Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 6 of 6
  1. MHB Master
    MHB Site Helper
    Peter's Avatar
    Status
    Offline
    Join Date
    Jun 2012
    Location
    Hobart, Tasmania
    Posts
    2,863
    Thanks
    2,625 times
    Thanked
    895 times
    Awards
    MHB Model User Award (2018)  

MHB Model User Award (2015)  

MHB Model User Award (Jul-Dec 2013)
    #1
    I am reading Micheal Searcoid's book: "Elements of Abstract Analysis" ... ...

    I am currently focused on understanding Chapter 1: Sets ... and in particular Section 1.2 Relations and Functions ...

    I need some help in fully understanding some remarks by Searcoid in his sub-section on products ...

    The subsection on products reads as follows:







    In the above text by Searcoid we read the following:

    " ... ... Suppose $ \displaystyle (X_i)_{ i \in I }$ is a family of sets . Then $ \displaystyle \bigcup (X_i)$ is a set by Axiom IV, whence $ \displaystyle ( \bigcup (X_i) )^I$ is a set by (1.2.11) and $ \displaystyle \{ x \in ( \bigcup (X_i) )^I \ \lvert \ \forall j \in I , x_j \in X_j \}$ ... ... "


    I do not understand the form and nature of $ \displaystyle ( \bigcup (X_i) )^I$ ... ... what is meant by this? Can someone please give a simple and clear explanation ... and perhaps an example or two ...

    Further, likewise I do not understand the form and nature of $ \displaystyle \{ x \in ( \bigcup (X_i) )^I \ \lvert \ \forall j \in I , x_j \in X_j \}$ ... ... again, what is meant by this? Can someone please give a simple and clear explanation ... and perhaps an example or two ...


    Help will be much appreciated ... ...

    Peter


    =========================================================================================


    Section 1.2.11 is mentioned in the above quote ... so I am proving the text of the same ... as follows:






    Searcoid's sub-section on families of sets may also be relevant so I am providing the same ... as follows:







    Hope someone can help ...

    Peter
    Last edited by Peter; November 13th, 2017 at 07:28.

  2. # ADS
    Circuit advertisement
    Join Date
    Always
    Location
    Advertising world
    Posts
    Many
     

  3. MHB Craftsman
    MHB Math Scholar
    castor28's Avatar
    Status
    Offline
    Join Date
    Oct 2017
    Location
    Brussels, Belgium
    Posts
    227
    Thanks
    19 times
    Thanked
    317 times
    Thank/Post
    1.396
    Awards
    MHB Challenges Solver Award (2017)
    #2
    Hi Peter,

    I will try to explain this at an intuitive, although not rigorous, level.

    If $A$ and $B$ are sets, $A^B$ represents the set of functions from $B$ to $A$. The notation is justified by the fact that, if both sets are finite, with $|A|=m$ and $|B|=n$, then $A^B$ is a set of $m^n$ elements.

    When we are trying to define the Cartesian product of a family of sets, we must take into account the fact that some of these sets may be identical. For example, if we look at $A\times A$, we cannot simply consider the set $\{A,A\}$, because this is just $\{A\}$ : in a set, each element occurs exactly once. We must somehow be able to distinguish the first $A$ from the second $A$.

    To do this, instead of considering simply a set of sets, we consider a family of sets. We must assume that each set in the family is uniquely identified by some label or index; the set of indexes, denoted by $I$ in this case, is called the indexing set. We write $X_i$ for the set corresponding to the index $i\in I$, and we denote the family by $(X_i)_{i\in I}$. Note that:


    • The same set may occur more than once in the family: we may have $X_i = X_j$ for $i\ne j$.
    • In many examples, you would have $I=\{1, 2, \dots\}$, but $I$ can be any set, even an uncountable set like $\mathbb{R}$. Here, to illustrate the concepts, I will only use the first case, but the possibility of the second case is one of the reasons for this elaborate definition.


    We want elements of the Cartesian product to be the analog of ordered tuples. This means that, for each element of the Cartesian product, we must take one element for each set in the family. Now, a set $X_i$ of the family is identified by its index $i$, and we will denote the corresponding element by $x_i$, with $x_i\in X_i$.

    This means that, for each index $i$, we choose exactly one element $x_i$. This is the definition of a function $f$ defined on the set $I$, such that $f(i)=x_i$.
    The co-domain of that function is the set of all possible elements to choose from, and this is $\bigcup_{i\in I}X_i$. This shows that we have a function $f:I\to\bigcup_{i\in I}X_i$, and the set of all these functions is $\left(\bigcup_{i\in I}X_i\right)^I$, as explained at the beginning of this answer.

