Apologies for the vague title - I couldn't come up with a single sentence that summarised this problem well. If you can, please edit or suggest a better one!

This question is also rather specific and contains lots of annoying technical detail. I must admit to not really expecting an answer unless there's an obvious solution I'm missing (which is very possible - I feel like any solution is either going to be obvious or very deep), but some pointers in plausible sounding directions would be greatly appreciated. I suspect the answer will depend on the combinatorics of $\omega_1$, which I know relatively little about.

Let $V$ be a normed space. For $A \subseteq V$, define $r(A) = \inf \{ r : \exists V, A \subseteq B(v, r) \}$. Define a bad sequence in $V$ to be a sequence $\{ v_\alpha : \alpha < \omega_1 \}$ with the properties that:

$\forall \beta, r(\{ v_\alpha : \alpha < \beta \}) \leq 1$

$\inf_\beta r(\{ v_\alpha : \alpha \geq \beta \}) > 1$

An example of a space with a bad sequence is $c_0(\omega_1)$ (the set of all bounded real-valued sequences of length $\omega_1$ such that $\{ \alpha : |x_\alpha| > 0 \}$ is countable). The sequence $2 * 1_{\{\alpha\}}$ is bad. The radius of any tail is $2$ because the center must be eventually 0. The radius of the initial segments is $\leq 1$ because the segment up to $\alpha$ is contained in the closed ball of radius 1 around $1_{[0, \alpha]}$, which is in $c_0(\omega_1)$ because $\alpha < \omega_1$.

I have two (three depending on how you count it) major examples of spaces which have no bad sequences:

  • Any separable space: you can choose centers to lie in the countable dense set, so one center must work as a radius for the initial segment for unboundedly many and thus for all $\alpha$.
  • Any space which has what I'm imaginatively calling the chain-radius condition: The union of a chain of sets of radius $\leq r$ has radius $\leq r$. This includes:
    • Any reflexive space: If $U_\alpha$ forms a chain, the sets $F_\alpha = \bigcap_{v \in U_\alpha} \overline{B}(v, r + \epsilon)$ form non-empty closed and bounded convex sets with the finite intersection property, so compactness in the weak topology implies they have non-empty intersection. Any element of the intersection contains the union of the chain in $\overline{B}(c, r + \epsilon)$
    • any space with the property that $\textrm{diam}(A) = 2 r(A)$ (in particular the $l^\infty$ space on any set) because it's clear that unions of chains of diameter $\leq 2r$ have diameter $\leq 2r$.

So... that's all the backstory for this question. Given that, my actual question is very simple: Does $C(\omega_1)$ contain a bad sequence?

I feel like the answer "must" be no. In particular note that the projection of any sequence onto the first $\alpha$ entries is not bad (because it's a sequence in a separable space) and that if you drop the restriction for continuity the answer is immediately yes. So it sits right between two classes of examples where there are no bad sequences, and I feel that one really should be able to take advantage of that. But on the other hand, functions in $C(\omega_1)$ are eventually constant, so maybe you can take advantage of that to construct some sets with arbitrary bad tails.

For bonus kudos, I'd love to know for what compact Hausdorff spaces $K$, $C(K)$ contains a bad sequence.

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    $\begingroup$ Can you explain why your example sequence is bad? Isn't it contained in the closed unit ball centered at 0? $\endgroup$ Apr 11, 2010 at 19:43
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    $\begingroup$ Apologies, I actually meant twice that sequence. The sequence as previously mentioned actually has each of the initial segments having radius $\leq \frac{1}{2}$ and the radius of the tails is 1. I've fixed the post now. $\endgroup$ Apr 11, 2010 at 19:50
  • $\begingroup$ In response to your comment below about citing MO, you may find the discussion at tea.mathoverflow.net/discussion/64 useful. $\endgroup$ Apr 13, 2010 at 18:17

3 Answers 3


I claim that there are no bad sequences in C(ω1).

Suppose to the contrary that xα is bad. For any countable ordinal β, there is rβ in C(ω1) such that the distance between rβ and xα for α < β is at most 1. For any countable ordinal β and any positive rational number ε, there is a smaller ordinal γ < β such that all rβ(α) are within ε of rβ(β) for α in (γ,β]. For fixed ε, this is a regressive function on the countable ordinals. Thus, by Fodor's Lemma, there is a stationary and hence unbounded set of ordinals on which the function has constant value, which we may call γε. Since there are only countably many ε, we may find a countable ordinal γ above all γε. This ordinal has the property that for all ordinals β above γ, we have rβ(α) = rβ(β) for all α in the interval (γ,β], since the values are within every ε of each other. That is, every rβ function is constant from the same fixed γ up to β.

