Author Topic: Concrete Sets - Subsets (Belonging to..) (Read 5410 times) Tweet Share

Yendall · « **on:** March 08, 2013, 05:28:22 pm »

Just need some clarification for my MATH-1074 class.

We are following the rule: $\varnothing \subseteq \mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q}\subseteq \mathbb{R}$

Where:
$\varnothing$ = Null Set
$\mathbb{N}$ = Natural Numbers
$\mathbb{Z}$ = Integers
$\mathbb{Q}$ = Rational Numbers
$\mathbb{R}$ = Rational and Irrational Numbers

As far as I can understand, Null belongs to N, N belongs to Z,Z belongs to Q, Q belongs to R

So when I get a question like: To which (if any) of the sets N, Z, Q and R does 0 belong?

I automatically say that 0 is a part of the Integer set, as it is not a positive counting number and cannot belong to the natural set N. However this isn't the correct answer.

Why isn't this the correct answer?

EDIT

I realised that the multiple choice quiz online actually requires you to state that it belongs to more than one set, rather than the direct set after it.

$\therefore 0 \subseteq \mathbb{Z} \subseteq \mathbb{Q}\subseteq \mathbb{R}$

Sorry for wasting a thread.

kamil9876 · « **Reply #1 on:** March 08, 2013, 07:21:53 pm »

I would like to point out your confusion between $\in$ and $\subset$

Quote from: Yendall on March 08, 2013, 05:28:22 pm

$\therefore 0 \subseteq \mathbb{Z} \subseteq \mathbb{Q}\subseteq \mathbb{R}$ [/center]

It should be

$0 \in \mathbb{Z} \subseteq \mathbb{Q}\subseteq \mathbb{R}$

or the following would also be true

$\{0 \} \subset \mathbb{Z} \subseteq \mathbb{Q}\subseteq \mathbb{R}$

For instance, let $S=\{ \{1,2\}, 3\}$

Then is...

$3 \subset S$ ?

$3 \in S$ ?

$1 \in S$ ?

$\{1,2\} \subset S$ ?

$\{1,2\} \in S$ ?

Yendall · « **Reply #2 on:** March 08, 2013, 10:01:40 pm »

Oh yes sorry, I meant to write that. 0 is an element of each set because it is an element of set Z.

I understand elements completely, I was just confused as to whether you needed to mention ALL sets when referring to belonging.

And I believe these three are only true:

$3 \in S$

$1 \in S$

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

I would also write it as:

$X = \{1,2\}$

$X \subseteq S$

kamil9876 · « **Reply #3 on:** March 09, 2013, 12:30:03 am »

The elements of $S$ are $\{1,2\}$ and $3$ . So $1 \in S$ is FALSE since it is not equal to $\{1,2\}$ or $3$ . However $\{1,2\} \in S$ is TRUE, this is an element of $S$ . However $\{1,2\} \subset S$ is FALSE, there are only four subsets of $S$ : the empty set, $\{ \{1,2\} \}$ , $\{3 \}$ and $S$ itself. So you can see that $\{1,2\}$ is not one of these four sets.

Yendall · « **Reply #4 on:** March 09, 2013, 07:31:23 am »

Oh sorry I completely misread that $\{1,2\}$ is a seperate element to $3$ .

So;

$\{1,2\} \in S$

$3 \in S$

So the three sets in S are:

$\varnothing \in S$

$\{1,2\} \in S$

$3 \in S$

However, would this make the cardinality of the set 3 or 2? Do you count the null set as an element or because it technically isn't real, is it excluded?

Also, if I were to denote these elements with letters would this work as well:

$x = \{1,2\}$

$y = 3$

$\therefore x \subseteq S$

$\therefore y \subseteq S$

kamil9876 · « **Reply #5 on:** March 09, 2013, 12:23:55 pm »

Quote

However, would this make the cardinality of the set 3 or 2? Do you count the null set as an element or because it technically isn't real, is it excluded?

The cardinality is the number of elements, there are two elements $\{1,2\}$ and $3$ in $S$ hence it is of cardinality $2$ . The empty set is NOT an element of $S$ , it is a subset of $S$ i.e $\varnothing \notin S$ but $\varnothing \subset S$ . Whether it is real or not does not matter, you can make up abstract sets where the elements are not real numbers (in fact $S$ is an example of one since the element $\{1,2\}$ of $S$ is NOT a real number)

Quote from: Yendall on March 09, 2013, 07:31:23 am

Also, if I were to denote these elements with letters would this work as well:

$x = \{1,2\}$

$y = 3$
$\therefore x \subseteq S$

$\therefore y \subseteq S$

But you were doing so well before (only mistake was $\varnothing \in S$ which is FALSE). Why does renaming the elements suddenly change the answer? It shouldn't of course, you can't make $2+2=5$ by letting $2=x, 2=y, 5=z$

Yendall · « **Reply #6 on:** March 09, 2013, 05:32:08 pm »

Okay I think i've just completely confused myself. I'll start from the beginning again and see where I go wrong haha:

Set:

$S = \{\{1,2\}, 3\}$

Elements:

$\{1,2\} \in S$

$3 \in S$

Subsets:

$\varnothing \subseteq S$

I believe that is all there is to it! Maybe haha

kamil9876 · « **Reply #7 on:** March 10, 2013, 01:38:52 am »

All those statements you have made are correct. (Although there are other subsets of $S$ , in fact four altogether as I have stated in my previous post).

Yendall · « **Reply #8 on:** March 10, 2013, 09:20:09 am »

Quote from: kamil9876 on March 10, 2013, 01:38:52 am

All those statements you have made are correct. (Although there are other subsets of $S$ , in fact four altogether as I have stated in my previous post).

Oh yes that's right! I think i understand it all now, thanks for your help! Sorry if I was a bit confusing haha

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Author Topic: Concrete Sets - Subsets (Belonging to..) (Read 5410 times) Tweet Share

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Concrete Sets - Subsets (Belonging to..)

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