I think the thread can be suitably be labeled as "advanced".
Yes, I have also learned a number of things from the thread since my knowledge of the topic is limited to reading first six chapters derek goldrei's book on sets (without doing any of the exercises) about seven years ago. Since its been a long time and I didn't make any notes while studying I can't re-call a good number of things that I read. But some of the more basic ideas, I can re-call.
That and skimming a large number of questions/answers on MSE/MO (for the most part somewhat randomly last year in '19) to try to understand whatever small bits and pieces that I could get (to make some sense of).
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I think you would better wait from a more informed reply to be sure. Here is my (informal) thought upon looking at this (I will use "##\omega##" instead of "##\aleph##").
First of all, I think it helps to forget about models momentarily and just look at an analogous function ##G:\omega_2 \times \omega \rightarrow\{0,1\}##. Now I think it is reasonably clear that this function ##G## encodes ##\omega_2## many functions ##G_\alpha:\omega \rightarrow\{0,1\}## (with ##\alpha<\omega_2##). For any arbitrary ##\alpha<\omega_2## we can define ##G_\alpha## as:
##G_\alpha(x)=G(\alpha,x)## for all ##x \in \omega##
Now to ensure that we have actually encoded ##\omega_2## many functions we also have to verify that (here we have ##\alpha,\beta<\omega_2##):
##G_\alpha \neq G_\beta## whenever ##\alpha \neq \beta##
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Now let's come back to the function ##F:\omega^M_2 \times \omega \rightarrow \{0,1\}##. In an analogous way, we can define an ##F_\alpha## for all ##\alpha<\omega^M_2##. This function ##F## encodes functions ##F_\alpha:\omega \rightarrow \{0,1\}## (with ##\alpha<\omega^M_2##). Now for any arbitrary ##\alpha<\omega^M_2## we can define:
##F_\alpha(x)=F(\alpha,x)## for all ##x\in\omega##
Now we want a similar condition as before to be true (here we have ##\alpha,\beta<\omega^M_2##):
##F_\alpha \neq F_\beta## whenever ##\alpha \neq \beta##
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Now coming back to what you wrote:
nomadreid said:
I am a little puzzled by that last “interpret”: in which model? If we are interpreting it in M, then the Cartesian product would have cardinality ##\aleph##2M, which in M would not be the same as ##\aleph##0. Of course, from a more powerful model, ##\aleph##2M has the same cardinality as ##\aleph##0, but I don’t see why he would be saying that we would be interpreting it in a more powerful model.
Assuming that we would have ##\omega^M=\omega## (it is quite unclear to me whether this has to be kept as a necessary condition or not ... the text in linked document doesn't seem to mention this), as long as ##F## is contained in ##M## then
[within ##M##
] ##F## would encode ##\omega^2_M## many distinct functions
from ##\omega^M##
to ##\{0,1\}##. So, in that case, CH will be false in ##M##.
