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Math Help - Union of two sets

  1. #1
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    Union of two sets

    \{\emptyset\}\cup A=?

    I say \{\emptyset\}\cup A by definition is \emptyset \in \{\emptyset\} or \emptyset \in \ A.
    Since \emptyset \notin \ A, then
    \{\emptyset\}\cup A=\{\emptyset\}

    I don't know whether this is correct.
    How do you draw a Venn Diagram for \{\emptyset\}\cup A ?
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  2. #2
    Senior Member Dinkydoe's Avatar
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    A\cup \left\{\emptyset\right\} Is the union of A with an extra element " \emptyset".

    Or, you can also turn the picture around: It is the container \left\{\emptyset\right\} united with all elements of A.

    After the union we have: \left\{\emptyset\right\}\subset (A\cup \left\{\emptyset\right\}) and A\subset (A\cup  \left\{\emptyset\right\}).

    I say by definition is or .
    Since , then
    This is very wrong. It seems you're rather talking about the intersection: A\cap \left\{\emptyset\right\}

    In that case your example is still incorrect. Provided that \emptyset \notin A we have A\cap \left\{\emptyset\right\}= \emptyset

    If \emptyset \in A then A\cap \left\{\emptyset\right\} = \left\{\emptyset\right\}
    Last edited by Dinkydoe; January 28th 2010 at 07:03 AM.
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    Quote Originally Posted by Dinkydoe View Post
    A\cup \left\{\emptyset\right\} Is the union of A with an extra element " \emptyset".

    Or, you can also turn the picture around: It is the container \left\{\emptyset\right\} united with all elements of A.

    After the union we have: \left\{\emptyset\right\}\subset (A\cup \left\{\emptyset\right\}) and A\subset (A\cup \left\{\emptyset\right\}).


    You painted a good picture using \left\{\emptyset\right\}\subset (A\cup \left\{\emptyset\right\}) and A\subset (A\cup \left\{\emptyset\right\}). I understood this part, but I am still not sure I have a full grasp of the concept.

    Let's see: Assuming that A is a nonempty container and x is some candy, by definition

     A \cup \{\emptyset\}=\{x: x \in \{\emptyset\} or x \in A\}.

    Since x \notin \{\emptyset\}, I can conclude A \cup \{\emptyset\}= A.

    I am thinking since there isn't any candy in the empty container, \{\emptyset\}, and that A is a nonempty container, I may conclude that the candy is in A.

    Quote Originally Posted by Dinkydoe View Post
    This is very wrong. It seems you're rather talking about the intersection: A\cap \left\{\emptyset\right\}

    In that case your example is still incorrect. Provided that \emptyset \notin A we have A\cap \left\{\emptyset\right\}= \emptyset

    If \emptyset \in A then A\cap \left\{\emptyset\right\} = \left\{\emptyset\right\}
    For this, suppose I make the question like \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\}\cap A=?.

    I say since \emptyset, \{\emptyset\}, \{\{\emptyset\}\} \notin A, thus \{\emptyset, \{\emptyset\},\{\{\emptyset\}\}\}\cap A=\emptyset.
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  4. #4
    Senior Member Dinkydoe's Avatar
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    Let's see: Assuming that is a nonempty container and is some candy, by definition

    or .
    I can better paint the picture by making it more concrete: Let A = \left\{a_1,a_2,\cdots \right\}. Then A\cup \left\{\emptyset\right\}=\left\{\emptyset, a_1,a_2,\cdots \right\}

    Since , I can conclude .
    Wrong!

    \emptyset \in \left\{\emptyset\right\}. The empty set is an element of this set. You can write the emptyset as: \emptyset = \left\{\right\}. It's a container without elements. But a container without elements, can still be an element of another container! Namely: \left\{\emptyset\right\} = \left\{\left\{\right\}\right\}

    And how do you conclude \emptyset \not\in A? The empty-set may, or may not be in A. It depends on how we define A to be.

    For this, suppose I make the question like

    I say since , thus
    \emptyset, \left\{\emptyset\right\}, \left\{\left\{\emptyset\right\}\right\} all may, or may not be contained in A. We didn't specify A, so we don't know.
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    Quote Originally Posted by Dinkydoe View Post
    I can better paint the picture by making it more concrete: Let A = \left\{a_1,a_2,\cdots \right\}. Then A\cup \left\{\emptyset\right\}=\left\{\emptyset, a_1,a_2,\cdots \right\}

    Wrong!