Let Cβ be the closed interval of values s such that the constant sequence s of length β lies within 1 of all xη(α) for all η ≤ β and all γ < α ≤ β. These are nested and not empty, since rβ(β) is in Cβ. By compactness, there is a value s in all Cβ. Thus, the number s is within xη(α) for all η and all α above γ.

Thus, we may form the desired sequence r by finding a center that works for the sequences up to stage γ, using the separability idea in your question, augmented with the constant value s at the stages above γ up to ω1. That is, we solve the problem separately on the first γ many coordinates, and then append the constant s sequence up to ω1. This sequence is continuous, and it lies within 1 of every xη, as desired.

  • $\begingroup$ Thanks. I'm pretty sure I should have spotted something like that (it looks very similar to the proof which got me to this problem in the first place), but I'm not sure I would have! What's the etiquette/convention for citing a mathoverflow answer in a paper? $\endgroup$ Apr 12, 2010 at 7:44
  • $\begingroup$ Oh, by the way, it doesn't actually damage the proof, but your interval is slightly wrong. I think you actually mean $(\gamma, \beta]$. When $\beta$ is a successor ordinal you can't guarantee that there are any smaller $\gamma$ with value close to it. $\endgroup$ Apr 12, 2010 at 7:49
  • $\begingroup$ Another wrong but fixable detail: I don't believe it works to use $r_\gamma$ up to and including stage $\gamma$. The problem is that later elements of the sequence may add variation before $\gamma$ even while being well behaved after it. However, $C([0, \gamma])$ is separable, so has no bad sequences, so we can still find a center that works for all of $[0, \gamma]$ and then use the constant center after that as per your argument. $\endgroup$ Apr 12, 2010 at 8:14
  • $\begingroup$ Actually... I'm starting to have serious doubts about the core idea of this proof. It doesn't seem to use anything about the $r_\alpha$ other than that they're continuous, and it's certainly not the case that every $\omega_1$ sequence of continuous functions is mutually constant after some point. Why, for example, would this not work with the following: Let $x_\alpha = 1_{[0, \alpha]}$ and let $r_\alpha = x_\alpha$. Obviously this isn't a bad sequence, but there's no common point above which the $r_\alpha$ are all constant. $\endgroup$ Apr 12, 2010 at 8:21
  • $\begingroup$ David, my proof wasn't claiming that the r_beta are simultaneously constant from gamma out to omega_1, but only that r_beta is constant on the interval [gamma,beta]. In the example of your comment, gamma=0 works for this, since every r_alpha is constant from [0,alpha] in your example. And the sequence I build would have s=1. So, the sequence provided by my solution has value 1 from gamma=0 onwards, which works for this example. $\endgroup$ Apr 12, 2010 at 12:30

I can partially answer the second question. If $X$ is a compact Hausdorff space whose topology has a countable base at every point [Edit: $\omega_1$ has this property but not compact], then there are no bad sequences. Moreover the following holds:

If $F\subset C(X)$ is a family such that every countable subfamily has radius $\le 1$, then $r(F)\le 1$.

Define a function $S=S_F:X\to\widehat{\mathbb R}=[-\infty,+\infty]$ (the "essential supremum" of $F$) as follows: $S(x)$ is the maximum $t$ such that for every neighborhood $U$ of $x$ one has $\sup\{f(y):f\in F,y\in U\}\ge t$. Observe that $S$ is upper semi-continuous: for every $t\in\widehat{\mathbb R}$, the set $\{x\in X:S(x)\ge t\}$ is closed.

Define the essential infimum $I_F$ similarly, this function is lower semi-continuous.

For every $x\in X$ there is a countable family $G\subset F$ such that $S_G(x)=S_F(x)$ and $I_G(x)=I_F(x)$. Indeed, using countable base at $x$, one can realize $S(x)$ by a sequence $x_i:i\in\mathbb N$ converging to $x$ and functions $f_i\in F$ such that $f_i(x_i)\to S(x)$.

It follows that $S(x)\le I(x)+2$ for all $x\in X$. Indeed, take $G$ as above, it is contained in a $(1+\epsilon)$-ball centered at some $f\in C(X)$, then $S_G(x)\le f(x)+1+\epsilon$ and $I_G(x)\ge f(x)-1-\epsilon$.

Fix $\epsilon>0$ and let us prove that $F$ is contained in a $(1+\epsilon)$-ball. For $x\in X$, define $C_x=\frac12(S(x)+I(x))$. Note that $S(x)\le C_x+1$ and $I(x)\ge C_x-1$. By semi-continuity, there is a neighborhood $U_x$ of $x$ such that $S(y)<C_x+1+\epsilon$ and $I(y)>C_x-1-\epsilon$ for all $y\in U_x$. Choose a finite subcovering $V_i=U_{x_i}:i\le N$.

