"Continuing in this manner" looks rather hard when each step involves a choice (in this case a particular open set). Am I missing something?

Actually, choosing open sets might be pretty easy: $U_i = \{x_k : i < k\}$. Then you can choose the connected component of the first element of $X \setminus \overline U$.

@dfeuer Yes, I think so... maybe my proof requires some different phrasing but I doubt that true AC is needed. Of course anyone is free to let me know if I'm wrong!

No, I think I mixed that up.... we need an open set that contains $x_i$. Let me see if I can think of anything.

By Hausdorff property of $X$ we have $\{U\in \tau:x^*\in U \text{ and }C\setminus \overline U\neq\varnothing\}\neq\varnothing$... so let $U$ be an element of this set.

You say "so let $U$ be an element of this set", but you don't say which element.

@dfeuer I thought choice is only needed to choose one element from each set in an infinite collection of sets. I still think the proof will hold without any choice. But I have to admit I don't think much about using choice in my daily work. Perhaps a set theory expert here will chime in.

@dfeuer Also back to one of your deleted comments, proving a compact Hausdorff space is regular (or even normal) does not require choice. And using normality, one can prove each $C_n$ must be non-degenerate.

The standard proof uses choice. How do you avoid it?

Let $A$ be a closed subset of $X$, and let $x\in X\setminus A$.

are you using chatjax?

Let $\mathcal U=\{U\in \tau:U\cap A\neq\varnothing \text{ and } x\notin \overline U\}$

can you read this stuff okay?

Since $X$ is Hausdorff, $\mathcal U$ is an open cover of $A$.

No, I don't have chatjax. I can read this a bit awkwardly.

Do you see how to take the proof from here?

The usual proof requires choice because you say "for each y in A let U_y be such that..."

One sec.

OK, ChatJax is running now.

Choice can be avoided by explicitly constructing a set like that

One sec....

So $\mathcal U$ is an open cover with a finite subcover, and so its complement ends up open.

OK.

yeah then it goes the same way

What goes the same way?

Sorry ,not its complement... but whatever.

Jeez, I'm rusty.

yeah you see

you mean the intersection of the complements of closures of sets in the finite subcover :)

that will be the open set containing $x$

and the union of the finite subcover is the one containing $A$

Yes, I think I see that.

But now how'd that relate to the original problem?

And you can get normal similarly.

Well, because I need normality to prove each $C_n$ is non-degenerate.

I still don't see how you're getting the $U_n$s.

At all.

I'm only choosing one set at a time there.

at stage $n$ of the recursion, I pick one open set $U_n$

Yes, but that doesn't let you get an infinite number, only as big a finite number as you want.

I can define an infinite collection of sets without choice

now, if I wanted to choose one element from each element of the sequence $(U_n)$, it would require choice

Well, sometimes, yes. But you still have to say how you're defining that collection.

Let $z\in X\setminus \{x_0\}$.

Let $U_0$ be an open set such that $x_0\in U_0$ and $z\notin \overline{U_0}$

By the Hausdorff property such a set exists.

I do not have to say how to define it beyond that.

Sure. $U_0$ is no problem.

Okay, then $U_n$ is similar.

Indeed, for any $n$ you can magic up a finite sequence $U_0, U_1, \hellip, U_n$.

Only now $x_0$ is replaced with $x^*$, and $z$ needs to be in $C_{n-1}\setminus \{x^*\}$

But you can't just assume that there's a suitable infinite sequence.

Right, but I am defining the sequence

recursively

It looks to me like you need Dependent Choice to do it that way, and that's actually stronger than Countable Choice. See proofwiki.org/wiki/Axiom:Axiom_of_Dependent_Choice

I'm not saying "let $(U_n)$ be a sequence of open sets such that $x_n\in U_n$". That would require choice.

You're not defining it recursively, because each recursive step requires a choice.

Dependent Choice is not needed to say there is a strictly increasing sequence of numbers.

Because we can construct the sequence. 1,2,3,...

Of course.

We have a specific way to form each number from the last.

But you're saying "At each step, there must exist an open set satisfying this property. Pick one of them."

