set theory-partition

**sah_mat** · November 22nd 2008, 01:21 PM

the family set of{Ai,iEI}is a partition of A,
the family set of{Bj,jEJ}is a partition of B,
than prove {AixBj,iEI and jEJ}is a parititon of AxB.
thanks for your helps.

**Plato** · November 22nd 2008, 02:12 PM

First we show that each cell is not empty.
Given $A_c \times B_d$ because of partitions $\left( {\exists x \in A_{\,c} } \right)\left( {\exists y \in B_d } \right) \Rightarrow \quad \left( {x,y} \right) \in A_c \times B_d$ .
Does that prove the first requirement?

Now disjoint cells.
If $\left( {x,y} \right) \in \left( {A_c \times B_d } \right) \cap \left( {A_j \times B_k } \right)$ then $x \in A_c \cap A_j \,\& \,y \in \left( {B_d \times B_k } \right)<br />$ WHY?
WHY does mean that $\left( {A_c = A_j } \right)\,\& \,\left( {B_d = B_k } \right)$ ?
Now can you explain how this shows that different cells are disjoint?

Now the covering property.
If $\left( {a,b} \right) \in \left( {A \times B} \right) \Rightarrow \quad \left( {\exists j} \right)\left[ {a \in A_j } \right]\,\& \,\left( {\exists k} \right)\left[ {b \in B_k } \right]\,$ . How are we sure of this?
This means that $\left( {a,b} \right) \in \left( {A_j \times B_k } \right)$ .
How does that show that $\left( {A \times B} \right) = \bigcup\limits_{j,k} {\left( {A_j \times B_k } \right)}$ ?

**sah_mat** · November 22nd 2008, 02:14 PM

thanks ı call you as saint plato thanks for your helps

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