100% correct!
Hello everyone. I would greatly appreciate if someone could validate whether or not the following solution is correct.
Problem: Is it possible for a countably infinte set to have an uncountable collection of subsets such that the intersection of any two is finite?
Answer: Yes it is.
Proof:
Define to be a rational sequence converging to and and let . It is clear that if that is finite, otherwise both sequences would converge to the same value. Particularly this implies that thus . Lastly noting that we may conclude that is an uncountable collection of subsets of a countable set such that the intersection of any two is finite.
What do you think?
Almost. I made a typo. Here was my final solution, with all the accrutements.
Problem: Does there exist an uncountable collection of subsets of a countably infinite set such that the intersection of any two is finite?
Claim: Yes there does.
Proof:
Lemma: Let be distinct real numbers. Furthermore let be rational sequences converging to respectively. Next define . Then is finite.
Proof: WLOG assume that . Then since and it can be seen that . This implies that such that . Manipulating this yields . Taking and focusing on the first two terms in teh above inequality tells us such that . And since is finite the conclusion follows
Now because both the rationals are dense in the reals for every real number there exists some rational sequence such that . Define as above and let . Since the above in particular implies that we see that the mapping given by is injective. Therefore . Lastly noting that we may conclude that is an uncountable collection of subsets of a countable set such that the intersection of any two is finite.
Remark: Although there must exist a rational sequence such that finding them is not altogether obvious. A simple example would be to define
I also came up with an alternate solution. Granted, I was given the hint 'lattice points'.
Problem: Does there exist an uncountable collection of subsets of a countable set such that the intersection of any two is finite?
Claim: Yes there is.
Proof: Let and let where is any point on the line . Clearly any two lines will only intersect at the origin . Next note that these two lines will eventually (no matter how close is) move one unit apart. This will occur when for some finite constant . This implies that is finite for any dinstinct . Next we can see that , which is clear from the previous observation. Therefore if we define that the mapping given by is injective. Thus . Lastly, notice that since consists entirely of lattice points that . Putting this all together we see that is an uncountably infite collection of subsets of a countably infinite set such that the intersection of any two is finite.
Hello. I've got an another idea. How about this:
Let be the prime numbers. For each , let . Then the set satisfies the condition required in the statement. It is easy to see that uniqueness of prime factorization gives us the finiteness of the intersection of any two distinct members of and that Cantor's diagonal slash gives us uncountability of , namely .
Thank you.
Misako Kawasoe