I need some help with part (b) of the question. Would appreciate feedback on whether my solution to part (a) is correct too.
For each 𝑥 ∈ ℝ, define 𝐴𝑥 = {𝑥 + 𝑘 ∶ 𝑘 ∈ ℤ}. Let 𝒞 = {𝐴𝑥 ∶ 𝑥 ∈ ℝ}.
(a) Prove that 𝐴𝑥 is countable for every 𝑥 ∈ ℝ.
(b) Prove that 𝒞 is uncountable.
You may use without proof the fact that a set 𝐵 is countable if and only if there is a sequence
𝑏0, 𝑏1, 𝑏2, … ∈ 𝐵 in which every element of 𝐵 appears.
For part (a), I proved that for all 𝑥 ∈ ℝ, 𝐴𝑥 can be written in a sequence defined as below:
c2i = x – i
c2i+1 = x + i + 1
i.e. 𝐴𝑥 = {x, x + 1, x – 1, x + 2, x – 2, x + 3, x – 3, ….}
For part (b), however, I am stuck on proving 𝒞.
I believe it is probably something to do with cardinality of Unions? Since 𝒞 is just a Union of 𝐴1, 𝐴2, 𝐴3 …
However, in my current syllabus, one theorem I am taught is that:
"Let A,B be countable infinite sets. Then A U B is countable."
Thank you for taking the time to read this and I appreciate all feedback! Thank you
Best Answer
No! 𝒞 is a set of sets, not a union of sets. The cardinality of 𝒞 is just the number of sets it contains. Naively you might think that this is the same as the cardinality of $\{x:x\in\Bbb R\}$, which of course is just $\mathfrak{c}$. But there are duplications in there, so you have to be a bit careful.