Set confusion

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• Jan 26th 2010, 07:16 AM
novice
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.
• Jan 26th 2010, 07:23 AM
Manx
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?
• Jan 26th 2010, 07:25 AM
Plato
Quote:

Originally Posted by novice
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 $\displaystyle S$.
• Jan 26th 2010, 07:32 AM
novice
Quote:

Originally Posted by Manx
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.
• Jan 26th 2010, 08:08 AM
Manx
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?
• Jan 26th 2010, 08:17 AM
novice
Quote:

Originally Posted by Plato
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. (Clapping)(Bow)
• Jan 26th 2010, 08:22 AM
novice
Quote:

Originally Posted by Manx
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?(Nod)
• Jan 26th 2010, 08:36 AM
Manx
not sure what you mean.
I used ( in place of { in one set.
• Jan 26th 2010, 09:20 AM
novice
Quote:

Originally Posted by Manx
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 $\displaystyle A \subseteq B$.
• Jan 26th 2010, 09:47 AM
Manx
are we communicating?
1 is an element of S so {1} is one of several subsets of S.
• Jan 27th 2010, 09:29 AM
novice
Quote:

Originally Posted by Manx
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?
• Jan 27th 2010, 11:21 AM
Manx
no
If 1 is an element of S, then {1} is one subset of S.
• Jan 27th 2010, 03:04 PM
novice
Quote:

Originally Posted by Manx
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.(Clapping)(Clapping)(Clapping)
• Jan 27th 2010, 05:05 PM
Drexel28
Quote:

Originally Posted by novice
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.(Clapping)(Clapping)(Clapping)

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? $\displaystyle E\subseteq S \text{ iff }x\in E\implies x\in S$.
• Jan 28th 2010, 05:23 AM
canyiah
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.(Clapping)(Clapping)(Clapping)[/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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