In Abbott's Understanding Analysis 2e, part of exercise 1.5.10 asks

Let $C \subset [0,1]$ be uncountable, let $A := \{ a \in (0,1): C \cap [a,1] \text{ is uncountable} \}$, and let $\alpha := \sup A$. Is $C \cap [\alpha,1]$ uncountable?

Presumably the question is whether $C \cap [\alpha,1]$ is necessarily, always uncountable and the answer to this is a clear NO (to me at least; just take $C$ a sub-interval). What interests me is, can it be unclear that it is clear? In other words, I ask

Does $C \subset [0,1]$ even exist such that $C \cap [\alpha,1]$ as above is uncountable?

I fiddled with Cantor-like sets and everything I knew and have thoroughly confused myself.

  • On the one hand, I constructed no examples. No, I don't know what 'constructively' or any foundations mean, but here I don't mean 'constructed' literally. I would be content with a semi-explicit $C$ of sorts, and I could not even describe that.
  • On the other hand, I vaguely imagine there being such examples -- $C$ couldn't be (the image of) a sequence converging to say $0$ from the right, but perhaps of a thicker cloud that still contains no interval $(0,b)$ -- but a little of my intuition is a dangerous thing.
  • On the third hand, searching for the phrase 'single condensation point' produces a few articles mentioning models of ZF and the rest about the other, literal kind of condensation, which makes me fear that my point above, about what counts as a construction, might matter.

And finally, this same exercise previously asked to show $A \neq \emptyset$ (so that the sup is useful after all), but this seems to rule out my idea of a cloud as in the second bullet saturating the right side of $0$: I wanted to make $C$ so its 'uncountability' happens only at $0$ and so cutting off any $(0,a)$ leaves only countable dust, but the very meaning of $A \neq \emptyset$ is that one can move to the right and start at some $a>0$ such that $C \cap [a,1]$ remains uncountable. So this would suggest no such $C$ exists.

TL;DR -- My question is the second box above. I initially looked for an explicit example and in view of references to foundational issues have prepared for disappointment, but an answer might be obvious to someone else and just eluding me.


Suppose there exists such a set $C$. We will show that $C\cap(\alpha,1]$ is countable, which contradicts the assumption $C\cap[\alpha,1]$ is uncountable, since the latter contains at most one more point.

First we note that for any $a>\alpha$, the intersection $C\cap[a,1]$ is (at most) countable. We can write $$C\cap(\alpha,1]=C\cap\left(\bigcup_{n=1}^\infty\left[\alpha+\frac1n,1\right]\right)=\bigcup_{n=1}^\infty\left(C\cap\left[\alpha+\frac1n,1\right]\right).$$ Since a countable union of countable sets is countable, we see that $C\cap(\alpha,1]$ is countable.


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