# Thinking about equality of infinite sets

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• ago01
So in summary, if we have the following two sets:$$C = \\{x | x = 3r - 1, r \in Z\\}$$$$D = \\{x | x = 3s + 2, s \in Z \\}$$Then the proof can go as follows which I hamfisted using algebra:Note that $x = 3r - 1$ can be rewritten $x=3(s+1)-1 = 3s+2$. Since this is exactly the set $D$ for every s in Z, D will contain a c in C at s+1. Then $$C \subsetf #### ago01 TL;DR Summary Is there a better way to approach thinking about equality with infinite sets? I am reading an abstract algebra textbook and enjoying it. I am working through preliminaries some more to refine my knowledge on proofs with sets before really digging in. I understand that if$$X \subseteq Y$$and$$ Y \subseteq X$$Then$$ X = Y$$This makes sense to me. However, the following problem while simple caused problems with brain trying to establish a "bound" to make things make sense. Suppose we have the following two sets and wish to prove them equal:$$C = \\{x | x = 3r - 1, r \in Z\\}D = \\{x | x = 3s + 2, s \in Z \\}$$Then the proof can go as follows which I hamfisted using algebra: Note that x = 3r - 1 can be rewritten x=3(s+1)-1 = 3s+2. Since this is exactly the set D for every s in Z, D will contain a c in C at s+1. Then$$C \subseteq D$$. Conversely, note that x = 3s + 2 can be rewritten x = 3(r-1) + 2 = 3r - 1. Then for every r in Z, C will contain a d in D at r - 1. So D \subseteq C, and C = D. The arithmetic follows. The two equations are equal at r -1 and s+1 for any r and and s. I drew up a table and indeed the pattern is obvious. It is "off by one" so to speak, as indicated by the solutions. But this "off by one" property is exactly what is causing me confusion. There are two contradictory things in my head right now: 1. If we fix s and r to "run" for the same length it will always be off by one. So this tells me the sets aren't always equal. 2. However, if we see that we can choose s and r we can always choose s and r such that they produce the same value. In fact, this comes directly from the arithmetic. So I think the way to understand this is that the infinite sets are not really "built" incrementally, but rather by the set builder notation come into existing as a whole (if that makes sense). Then, no matter the case, we can always choose an s and r in these infinite sets proving that they're equal. But if I think about it like a person who might play with the sets first to see if the conjecture holds you will find it's always off by one. These two contradictory things I am trying to resolve and I can't split them enough in my head. Can anyone provide a better explanation? A set is a set in its entirety. It's not a process. • topsquark and FactChecker A set is a set in its entirety. It's not a process. So then it appears the best way to think about infinite sets is (2). Since we can choose a s and r in each set and produce an equal value for any s and r, then certainly the sets must be equal. The sets may be "out of order" but we can still choose two numbers that make them equal at any point even though one may appear "later" in the set. So then it appears the best way to think about infinite sets is (2). Since we can choose a s and r in each set and produce an equal value for any s and r, then certainly the sets must be equal. The sets may be "out of order" but we can still choose two numbers that make them equal at any point even though one may appear "later" in the set. A set is an unordered collection in the sense that there is no defined order to the set. E.g.$$\{1,2\} = \{2,1\}$$• ago01 and topsquark On a related point, I'll just throw this in because it may be instructive. Two sets have equal order if there exists a bijection between them. For infinite sets of equal order not every one to one mapping is a bijection. You can't show that two sets have different order, therefore, by finding a mapping between them that isn't a bijection. Instead, you would have to show that no bijection exists. It's the same idea in your example. It's not enough to compare the elements out of order- you can always do that. • ago01 and topsquark A set is an unordered collection in the sense that there is no defined order to the set. E.g.$$\{1,2\} = \{2,1\}

Ah yes, this is my computer science coming through and also why I am taking this class to improve my rigor. You don't "index" into sets. Sets either contain something or don't contain something. We can check the set from any point r, s in the sense that no matter what we will always move through the entire unbounded set of integers. Since if we fix r, we can always find an s that produces the same element in D, and vis-versa. Hence the reason we can reason about these infinitely large set. I was thinking too linearly.

I see I will mull over this some more. It's going to take some de-programming to start thinking of these like a mathematician.

• PeroK