I think I see what you are saying though, we need a more specific formula for the $U_n$'s, possibly.

Ah, but at the end I am not saying $(U_n)$ exits as a set.

I am, however, saying $(C_n)$ exists as a set...

This is like fruit of a poisoned tree or something ;)

And the $C_n$s. And I'm going to conjecture that it's impossible. Indeed, I would bet an ice cream cone that this theorem is at least as strong as, say the axiom of countable choice for finite sets. Find a way for an open cover of some space or other to represent a set of partial choice functions, or something like that.

But then again my intuition is extremely fallible.

So you're saying countable choice for finite sets is used in my proof?

mathoverflow.net/questions/38450/… looks pretty related (Asaph Karagila's answer specifically), but the uncountable vs. non-countable is tricky.

I'm saying your proof uses dependent choice, which is stronger than countable choice, which is stronger than countable choice for finite sets.

what is Dedekind-finite?

I think it means it's not in bijection with any proper subset of itself, but my memory is hazy.

I think maybe if I were more careful defining sets, choice could be avoided.

It's so strange though cause I use proof techniques like this all the time without thinking about choice.

How would you even go about showing that the Theorem implies CC?

Oh, I don't have anything concrete. Just an extremely vague shape. Something about closed sets in some topology being partial choice functions.

This would be a very strange equivalence I think.

To be honest, I don't even remember where I saw a proof that countable choice was sufficient to prove the theorem. This is all murky mists of time, sadly.

or dependent choice or whatever

the proof I gave should be the simplest. it's just a matter of whether it can be re-structured

it's still very subtle to me that choice was used

Choice is usually subtle.

Do you have any classical arguments that use DC?

Off the top of my head? No. But any time you say "Assume you have $x_i$. By some argument, there must exist a $y$ satisfying predicate $p_i(x_i, y)$. Let $x_{i+1} = y$, and thereby construct the infinite sequence $x$," you're implicitly using dependent choice. DC is apparently a very popular choice principle among analysts.

I still think the statement of DC is more than what I used. I used CC at most.

Each choice depended on the previous one.

You have a cycle, which you can think of as a sequence of $U_i \times C_i$, and you need the previous one to narrow down the requirements for the next.

I could always define the $C_n$ to be the component containing the least indexed member of $X\setminus \overline{U_n}$

OK, but you're still stuck with $U_i$.

so then it's not ambigious "existence"

Forget $C$.

Ok, so how can we define that differently

You have $U_i$. You can't calculate $U_{i+1}$, because a key step is "The space is Hausdorff, so there must exist an open set like this; pick one."

But we can say which one to pick

maybe

Well, if you can see which one to pick, that's another story.

Oh wait

Consider the set of all such $U$'s

(we are in a single state of the recursion)

Don't even define $U_n$.

That's a direction to attempt.

Then take the closure of the union of all continua which contain the least indexed point missing one of the U's

There's your $C_n$

All continua?

yes!

Oh no

which miss U

I can express this in set notation

I'm pretty sure

You'll have to write that up for me to be able to consider it.....

Right, I'll work on it now

give me just a minute

Here we go...

Let $x^*$ be the least indexed element of $C_{n-1}$.

Let $x^{**}$ be the element of $C_{n-1}$ with least index greater than $x^*$.

Let $\mathcal U=\{U\in\tau:x^*\in U\text{ and }x^{**}\in C_{n-1}\setminus \overline U\}$

So far so good.

Let $\mathcal C=\{C\subseteq C_{n-1}:C\text{ is connected }, x^{**}\in C, \text{ and there exists }U\in \mathcal U\text{ with }C\cap U=\varnothing\}$.

Then let $C_n=\overline {\bigcup \mathcal C}$

Done!

shall I update my answer?

Sounds plausible.

But does $C_n$ satisfy all the requirements?

Thanks for helping, I would never have recognized a problem.

Yes because $C_n$ is a continuum. And many of the elements of $\mathcal C$ are non-degenerate, so $C_n$ is non-degenerate.

