LHS=RHS in the following expresion

[Zetta]

I would be thankful if anyone could show me exactly why the LHS=RHS in the following expresion:


$$\bigcup_{i=0}^{\infty}{(\bigcap_{j=0}^{\infty}{A_{ij}})}=\bigcup\{(\bigcap_{i=0}^{\infty}{A_{ih(i)}})|h\in \mathbb{N}^{\mathbb{N}}\}$$

where we agree that:
$$\bigcup S=\{x|x \in y\text{ for some }y\in S\}$$

and

$$\bigcap S=\{x|x \in y\text{ for all }y\in S\}$$

$$A_{ij} \text{ are sets}$$

and furthermore why there is no countable

$$H \subseteq \mathbb{N}^{\mathbb{N}$$

such that:
$$\bigcap_{i=0}^{\infty}(\bigcup_{j=0}^{\infty})=\bigcup\{(\bigcap_{i=0}^{\infty}{A_{ih(i)}})|h\in H\}$$

Thank You

 

[Valkarie]

in Advance.To prove that the LHS is equal to the RHS, we need to show two things: 1. Every element of the LHS is also an element of the RHS. 2. Every element of the RHS is also an element of the LHS. 1. Let x be an arbitrary element of the LHS, i.e. x belongs to at least one of the Aij sets. Then, for any h in the set of all functions from N to N, x belongs to Aih(i). This means that x is also an element of the RHS. 2. Let x be an arbitrary element of the RHS, i.e. x belongs to Aih(i) for some function h in the set of all functions from N to N. Since h is a function from N to N, for each i, there exists some j such that h(i)=j. This implies that x belongs to at least one of the Aij sets, and thus, x is also an element of the LHS. This shows that the LHS is equal to the RHS. As for why there is no countable subset H of the set of all functions from N to N such that the RHS is equal to the LHS, it is because the set of all functions from N to N is uncountable. Since H is a subset of this set, it must also be uncountable.