Indexed Families of Sets .... Just and Weese .... Exercise 8 ....

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The discussion revolves around Exercise 8 from "Discovering Modern Set Theory. I The Basics" by Winfried Just and Martin Weese, specifically focusing on Cartesian Products and indexed families of sets. The user, Peter, details his approach to solving the exercise, which involves defining the sets \( A_\phi \) and \( A_{\{ \phi \}} \) and calculating the Cartesian product \( A_\phi \times A_{\{ \phi \}} \). He also seeks confirmation of his solution and clarification on demonstrating a one-to-one mapping from the product of indexed sets to the Cartesian product.

PREREQUISITES
  • Understanding of Cartesian Products in set theory
  • Familiarity with indexed families of sets
  • Knowledge of functions as sets of ordered pairs
  • Basic concepts from "Discovering Modern Set Theory. I The Basics" by Just and Weese
NEXT STEPS
  • Review the definitions of Cartesian Products and indexed families in set theory
  • Study the concept of one-to-one mappings in set theory
  • Examine the properties of functions defined on sets
  • Practice additional exercises from "Discovering Modern Set Theory. I The Basics" to reinforce understanding
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Students of set theory, mathematicians exploring foundational concepts, and anyone seeking to deepen their understanding of Cartesian Products and indexed families of sets.

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I am reading the book: "Discovering Modern Set Theory. I The Basics" (AMS) by Winfried Just and Martin Weese.

I am currently focused on Chapter 1 Pairs, Relations and Functions ... and I am in particular focused on Cartesian Products and indexed families of sets ...

I need some help with Exercise 8 and some remarks following the exercise ...

The relevant section from J&W is as follows:View attachment 7533
View attachment 7534

It is also worth noting that earlier (on page 12) J&W defined ordered pairs and Cartesian Products as follows:

https://www.physicsforums.com/attachments/7535

I worked Exercise 8 as follows:

Elements of the set $$A_\phi \times A_{ \{ \phi \} }$$


Now ... $$\{ A_\phi = \{ \phi \}$$ and $$ A_{ \{ \phi \} } = \{ \phi , \{ \phi \} \}$$ ...So $$ A_\phi \times A_{ \{ \phi \} } = \{ \langle a,b \rangle \ \mid \ a \in A_\phi \text{ and } b \in A_{ \{ \phi \} } \}
$$

$$= \{ \ \langle \phi , \phi \rangle \ , \langle \phi , \{ \phi \} \rangle \ \} $$

$$ = \{ \ \{ \phi , \{ \phi \} \} \ , \ \{ \phi , \{ \{ \phi \} \} \ \} $$

Elements of the set $$\prod_{ i = \{ \phi , \{ \phi \} } A_i $$


Let $$I = \{ \phi , \{ \phi \} \} $$
$$\prod_{ i \in I } A_i = \{ f \in ( \bigcup \{ A_i \ : \ i \in I \} )^I \ : \ \forall i \in I , \ ( f(i) \in A_i ) \} $$

where $$( \bigcup \{ A_i \ : \ i \in I \} )^I = \{ f \ : \ I \rightarrow \bigcup A_i \} $$

$$= \{ f \ : \ \{ \phi , \{ \phi \} \} \rightarrow \bigcup A_i \} $$
Now ... we have to consider the function(s) from the domain $$\{ \phi , \{ \phi \} \}$$ to

$$ \bigcup A_i = A_\phi \cup A_{ \{ \phi \} } $$

$$= \{ \phi , \{ \phi \} \}$$The only function satisfying the required conditions is the following function f:

$$f = \{ \langle \phi , \phi \rangle , \langle \{ \phi \} , \{ \phi \} \rangle \} $$

$$= \{ \{ \phi , \{ \phi \} \} , \{ \{ \phi \} , \{ \{ \phi \} \} \}$$
My questions as follows:

Can someone please either confirm my working as correct or point out the errors ...?

Further, can someone show me simply and explicitly how H is a one to one map from $$\prod_{ i = \{ \phi , \{ \phi \} \} } A_i $$ onto $$A_\phi \times A_{ \{ \phi \} }$$ ... ... ?

Help will be much appreciated ...

Peter
 
Last edited:
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Peter said:
I am reading the book: "Discovering Modern Set Theory. I The Basics" (AMS) by Winfried Just and Martin Weese.

I am currently focused on Chapter 1 Pairs, Relations and Functions ... and I am in particular focused on Cartesian Products and indexed families of sets ...

I need some help with Exercise 8 and some remarks following the exercise ...

The relevant section from J&W is as follows:It is also worth noting that earlier (on page 12) J&W defined ordered pairs and Cartesian Products as follows:
I worked Exercise 8 as follows:

Elements of the set $$A_\phi \times A_{ \{ \phi \} }$$


Now ... $$\{ A_\phi = \{ \phi \}$$ and $$ A_{ \{ \phi \} } = \{ \phi , \{ \phi \} \}$$ ...So $$ A_\phi \times A_{ \{ \phi \} } = \{ \langle a,b \rangle \ \mid \ a \in A_\phi \text{ and } b \in A_{ \{ \phi \} } \}
$$

$$= \{ \ \langle \phi , \phi \rangle \ , \langle \phi , \{ \phi \} \rangle \ \} $$

$$ = \{ \ \{ \phi , \{ \phi \} \} \ , \ \{ \phi , \{ \{ \phi \} \} \ \} $$

Elements of the set $$\prod_{ i = \{ \phi , \{ \phi \} } A_i $$


Let $$I = \{ \phi , \{ \phi \} \} $$
$$\prod_{ i \in I } A_i = \{ f \in ( \bigcup \{ A_i \ : \ i \in I \} )^I \ : \ \forall i \in I , \ ( f(i) \in A_i ) \} $$

where $$( \bigcup \{ A_i \ : \ i \in I \} )^I = \{ f \ : \ I \rightarrow \bigcup A_i \} $$

$$= \{ f \ : \ \{ \phi , \{ \phi \} \} \rightarrow \bigcup A_i \} $$
Now ... we have to consider the function(s) from the domain $$\{ \phi , \{ \phi \} \}$$ to

$$ \bigcup A_i = A_\phi \cup A_{ \{ \phi \} } $$

$$= \{ \phi , \{ \phi \} \}$$The only function satisfying the required conditions is the following function f:

$$f = \{ \langle \phi , \phi \rangle , \langle \{ \phi \} , \{ \phi \} \rangle \} $$

$$= \{ \{ \phi , \{ \phi \} \} , \{ \{ \phi \} , \{ \{ \phi \} \} \}$$
My questions as follows:

Can someone please either confirm my working as correct or point out the errors ...?

Further, can someone show me simply and explicitly how H is a one to one map from $$\prod_{ i = \{ \phi , \{ \phi \} \} } A_i $$ onto $$A_\phi \times A_{ \{ \phi \} }$$ ... ... ?

Help will be much appreciated ...

Peter
I have been reflecting on my solution to Just and Weese Exercise 8 ...

I know think I may have made an error in determining the set $$\prod_{ i = \{ \phi , \{ \phi \} } A_i $$

... ... so I am now attempting to give a correct solution ...
We have that ... $$ A_\phi = \{ \phi \}$$ and $$ A_{ \{ \phi \} } = \{ \phi , \{ \phi \} \}$$ ...and we let $$I = \{ \phi , \{ \phi \} \} $$
Then ... ... $${}^I \!\left( \bigcup \{ A_i \}\right) = \{ f \ : \ I \rightarrow \bigcup A_i \}$$ But ... ... $$\bigcup A_i = \bigcup \{ A_\phi, A_{ \{ \phi \} } \} = A_\phi \cup A_{ \{ \phi \} } = \{ \phi , \{ \phi \} \}
$$Therefore ... $${}^I \!\left( \bigcup \{ A_i \}\right) = \{ f \ : \ \{ \phi , \{ \phi \} \} \rightarrow \{ \phi , \{ \phi \} \}$$ ... ... (1)and $$\prod_{ i \in I } A_i = \{ f \in {}^I \!\left( \bigcup \{ A_i \}\right) \ : \ \forall i \in I \ (f(i) \in A_i \ ) \ \} $$ ... ... ... (2)So ... working from (1) and (2) we have that $$\prod_{ i \in I } A_i = \{ g, h \}$$where (treating functions as sets of ordered pairs ...)$$g = \{ \ \langle \phi , \phi \rangle \ , \ \langle \{ \phi \} , \phi \rangle \ , \} = \{ \ \{ \phi , \{ \phi \} \} \ , \{ \{ \phi \} , \{ \phi \} \} \}$$and $$h = = \{ \ \langle \phi , \phi \rangle \ , \ \langle \{ \phi \} , \{ \phi \} \rangle \ \} = \{ \ \{ \phi , \{ \phi \} \} \ , \{ \{ \phi \} , \{ \{ \phi \} \} \ \} $$... ... BUT ... ... where to from here ... hmmm ...really need some help/guidance ...In particular how exactly and explicitly do we demonstrate the one-to-one map H from $$\prod_{ i = \{ \phi , \{ \phi \} } A_i $$ to $$A_\phi \times A_{ \{ \phi \} }$$ ... .. Help will be much appreciated ...

Peter
 
Last edited:

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