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. 



Shape of long sequences in C(ω_1)
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.


