07:18
No, I am thinking about trying to find an uncountable well ordered subset of $\Bbb{I}$. I thought I can evade the rationals countability problem by having countably many of the irrationals to shift like a hilbert hotel fashion to fill in all the holes where the rationals were, thus for any subset of the irrationals wrt the irrationals, there is only irrationals between every irrational, thus evading the contradiction due to countable rationals,
but the above two attempts suggest that there is no order preserving bijection that I can use to fill in the holes, thus it might be impossible
but the above two attempts suggest that there is no order preserving bijection that I can use to fill in the holes, thus it might be impossible
18 hours ago, by user21820
@Secret As for embedding ω[1] in the reals, here's the easiest proof that just popped into my head.
18 hours ago, by user21820
There are ω[1] many pairs of consecutive ordinals in ω[1]; just take all the limit ordinals and their successors. There are ω[1] many of them otherwise their union will be a countable ordinal and hence less than some further limit ordinal.
18 hours ago, by user21820
If all these pairs embed into the reals, then the corresponding reals have each a rational between them, and all the sandwiched rationals are distinct.
18 hours ago, by Secret
and hence, you must have uncountably many rationals, which is a contradiction as the reals has only countably many
07:41
10 mins ago, by Secret
Actually, that seems weaker than being computable. What do you call some mathematical object that can be encoded in a finite string?
@Secret Definable, but as I keep saying it's dangerous to talk about definability until you're familiar with logic. If you want a rigorous formal definition of definable reals that isn't inherently inconsistent, see here:
10
I see, I like to make this point clear because that's usually what I meant by explicit when outside of predicative settings
in Mathematics, Nov 9 '17 at 16:32, by Akiva Weinberger
And also write $\Bbb Q\setminus\{0\}$ as:$$(-\infty,-1)\cup[-1,-\frac12)\cup [-\frac12,-\frac14)\cup[-\frac14,-\frac18)\cup\dots\\ \cup[\frac18,\frac14)\cup[\frac14,\frac12) \cup[\frac12,1)\cup [1,\infty)$$
in Mathematics, Nov 9 '17 at 14:32, by Secret
$$\left(\frac{b+a}{m+a}+\frac{\text{round}\left(10^{a+2}\left(2s-\frac{\text{round}\left(10^{a+1}s\right)}{10^{a+1}}\right)\right)}{10^{a+2}},\frac{b+a}{m+a} \frac{\text{round}\left(10^{a+2}\right)}{10^{a+2}}\right)$$
in Mathematics, Nov 9 '17 at 14:33, by Secret
$$\left(-\frac{b+a}{m+a}+\frac{\operatorname{round}\left(10^{a+2}s\right)}{10^{a+2}},-\frac{b+a}{m+a}\cdot\frac{\operatorname{round}\left(10^{a+2}\right)}{10^{a+2}}\right)$$
@Secret But if you look at the formal definition, you will realize that you will have severe trouble when talking about whether something is 'definable' unless it already has a computable encoding into the system. And you then need to do something like what I did, namely restricting to models that have the objects you want to talk about in the first place.
This might be a very stupid, and possibly philosophical question, but attempt to apply mathematics to everything plus inspired by this question caused me to ask this question
Is there any mathematical object that has been proved to exist but cannot be described in words?
If the answer is...
If ZFC has a model with the same reals, then many of them are not definable over ZFC. But it's possible that ZFC is consistent but proves that there is no model of ZFC with the same reals... That would be an incredible joke (we would have to abandon ZFC as a meaningful foundation), but we absolutely can't exclude that possibility.
4 hours later…
12:56
One can made as many such cuts as they can, and end up with pieces labelled with the natural numbers e.g. $S_1,S_2, S_3, ...$
Now, $S$ has the following bizarre property: Let $|S|$ be the size of $S$, defined by the natural number associated to $S$ (and hence a function $A \to \Bbb{N}$, where $A$ is some existing object)
Therefore, $S$ and all its fragments can take at most n-1 number of cuts without changing size, and then suddenly cannot be cut anymore
2. Quasifinite I: $\forall S \in \text{obj}(\exists n, m \in\Bbb{N}(((m<n)\implies (|S_m|=|S|))\land (|S_n|=0)))$
Quasifinite I: Exhaustible by taking things away one by one or any, but size remain unchanged until the number of things taken away matches its size
Strictly amorphous: When being partition the number of times equal to its grade, all those partitions are of size 1
Countable: Equinumerous as the natural numbers. Can be enumerated by counting. Will in theory be exhausted if the counting goes on indefinitely
Quasicountable: Can only be exhausted by counting exactly the same number of times as counting the naturals from the smallest to largest, otherwise there is no other way of counting (even if countable) can exhaust it
Infinite IIb: Can only be exhausted if taking one fragment of size 1 away, followed by one element of size 2 and so on
Tower infinite: There is an extremely fast growing function such that f(|S|) of it is equal to itself, otherwise for any g that grows slower than f, g(|S|) < |S|
Inaccessible: No function, however fast growing, can reach said size if not starting from it (This is much stronger than the notion of inaccessible cardinals in set theory which only requires 2^K < |S| for K < |S|)
Proper classes/containers: These are so large that no size can be meaningfully assigned to them, but all sizes can never reach it
Transinfinite: This trolling sounding notion refers to a size so indescribably large that had this were set theory, the set difference S-S is not empty, but a proper subset with the same size
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