# Set Theory.

• Oct 12th 2010, 08:39 AM
skamoni
Set Theory.
Hi, could anyone help me start off this proof:

Notation: For a set $\displaystyle X$ we define $\displaystyle \bigcup\ X \;=\;\{x \mid x\in y\ for\ some\ y\in X \}$

Then show if $\displaystyle X_{ij}$ for $\displaystyle i,j\in\mathbb{N}$ are sets then:

$\displaystyle \bigcap\limits^\infty_{i=0}(\bigcup\limits^\infty_ {j=0}X_{ij})\;=\;\bigcup\{( \bigcap\limits^\infty_{i=0}X_{ih(i)})\mid h\in\mathbb{N^{N}}\}$
• Oct 12th 2010, 09:01 AM
Plato
If $\displaystyle x\in LHS$ then $\displaystyle \left( {\forall i} \right)\left( {\exists j} \right)\left[ {x \in X_{i~j} } \right]$.
We are going to use that to define $\displaystyle h:\mathBB{N}\to\mathBB{N}$ by $\displaystyle h:i\mapsto j$.
So $\displaystyle x\in X_{i~h(i)}$.
• Oct 12th 2010, 10:09 AM
skamoni
Thanks Plato, I think understand what you're saying but doesn't this only give $\displaystyle \bigcap\limits^\infty_{i=0}(\bigcup\limits^\infty_ {j=0}X_{ij})\;\subseteq\;\bigcup\{( \bigcap\limits^\infty_{i=0}X_{ih(i)})\mid h\in\mathbb{N^{N}}\}$?
If so, how would I prove $\displaystyle \supseteq$