Undergrad Direction of logical implication in bijectively related sets

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The discussion revolves around the logical implications between two sets, X and A, under the assumption of a bijection. It is proposed that if an element in set A is false, the corresponding element in set X must also be false, and vice versa. However, participants highlight that a bijection does not inherently establish a causal relationship between the elements of the two sets. The need for precise definitions is emphasized, particularly regarding the nature of the events and their implications. Ultimately, the conversation suggests that without clear definitions and understanding of causality, the implications drawn may not hold true.
entropy1
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I have a hypothesis of which I wonder if it's sound. Perhaps you guys can advise me:

Suppose ##x_n\Rightarrow a_n## (logical implication) for some set X and set A. I think we have to assume a bijection.

Then, if ##a_m = False##, ##x_m## should be ##False##, right?

So, in case of a bijection, if ##a_n = True##, it follows ##x_n = True##.

Does that make sense?
 
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I 'm not sure what the ##n## means in ##X_n \implies A_n##. Do you mean ##(\forall x \in X_n)(\forall a \in A_n) x \implies a##, or do you mean ##X## and ##A## are ordered sets of size ##n## where ##x_i \implies a_i##?

A bijection is just a pairing of the elements. It doesn't say anything about how the elements are related. A simple counter example is a bijection for ##X=\{False\}, A=\{True\}.##

1601403486730.png
 
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Jarvis323 said:
I 'm not sure what the ##n## means in ##X_n \implies A_n##. Do you mean ##(\forall x \in X_n)(\forall a \in A_n) x \implies a##, or do you mean ##X## and ##A## are ordered sets of size ##n## where ##x_i \implies a_i##?
What I ment to say is that if event ##x_n## happens, that implies that event ##a_n## happens. That holds for every n.

##x_n \neq x_m## if ##n\neq m##. Same for ##a_n##.

Concluding that if ##a_m## doesn't happen, ##x_m## doesn't happen.

All ##a_n## must have a cause out of set X.

There can only be a single ##x_n## and a single ##a_n## true.

Then ##a_n \Rightarrow x_n##, but it almost seems trivial now. 🤣
 
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This also assumes that both the as and the xs cover the space of things that can happen. Let x_n be the event that a random integer is n, and a_n be the event that it's either n or -n. Then the implication only goes one way.
 
entropy1 said:
Then ##a_n \Rightarrow x_n##, but it almost seems trivial now. 🤣
I think you need an accurate and precise definition to begin with. Having two sets and a bijection doesn't seem to change anything. It's still just 2 events you need to look at. Using logical implication I guess you mean that if ##x## happens then ##a## will happen. And then because there are only two events you think that the reverse must be true as well? But that is not generally true, and the conditions and definitions you've given don't lead to this.

I'm guessing you are assuming that ##a## and ##x## are the only two potential events in the universe, every event must be caused by another event, and every event in the universe must happen.

One scenario is that the universe begins with ##x##, then ##a## happens, then the universe ends. But ##x## had no cause, which violates your assumption. So it must be that ##a## causes ##x##, and you have a loop, with a chicken and egg paradox.
 
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entropy1 said:
All ##a_n## must have a cause out of set X.
This is not the first time you have talked about 'cause' in relation to implication. There is no point in doing anything else until you have corrected this misunderstanding.
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

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