    If, for example, $I = \{1, 2, \dots\}$ an element of the Cartesian product would be a function defined on $I$ such that $f(i) = x_i\in X_i$, and we can write that simply as a sequence $(x_1, x_2,\dots)$.

    Note that, altough the co-domain of $f$ is the whole set $\bigcup_{i\in I}X_i$ (the set of all possible choices for the $x_i$), the image of $f$ may be smaller if the sets are not all the same. Indeed, we can only take $x_i$ in the corresponding set $X_i$, and this explains the definition $\{ x \in ( \bigcup (X_i) )^I \ \lvert \ \forall j \in I , x_j \in X_j \}$.

    Does this begin to clarify things ? Feel free to write back if you require further help.

  4. MHB Master
    MHB Site Helper
    Peter's Avatar
    Status
    Offline
    Join Date
    Jun 2012
    Location
    Hobart, Tasmania
    Posts
    2,863
    Thanks
    2,625 times
    Thanked
    895 times
    Awards
    MHB Model User Award (2018)  

MHB Model User Award (2015)  

MHB Model User Award (Jul-Dec 2013)
    #3 Thread Author
    Quote Originally Posted by castor28 View Post
    Hi Peter,

    I will try to explain this at an intuitive, although not rigorous, level.

    If $A$ and $B$ are sets, $A^B$ represents the set of functions from $B$ to $A$. The notation is justified by the fact that, if both sets are finite, with $|A|=m$ and $|B|=n$, then $A^B$ is a set of $m^n$ elements.

    When we are trying to define the Cartesian product of a family of sets, we must take into account the fact that some of these sets may be identical. For example, if we look at $A\times A$, we cannot simply consider the set $\{A,A\}$, because this is just $\{A\}$ : in a set, each element occurs exactly once. We must somehow be able to distinguish the first $A$ from the second $A$.

    To do this, instead of considering simply a set of sets, we consider a family of sets. We must assume that each set in the family is uniquely identified by some label or index; the set of indexes, denoted by $I$ in this case, is called the indexing set. We write $X_i$ for the set corresponding to the index $i\in I$, and we denote the family by $(X_i)_{i\in I}$. Note that:


    • The same set may occur more than once in the family: we may have $X_i = X_j$ for $i\ne j$.
    • In many examples, you would have $I=\{1, 2, \dots\}$, but $I$ can be any set, even an uncountable set like $\mathbb{R}$. Here, to illustrate the concepts, I will only use the first case, but the possibility of the second case is one of the reasons for this elaborate definition.


    We want elements of the Cartesian product to be the analog of ordered tuples. This means that, for each element of the Cartesian product, we must take one element for each set in the family. Now, a set $X_i$ of the family is identified by its index $i$, and we will denote the corresponding element by $x_i$, with $x_i\in X_i$.

    This means that, for each index $i$, we choose exactly one element $x_i$. This is the definition of a function $f$ defined on the set $I$, such that $f(i)=x_i$.
    The co-domain of that function is the set of all possible elements to choose from, and this is $\bigcup_{i\in I}X_i$. This shows that we have a function $f:I\to\bigcup_{i\in I}X_i$, and the set of all these functions is $\left(\bigcup_{i\in I}X_i\right)^I$, as explained at the beginning of this answer.

    If, for example, $I = \{1, 2, \dots\}$ an element of the Cartesian product would be a function defined on $I$ such that $f(i) = x_i\in X_i$, and we can write that simply as a sequence $(x_1, x_2,\dots)$.

    Note that, altough the co-domain of $f$ is the whole set $\bigcup_{i\in I}X_i$ (the set of all possible choices for the $x_i$), the image of $f$ may be smaller if the sets are not all the same. Indeed, we can only take $x_i$ in the corresponding set $X_i$, and this explains the definition $\{ x \in ( \bigcup (X_i) )^I \ \lvert \ \forall j \in I , x_j \in X_j \}$.

    Does this begin to clarify things ? Feel free to write back if you require further help.

    Castor28 ... I am most grateful for this helpful and informative post ... thanks!!!

    I am now reflecting on what you have written ...

    ... will certainly write back if I have further issues ...

    Thanks again for your help!

    Peter

  5. MHB Master
    MHB Site Helper
    Peter's Avatar
    Status
    Offline
    Join Date
    Jun 2012
    Location
    Hobart, Tasmania
    Posts
    2,863
    Thanks
    2,625 times
    Thanked
    895 times
    Awards
    MHB Model User Award (2018)  

MHB Model User Award (2015)  

MHB Model User Award (Jul-Dec 2013)
    #4 Thread Author
    Quote Originally Posted by castor28 View Post
    Hi Peter,

    I will try to explain this at an intuitive, although not rigorous, level.