On each neighborhood $V_i$ we have a constant function $f_i:=C_{x_i}$ which works as a center within this neighborhood. It suffices to construct a function $g\in C(X)$ such that for every $x\in X$, $g(x)$ is between the minimum and maximum of these partially defined constant functions at $x$. This is easy to do by induction in the number of sets in the covering. Suppose we have already defined $g=g_{n-1}$ that works on $\bigcup_{i<n} V_i$. By Urysohn's lemma there is $\phi:X\to[0,1]$ such that $\phi=0$ on $X\setminus V_n$ and $\phi=1$ on $X\setminus\bigcup_{i\ne n} V_i$. Then $g_n:=\phi f_n+(1-\phi)g_{n-1}$ works on $\bigcup_{i\le n}V_i$.

  • $\begingroup$ Believe it or not, my answer was not inspired by this, even though they are quite similar in essence. It occurred to me while I was on the train and I wrote it up before I saw this one. $\endgroup$ Apr 12, 2010 at 9:38
  • $\begingroup$ Sorry, that sounded a bit ungrateful. Thanks for the answer! $\endgroup$ Apr 12, 2010 at 9:44
  • $\begingroup$ For some reason, I cannot edit my answer. To clarify, it does not work for $\omega_1$: it has countable base at every point but it is not compact. $\endgroup$ Apr 12, 2010 at 12:05
  • $\begingroup$ Hm. I was about to say that that wasn't a problem because $C(\omega_1 + 1) = C(\omega_1)$, but of course that's compact but doesn't have a countable base at every point. Vexing. $\endgroup$ Apr 12, 2010 at 12:15
  • $\begingroup$ I tricked this stupid software to let me edit the post. By the way, you don't really need compactness, only normality and that every countable covering has a finite subcovering. But I don't know whether $\omega$ is normal or not. $\endgroup$ Apr 12, 2010 at 12:25

Edit: This proof is wrong, but preserving for posterity.

Oh! In a very wizard of oz manner, the answer was within my power all along (it's within $\epsilon$ of a proof I'd already written for something else).

Here's a cute little proof that C(K) has the property that diam(A) = 2 * r(A), and thus has the chain-radius condition and thus has no bad sequences.

Let $A \subseteq C(K)$ be non-empty. Define

$ g(x) = \sup_{f \in A} f(x)$

$ h(x) = \inf_{f \in A} f(x)$

$g$ is upper semicontinuous: $g(x) > a$ iff there exists $f \in A$ such that $f(x) > a$. Similarly $h$ is lower semicontinuous.

Further, $g(x) - h(x) \leq \textrm{diam}(A)$.

Therefore $g(x) - \frac{1}{2}\textrm{diam}(A) \leq h(x) + \frac{1}{2}\textrm{diam}(A)$

But now we have an upper semicontinuous function which is $\leq$ a lower semicontinuous function. Thus by the katetov tong insertion theorem there is a continuous function $f$ with

$g(x) - \frac{1}{2}\textrm{diam}(A) \leq f \leq h + \frac{1}{2}\textrm{diam}(A)$

But this means that $A \subseteq B(f, \frac{1}{2}\textrm{diam}(A))$. Therefore $r(A) \leq \frac{1}{2}\textrm{diam}(A)$.

But we already know that $r(A) \geq \frac{1}{2}\textrm{diam}(A)$, so the two are equal and the result is proved.

  • $\begingroup$ The supremum is lower semi-continuous, not upper. On $K=[-1,1]$, the supremum can be like this: $g(x)=0$ if $x\le 0$ and $g(x)=1$ if $x>0$. And the infimum can be like this: $h(x)=0$ if $x\ge 0$, and $h(x)=-1$ if $x<0$. While $0\le g-h\le 1$, there is no continuous funtion within the distance $1/2+\epsilon$ from them. $\endgroup$ Apr 12, 2010 at 12:01
  • $\begingroup$ Hm. You're quite right. Got the definitions backwards. Sigh. $\endgroup$ Apr 12, 2010 at 12:17
  • $\begingroup$ By the way, what are the assumptions of the insertion theorem? If it does apply to $\omega_1$, you can use "essential supremum" from my answer. $\endgroup$ Apr 12, 2010 at 12:28
  • $\begingroup$ It does apply to $\omega_1$ yes. It works for any normal space. $\endgroup$ Apr 12, 2010 at 12:30
  • $\begingroup$ Ok. I see why I got so confused. In the original case I applied this proof (which is to do with approximating discontinuous functions) the definitions of g and h were actually essentially those of your essential supremum and infimum! In my case it was for a single function which wasn't necessarily continuous, but the details work out much the same. So, using those definitions and the insertion theorem we can drop the conditions of compactness and locally countable bases. $\endgroup$ Apr 12, 2010 at 12:40

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