# Set confusion

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• Jan 26th 2010, 08:16 AM
novice
Set confusion
Let $S$ = {1,{2},{1,2}}.

My book says the following are subsets of $S$: {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 $S$, but I have trouble undestanding why {1}, {1,{2}}, and {{1,2}} are being subsets of $S$. The sets {1}, {1,{2}}, and {{1,2}} seem to me are the subsets of the power set of $S$, $P(S)$.

I am so confused.
• Jan 26th 2010, 08:23 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, 08:25 AM
Plato
Quote:

Originally Posted by novice
Let $S$ = {1,{2},{1,2}}.

My book says the following are subsets of $S$: {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 $\color{red}S$, but I have trouble undestanding why {1}, {1,{2}}, and {{1,2}} are being subsets of $S$. The sets {1}, {1,{2}}, and {{1,2}} seem to me are the subsets of the power set of $S$, $P(S)$.

The part in red is false.
$2\notin S$ therefore neither $\{2\}\text{ nor }\{1,2\}$ could be a subset of $S$.
The both contain an element not in $S$.
• Jan 26th 2010, 08:32 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 $A\subset B\subset C$

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

Still can't picture what you are trying to say.
• Jan 26th 2010, 09:08 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, 09:17 AM
novice
Quote:

Originally Posted by Plato
The part in red is false.
$2\notin S$ therefore neither $\{2\}\text{ nor }\{1,2\}$ could be a subset of $S$.
The both contain an element not in $S$.

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

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

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

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

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

Thanks you, Plato. Nice to know that you are not too far to reach. (Clapping)(Bow)
• Jan 26th 2010, 09:22 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, 09:36 AM
Manx
not sure what you mean.
I used ( in place of { in one set.
• Jan 26th 2010, 10:20 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 $S$; it's not a subset of $S$.

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

Definition of subset:

For sets $A$ and $B$, if every element in $A$ belongs to $B$, then $A \subseteq B$.
• Jan 26th 2010, 10:47 AM
Manx
are we communicating?
1 is an element of S so {1} is one of several subsets of S.
• Jan 27th 2010, 10:29 AM
novice
Quote:

Originally Posted by Manx
1 is an element of S so {1} is one of several subsets of S.

Since $S$={1,{2},{1,2}}, 1 $\in S$.

$S$ 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, 12:21 PM
Manx
no
If 1 is an element of S, then {1} is one subset of S.
• Jan 27th 2010, 04:04 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 $\in S$ and {1} $\in S$ imply {1} $\subset S$

Manx, you are a good friend because you put up with me.(Clapping)(Clapping)(Clapping)
• Jan 27th 2010, 06:05 PM
Drexel28
Quote:

Originally Posted by novice
Got it now after plenty of sleep. 1 $\in S$ and {1} $\in S$ imply {1} $\subset S$

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 $a\in S$ THEN $\{a\}\subseteq S$. You seem to be having an awful lot of confusion. Have you tried looking at what subset means abstractly? $E\subseteq S \text{ iff }x\in E\implies x\in S$.
• Jan 28th 2010, 06:23 AM
canyiah
correction
[quote=novice;446392]Got it now after plenty of sleep. 1 $\in S$ and {1} $\in S$ imply {1} $\subset S$

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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