    If $A$ and $B$ are sets, $A^B$ represents the set of functions from $B$ to $A$. The notation is justified by the fact that, if both sets are finite, with $|A|=m$ and $|B|=n$, then $A^B$ is a set of $m^n$ elements.

    When we are trying to define the Cartesian product of a family of sets, we must take into account the fact that some of these sets may be identical. For example, if we look at $A\times A$, we cannot simply consider the set $\{A,A\}$, because this is just $\{A\}$ : in a set, each element occurs exactly once. We must somehow be able to distinguish the first $A$ from the second $A$.

    To do this, instead of considering simply a set of sets, we consider a family of sets. We must assume that each set in the family is uniquely identified by some label or index; the set of indexes, denoted by $I$ in this case, is called the indexing set. We write $X_i$ for the set corresponding to the index $i\in I$, and we denote the family by $(X_i)_{i\in I}$. Note that:


    • The same set may occur more than once in the family: we may have $X_i = X_j$ for $i\ne j$.
    • In many examples, you would have $I=\{1, 2, \dots\}$, but $I$ can be any set, even an uncountable set like $\mathbb{R}$. Here, to illustrate the concepts, I will only use the first case, but the possibility of the second case is one of the reasons for this elaborate definition.


    We want elements of the Cartesian product to be the analog of ordered tuples. This means that, for each element of the Cartesian product, we must take one element for each set in the family. Now, a set $X_i$ of the family is identified by its index $i$, and we will denote the corresponding element by $x_i$, with $x_i\in X_i$.

    This means that, for each index $i$, we choose exactly one element $x_i$. This is the definition of a function $f$ defined on the set $I$, such that $f(i)=x_i$.
    The co-domain of that function is the set of all possible elements to choose from, and this is $\bigcup_{i\in I}X_i$. This shows that we have a function $f:I\to\bigcup_{i\in I}X_i$, and the set of all these functions is $\left(\bigcup_{i\in I}X_i\right)^I$, as explained at the beginning of this answer.

    If, for example, $I = \{1, 2, \dots\}$ an element of the Cartesian product would be a function defined on $I$ such that $f(i) = x_i\in X_i$, and we can write that simply as a sequence $(x_1, x_2,\dots)$.

    Note that, altough the co-domain of $f$ is the whole set $\bigcup_{i\in I}X_i$ (the set of all possible choices for the $x_i$), the image of $f$ may be smaller if the sets are not all the same. Indeed, we can only take $x_i$ in the corresponding set $X_i$, and this explains the definition $\{ x \in ( \bigcup (X_i) )^I \ \lvert \ \forall j \in I , x_j \in X_j \}$.

    Does this begin to clarify things ? Feel free to write back if you require further help.



    Hi Castor28 ... thanks again for your help ... you clarified things a great deal ...


    Now ... based on what I have understood from your post we have ... (I think ...? ... please comment ,,,)


    $ \displaystyle ( \bigcup ( X_i ))^I = \{ x \ \mid \ I \rightarrow \bigcup ( X_i ) \}$

    ... ... where $ \displaystyle x$ is a function or a set of ordered pairs no two of which have the same first member ... ...


    Now ... in the set $ \displaystyle \prod_i X_i = \{ x \in ( \bigcup ( X_i ))^I \ \mid \ \forall j \in I , \ x_j \in X_j \} $


    we have that ...


    $ \displaystyle x_j = x(j)$ is the value of the function $ \displaystyle x$ at $ \displaystyle j$


    so we would have the following (for a finite Cartesian Product ... and where, for example we take $ \displaystyle I $ as the set $ \displaystyle \{1,2, \ ... \ ... \ n \}$ )


    $ \displaystyle x(1) = x_1 \in X_1$, $ \displaystyle x(2) = x_2 \in X_2$, ... ... ... $ \displaystyle x(n) = x_n \in X_n$.


    Therefore the set/function $ \displaystyle x$ is the following set of ordered pairs ... ...


    $ \displaystyle x = \{ (1, x_1), (2, x_2), \ ... \ ... \ ... \ (n, x_n) \}$


    and the function $ \displaystyle x$ constitutes one element belonging to $ \displaystyle \prod_i X_i$ ... ...


    Other elements come from other functions $ \displaystyle y, z,$ ... ... belonging to the set $ \displaystyle ( \bigcup ( X_i ))^I$ ... ...


    Is the above interpretation of Searcoid correct?


    Somehow I think it would have been more 'natural' if we had arrived at an element $ \displaystyle x$ of $ \displaystyle \prod_i X_i$ having the form of a $ \displaystyle n$-tuple ... ... $ \displaystyle (x_1, x_2, \ ... \ ... \ ... \ x_n)$ rather than a set of ordered pairs ... can you comment ...

    ... BUT ... ... the aim of the exercise, I think, is to show that the Cartesian Product is a set ...is that a correct interpretation of what Searcoid is doing? ... can you comment ...


