Shape of long sequences in C(ω_1)

Question

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.

Answer 1

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

Suppose to the contrary that x_α is bad. For any countable ordinal β, there is r_β in C(ω₁) such that the distance between r_β and x_α 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 r_β(α) are within ε of r_β(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 γ_ε. Since there are only countably many ε, we may find a countable ordinal γ above all γ_ε. This ordinal has the property that for all ordinals β above γ, we have r_β(α) = r_β(β) for all α in the interval (γ,β], since the values are within every ε of each other. That is, every r_β function is constant from the same fixed γ up to β.

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

Answer 2

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 $\bigcup_{i\le n}V_i$.

Answer 3

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.