{k_3},E_{k_5},\ldots$ Par conséquent la suite infinie d'ensembles $E_{k_1},E_{k_2},E_{k_3},\ldots$ n'est pas convergente. La suite $E_1,E_2,\ldots$ ne contient donc aucune sous-suite convergente, c.q.f.d.

Roughly:

Since the metric space $M$ is non-separable, there exist $d > 0$ and a sequence $\{p_\xi\}_{\xi<\omega_{1}}$ of points in $M$ such that $\varrho(p_\xi,p_\eta)\ge d$ for $\xi<\eta<\omega_{1}$, where $\varrho(x,y)$ is the metric on $M$.

Using CH, list all the sequences $s^\xi, \xi < \omega_{1}$ of natural numbers, $s^\xi = \langle n_1^\xi,n_2^\xi,n_3^\xi,\ldots\rangle$; for a given $k \in \mathbb{N}$, let $E_k$ be the set of all $p_\xi$ such that $k\in \{n_1^\xi,n_2^\xi,\ldots\}$. I claim that the sequence $E_1,E_2,E_3,\ldots$ does not contain any convergent subsequence. For, if $E_{k_1}, E_{k_2},\ldots$ where $k_1<k_2<\cdots$ is an arbitrary subsequence of $E_1,E_2,\ldots$, then there exists $\alpha<\omega_{1}$, such that $s^\alpha$ is the sequence $k_2, k_4, k_6, \ldots$, and so $n_i^\alpha=k_{2i}$ for $i=1,2,\ldots$ Since $k_1<k_2<k_3<\cdots$, it follows $k_{2i}\in\{n_1^\alpha,n_2^\alpha,\ldots\}$ and $k_{2i-1} \notin \{n_1^\alpha,n_2^\alpha,\ldots\}$, so $p_\alpha\in E_{k_{2i}}$ and $p_\alpha \notin E_{k_{2i-1}}$ for $i=1,2,\ldots$ The open ball with centre $p_\alpha$ and radius $d$ (which is actually just the singleton $p_\alpha$) intersects each of $E_{k_2},E_{k_4},E_{k_6},\ldots$ non-trivially but is disjoint from $E_{k_1},E_{k_3},E_{k_5},\ldots$ Consequently, $E_{k_1},E_{k_2},E_{k_3},\ldots$ is not convergent. So the sequence $E_1,E_2,\ldots$ contains no convergent subsequence, q.e.d.

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    $\begingroup$ Thanks very much! But is there any chance to translate, for those of us whose French is weak? $\endgroup$ – Joel David Hamkins Sep 19 '14 at 1:59
  • $\begingroup$ Added a rough translation of Sierpinski's argument in Timothy's answer. $\endgroup$ – Avshalom Sep 19 '14 at 15:21

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