This is by the so-called "boundary bumping theorem" which uses normality

I think you've probably got it. Of course, this is about "non-countable" rather than "uncountable", but it sounds like the latter is simply impossible, so assuming I don't find any further gaps or errors, I'll accept the repaired and fleshed-out answer.

what is the difference again?

I showed $X$ cannot be put into one-to-one correspondence with the natural numbers

or a subset

Yep.

But I don't see the edited proof yet.

It's coming

No rush. Just making sure you didn't think you'd submitted when you hadn't.

Think I fixed it

Lemme look.

But I should make the uncountable/non-countable distinction.

what does uncountable mean ?

1 sec.

Can't you just define $C_n$ as the connected component of $x^{**}$ in $C_{n-1} \setminus U$?

There are several different notions of "countable" and "uncountable" in set theory without choice. I don't know the canonical names of all of them.

No because U is not fixed

What does $U$ range over? I don't see it explained there.....

$U\in \tau$ where $\tau$ is the topology

let me edit

No, that line waws clear before.

I mean the next line, where you define $\mathcal C_n$.

ah let me fix that

ok

I'm going to have to dig into details to see what compactness and normality have to do with quasicomponents, and how those relate to boundaries.

I can explain quickly

The quasi-component of a point is the intersection of all clopen sets containing the point.

Yes.

Now, in a compact Hausdorff space each quasi-component is connected.

Otherwise it would be the union of two disjoint closed sets $A$ and $B$

By normality, we can put these into disjoint open sets $U$ and $V$

Notice that clopen sets are closed under finite intersections

And so by compactness, there must exist a clopen set contained in $U\cup V$

But this clopen set intersected with $U$ shows the quasi-component is smaller than it really is

so contradiction

Now, if the quasi-component misses the boundary, then by compactness a clopen set misses the boundary, and we get a proper clopen subset of $X$.

So this contradicts connectedness of $X$.

All these need is normality and "finite intersection property" compactness.

What do clopen sets being closed under finite intersection and compactness have to do with a clopen set contained in $U \cup V$?

because if every clopen set meets $X\setminus (U\cup V)$, then the intersection of all does.

the intersection of a collection of compact sets with FIP is nonempty

that's an alternative definition

of compactness

Closed sets, I think, not compact.

yes, but closed subsets of compact spaces are compact

And choice is not needed for this!

Yes, but without choice I don't think the compact subspaces of a Hausdorff space have to be closed. Anyway, give me a minute.

I mean, technically this all makes for a 3 page paper and not a MO answer

in order to explain all the details along the way

and convince the reader that choice is not used.

Oh, I see. You're looking at finite subsets of the set $\{ X \setminus (U \cup V) \} \cup \{ Q : \text{$Q$ is a clopen neighborhood of $x$} \}$, where $x$ is the point whose quasicomponent we're considering.

Yes

That has the finite intersection property if each clopen set meets the complement of $U\cup V$

So what is the difference between uncountable and non-countable?

And could I possibly strengthen my argument?

I'll have to read in more detail later and expand it out to my own level of understanding. You asked about different notions of countability. Here are a few: 1. A set $S$ is countable if there is an injection $S -> N$. 2. A set is countable if there is a surjection $N -> S$.

3. A set is uncountable if there is an injection from $N -> S$ but no injection from $S -> N$. Etc., etc. I don't know exactly what definitions are common, but in constructive mathematics of different flavors, some of these can be different from each other.

Bleh. Forgot my arrows.

I think it might be worth adding the choice-free argument that a compact Hausdorff space is normal, since there is a common version that uses choice.

Yes, I may do that.

So to prove uncountable, I must show more?

I must show there is an injection from $N$ to $X$?

Gimme a sec...

This is all misty.

Wikipedia says this: en.wikipedia.org/wiki/Cardinality#Comparing_sets

So right, you'd need an injection $N \rightarrow X$. But based on Asaf Karagila's answer, that seems impossible.

But we know $X$ is infinite.

So that seems automatic.

Heh. You might think so.

are you saying choice is required?

I guess it is

I would guess DC; not sure.

yeah

But my definition of countable will be: There is an injection $X\to N$.

Then I proved the theorem :)

Sounds good to me.

Thanks for the help!

You're welcome; thanks for your feedback also.