    \emptyset \in \left\{\emptyset\right\}. The empty set is an element of this set. You can write the emptyset as: \emptyset = \left\{\right\}. It's a container without elements. But a container without elements, can still be an element of another container! Namely: \left\{\emptyset\right\} = \left\{\left\{\right\}\right\}

    And how do you conclude \emptyset \not\in A? The empty-set may, or may not be in A. It depends on how we define A to be.



    \emptyset, \left\{\emptyset\right\}, \left\{\left\{\emptyset\right\}\right\} all may, or may not be contained in A. We didn't specify A, so we don't know.
    You have an incredible mind and patience.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by novice View Post
    You have an incredible mind and patience.
    Novice, remember the bag analogy? If you have a bag with stuff and an empty bag, and you pour them both into a new bag....then...
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    Quote Originally Posted by Drexel28 View Post
    Novice, remember the bag analogy? If you have a bag with stuff and an empty bag, and you pour them both into a new bag....then...
    Oh, yes, I remember now. That was a good one.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by novice View Post
    Oh, yes, I remember now. That was a good one.
    I think you are making this much harder on yourself than it is supposed to be. You are doing what is called Naive Set Theory. In other words, it's naive because the concepts are supposed to be taken axiomatically and are intuitive. You are not rigorously defining what A\cup B means, you are just supposed to realize that A\cup B means "take everything in A and everything in B and put it in this new set A\cup B while making sure to take out duplicates."
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  9. #9
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Dinkydoe View Post

    And how do you conclude \emptyset \not\in A? The empty-set may, or may not be in A. It depends on how we define A to be.
    .
    Isn't it the case that the empty set is a subset of any set?
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    Isn't it the case that the empty set is a subset of any set?
    Yes, it is true that \varnothing\subseteq E for any set E...but \subseteq{\color{red}\ne}\in. Contemplate the differences.
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  11. #11
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    set identities

    [IMG]file:///C:/Users/VCOX/AppData/Local/Temp/moz-screenshot-3.png[/IMG]i attached some set identities. a book by rosen discrete math and its applications is a really good book. there is a wikiibook on discrete math to although i havent really looked at it. did your problem come with what elements are in A I would think that you would need that information to solve for A U "set of nothing".

    also an set of "nothing" is NOT equal to a set of a set of "nothing" which is what {0} represents. 0 means the empty set and {0} means a set of an empty set (NOTE: zero represents the empty set i didnt want to open word and have to print screen add to paint then insert the picture)
    "nothing" is an element of any set but a set of "nothing" isnt an element of every set. sets and elements are different a set has elements in them(group of things ie the entire alphabet is a set "a to z"). elements goes into sets(things ie individual alphabets "a" are considered elements).
    Attached Thumbnails Attached Thumbnails Union of two sets-setid.jpg  
    Last edited by canyiah; January 29th 2010 at 03:13 AM.
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  12. #12
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    Quote Originally Posted by novice View Post
    \{\emptyset\}\cup A=?

    I say \{\emptyset\}\cup A by definition is \emptyset \in \{\emptyset\} or \emptyset \in \ A.
    Since \emptyset \notin \ A, then
    \{\emptyset\}\cup A=\{\emptyset\}

    I don't know whether this is correct.
    How do you draw a Venn Diagram for \{\emptyset\}\cup A ?
    Be careful with your notation! \phi and {} are often used to mean the empty set. But \{\phi\} is NOT empty. It contains one member- the empty set.
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  13. #13
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    Quote Originally Posted by Drexel28 View Post
    You are doing what is called Naive Set Theory. In other words, it's naive because... "
    I sounds very funny.

    Quote Originally Posted by Drexel28 View Post
    A\cup B means "take everything in A and everything in B and put it in this new set A\cup B while making sure to take out duplicates."
    That is a clever definition. It would be nice if that was in all the math books.

    I told a friend that I was bothered by the fact that the set A wasn't being defined. He told me the point was to show \{\emptyset\} \cup A is no longer two sets but one---just as you explained it.

    I think I understand the concept thoroughly this time.
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  14. #14
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    Rosen Discrete Math link

    here is a link to rosens book site has alot of interactive stuff on it you can practice problems and other things
    Discrete Mathematics and Its Applications
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