    Thanks again for your help ...

    Peter
    Last edited by Peter; November 14th, 2017 at 05:09.

  6. MHB Craftsman
    MHB Math Scholar
    castor28's Avatar
    Status
    Offline
    Join Date
    Oct 2017
    Location
    Brussels, Belgium
    Posts
    227
    Thanks
    19 times
    Thanked
    317 times
    Thank/Post
    1.396
    Awards
    MHB Challenges Solver Award (2017)
    #5
    Hi Peter,

    Yes, your interpretation is correct.

    Quote Quote:
    Somehow I think it would have been more 'natural' if we had arrived at an element $ \displaystyle x$ of $ \displaystyle \prod_i X_i$ having the form of a $ \displaystyle n$-tuple ... ... $ \displaystyle (x_1, x_2, \ ... \ ... \ ... \ x_n)$ rather than a set of ordered pairs ... can you comment ...
    The interpretation as ordered $n$-tuples is more natural, but you need a formal definition of an $n$-tuple to prove that is it a set. The usual way to define the ordered pair $(a,b)$ is $\{\{a\},\{a,b\}\}$, which is already cumbersome; this could be extended to a finite $n$-tuple, but it would not be very practical.

    The alternative way to define an $n$-tuple is as a function, which is what we did here. In the case of sequences (finite or infinite), these definitions are equivalent. However, as I mentioned, you can define Cartesian products of uncountable families. In that case, the function approach is still valid, but the ordered $n$-tuple representation does not make sense. For example, $\prod_{i\in\mathbb{R}}X_i$, where $X_i= \mathbb{R}$ is the Cartesian product of copies of $\mathbb{R}$ indexed by $\mathbb{R}$ (this is just the set of functions from $\mathbb{R}$ to itself). You cannot represent an element as an ordered $n$-tuple or an infinite sequence.

    Quote Quote:
    ... BUT ... ... the aim of the exercise, I think, is to show that the Cartesian Product is a set ...is that a correct interpretation of what Searcoid is doing? ... can you comment ...
    I looks like it. However, as I don't have the book, I cannot guess the intentions of the author: this should be interpreted in the context of the rest of the book.

  7. MHB Master
    MHB Site Helper
    Peter's Avatar
    Status
    Offline
    Join Date
    Jun 2012
    Location
    Hobart, Tasmania
    Posts
    2,863
    Thanks
    2,625 times
    Thanked
    895 times
    Awards
    MHB Model User Award (2018)  

MHB Model User Award (2015)  

MHB Model User Award (Jul-Dec 2013)
    #6 Thread Author
    Quote Originally Posted by castor28 View Post
    Hi Peter,

    Yes, your interpretation is correct.



    The interpretation as ordered $n$-tuples is more natural, but you need a formal definition of an $n$-tuple to prove that is it a set. The usual way to define the ordered pair $(a,b)$ is $\{\{a\},\{a,b\}\}$, which is already cumbersome; this could be extended to a finite $n$-tuple, but it would not be very practical.

    The alternative way to define an $n$-tuple is as a function, which is what we did here. In the case of sequences (finite or infinite), these definitions are equivalent. However, as I mentioned, you can define Cartesian products of uncountable families. In that case, the function approach is still valid, but the ordered $n$-tuple representation does not make sense. For example, $\prod_{i\in\mathbb{R}}X_i$, where $X_i= \mathbb{R}$ is the Cartesian product of copies of $\mathbb{R}$ indexed by $\mathbb{R}$ (this is just the set of functions from $\mathbb{R}$ to itself). You cannot represent an element as an ordered $n$-tuple or an infinite sequence.



    I looks like it. However, as I don't have the book, I cannot guess the intentions of the author: this should be interpreted in the context of the rest of the book.


    Thanks castor28 ... really appreciate your help and guidance ...

    I now have a much better understanding of the issues of my initial post ...

    Thanks again.

    Peter

Similar Threads

  1. Proving a Set Theory Statement Regarding Families of Sets
    By logan3 in forum Discrete Mathematics, Set Theory, and Logic
    Replies: 1
    Last Post: July 26th, 2016, 19:25
  2. Replies: 2
    Last Post: May 23rd, 2016, 00:47
  3. Why doesn't it come from a cartesian product of sets?
    By evinda in forum Discrete Mathematics, Set Theory, and Logic
    Replies: 4
    Last Post: November 12th, 2014, 15:59
  4. Tensor Products - Dummit and Foote Section 10.4, Example 2, page 363
    By Peter in forum Linear and Abstract Algebra
    Replies: 4
    Last Post: February 26th, 2014, 08:34
  5. Replies: 1
    Last Post: January 1st, 2013, 11:17

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards