Forever Mozart

so that's been my day so far

math is hard

misery

no proof

and came up empty

down the rabbit hole

today I went deep

You're welcome; thanks for your feedback also.

Then I proved the theorem :)

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

yeah

I guess it is

are you saying choice is required?

So that seems automatic.

But we know $X$ is infinite.

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

So to prove uncountable, I must show more?

Yes, I may do that.

And could I possibly strengthen my argument?

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

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

Yes

and convince the reader that choice is not used.

in order to explain all the details along the way

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

And choice is not needed for this!

yes, but closed subsets of compact spaces are compact

of compactness

that's an alternative definition

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

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

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

So this contradicts connectedness of $X$.

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 contradiction

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

And containing this quasi-component

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

Notice that clopen sets are closed under finite intersections

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

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

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

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

I can explain quickly

ok

ah let me fix that

let me edit

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

No because U is not fixed

what does uncountable mean ?