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Thread: Set confusion

  1. #1
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    Set confusion

    Let $\displaystyle S$ = {1,{2},{1,2}}.

    My book says the following are subsets of $\displaystyle S$: {1}, {1,{2}}, {{1,2}}

    I know that 1 , {2}, {1,2} are elements of S and that {2} and {1,2} are subsets of $\displaystyle S$, but I have trouble undestanding why {1}, {1,{2}}, and {{1,2}} are being subsets of $\displaystyle S$. The sets {1}, {1,{2}}, and {{1,2}} seem to me are the subsets of the power set of $\displaystyle S$, $\displaystyle P(S)$.

    I am so confused.
    Someone please help.
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  2. #2
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    elements of sets can be sets

    Set S has three elements. If you re-label {2} as A and {1,2} as B, then S={1,A,B}
    Form the subsets of S from these symbols.
    Then ...


    replace A and B with {2} and {1,2} respectively. What do you have?
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  3. #3
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    Quote Originally Posted by novice View Post
    Let $\displaystyle S$ = {1,{2},{1,2}}.

    My book says the following are subsets of $\displaystyle S$: {1}, {1,{2}}, {{1,2}}

    I know that 1 , {2}, {1,2} are elements of S and that {2} and {1,2} are subsets of $\displaystyle \color{red}S$, but I have trouble undestanding why {1}, {1,{2}}, and {{1,2}} are being subsets of $\displaystyle S$. The sets {1}, {1,{2}}, and {{1,2}} seem to me are the subsets of the power set of $\displaystyle S$, $\displaystyle P(S)$.
    The part in red is false.
    $\displaystyle 2\notin S$ therefore neither $\displaystyle \{2\}\text{ nor }\{1,2\}$ could be a subset of $\displaystyle S$.
    The both contain an element not in $\displaystyle S$.
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  4. #4
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    Quote Originally Posted by Manx View Post
    Set S has three elements. If you re-label {2} as A and {1,2} as B, then S={1,A,B}
    Form the subsets of S from these symbols.
    Then ...


    replace A and B with {2} and {1,2} respectively. What do you have?
    I could only see $\displaystyle A\subset B\subset C$

    Using your notation for the answer in my book I got these:
    {1} $\displaystyle \subset S$ ,
    $\displaystyle A\subset$ {1, A} $\displaystyle \subset S$,
    and $\displaystyle {{B}} \subset S$.

    Still can't picture what you are trying to say.
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  5. #5
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    subset list

    S={1,A,B}

    3 elements, 8 subsets

    { }
    {1}, {A}, {B}
    {1,A}, {1,B}, {A,B}
    {1,A,B}


    replace A and B
    (extra spaces added for emphasis -- look carefully at brackets.)
    { }
    {1}, ( {2} }, { (1,2} }
    {1, {2} }, {1, {1,2} } , { {2} , {1,2} }
    { 1, {2}, {1,2} }

    Do you see the sets you are looking for?
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  6. #6
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    Quote Originally Posted by Plato View Post
    The part in red is false.
    $\displaystyle 2\notin S$ therefore neither $\displaystyle \{2\}\text{ nor }\{1,2\}$ could be a subset of $\displaystyle S$.
    The both contain an element not in $\displaystyle S$.
    I understand it fully now.

    My problem being that I did not look at the definition carefully.

    Following the definition, I piece them together as follows:

    {1} $\displaystyle \subseteq S$, since 1 $\displaystyle \in$ {1} and 1 $\displaystyle \in S$,

    Next, {1,{2}} $\displaystyle \subseteq S$, since 1, {2} $\displaystyle \in ${1,{2}} and 1, {2} $\displaystyle \in S$,

    Last, {{1,2}} $\displaystyle \subseteq S$ , since {1,2} $\displaystyle \in ${{1,2}} and {1,2} $\displaystyle \in S$.

    I realized that the above is not exhaustive, since I can come up with more subsets, such as

    $\displaystyle S \subset S$
    {{2}{1,2}} $\displaystyle \subseteq S$.

    Thanks you, Plato. Nice to know that you are not too far to reach.
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  7. #7
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    Quote Originally Posted by Manx View Post
    S={1,A,B}

    3 elements, 8 subsets

    { }
    {1}, {A}, {B}
    {1,A}, {1,B}, {A,B}
    {1,A,B}


    replace A and B
    (extra spaces added for emphasis -- look carefully at brackets.)
    { }
    {1}, ( {2} }, { (1,2} }
    {1, {2} }, {1, {1,2} } , { {2} , {1,2} }
    { 1, {2}, {1,2} }

    Do you see the sets you are looking for?
    I assume {1} is a typo. Yah?
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  8. #8
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    not sure what you mean.
    I used ( in place of { in one set.
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  9. #9
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    Quote Originally Posted by Manx View Post
    not sure what you mean.
    I used ( in place of { in one set.
    1 is an element in $\displaystyle S$; it's not a subset of $\displaystyle S$.

    Here is the definition I read and reread till it sunk in:

    Definition of subset:

    For sets $\displaystyle A$ and $\displaystyle B$, if every element in $\displaystyle A$ belongs to $\displaystyle B$, then $\displaystyle A \subseteq B$.
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  10. #10
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    are we communicating?

    1 is an element of S so {1} is one of several subsets of S.
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  11. #11
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    Quote Originally Posted by Manx View Post
    1 is an element of S so {1} is one of several subsets of S.
    Since $\displaystyle S$={1,{2},{1,2}}, 1 $\displaystyle \in S$.

    $\displaystyle S$ contains there elements: 1, {2}, {1,2}, two of which are sets, namely {2}, {1,2}. So we know {1} is not one of them.

    Yah?
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  12. #12
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    no

    If 1 is an element of S, then {1} is one subset of S.
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  13. #13
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    Quote Originally Posted by Manx View Post
    If 1 is an element of S, then {1} is one subset of S.
    Got it now after plenty of sleep. 1 $\displaystyle \in S$ and {1} $\displaystyle \in S$ imply {1}$\displaystyle \subset S$

    Manx, you are a good friend because you put up with me.
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  14. #14
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    Quote Originally Posted by novice View Post
    Got it now after plenty of sleep. 1 $\displaystyle \in S$ and {1} $\displaystyle \in S$ imply {1}$\displaystyle \subset S$

    Manx, you are a good friend because you put up with me.
    I don't think what you are saying is correct. If $\displaystyle a\in S$ THEN $\displaystyle \{a\}\subseteq S$. You seem to be having an awful lot of confusion. Have you tried looking at what subset means abstractly? $\displaystyle E\subseteq S \text{ iff }x\in E\implies x\in S$.
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  15. #15
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    correction

    [quote=novice;446392]Got it now after plenty of sleep. 1 $\displaystyle \in S$ and {1} $\displaystyle \in S$ imply {1}$\displaystyle \subset S$

    Manx, you are a good friend because you put up with me.[/quote


    technically saying {1} E S is wrong. 1 E S given S {1,{2}, {1,2}}. {1} isnt an element of S. but i think i know where your confusion comes from. a subset gets {} regardless so {1} represents the 1 in S. {{2}} represents the {2} in S and {1,{2}} represents the 1,{2} in S. on a test if you say {1} E S that would be wrong because there is no element {1} in S there IS an element of 1 in S. Saying that there is a {1} E S means that this would be S {{1},{2},{1,2}} saying 1 E S is saying there is a 1 in S which there is {1,{2},{1,2}}
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