Union of two sets

**novice** · January 28th 2010, 04:36 AM

$\{\emptyset\}\cup A$ =?

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 $\{\emptyset\}\cup A$ ?

**Dinkydoe** · January 28th 2010, 05:06 AM

$A\cup \left\{\emptyset\right\}$ 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 $A\cap \left\{\emptyset\right\} = \left\{\emptyset\right\}$

**novice** · January 28th 2010, 07:17 AM

Originally Posted by Dinkydoe

$A\cup \left\{\emptyset\right\}$ 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 $A$ .

Originally Posted by Dinkydoe

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 $\{\emptyset, \{\emptyset\},\{\{\emptyset\}\}\}\cap A=\emptyset.$

**Dinkydoe** · January 28th 2010, 07:44 AM

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.

**novice** · January 28th 2010, 07:54 AM

Originally Posted by Dinkydoe

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.

**Drexel28** · January 28th 2010, 02:30 PM

Originally Posted by novice

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...

**novice** · January 28th 2010, 08:30 PM

Originally Posted by Drexel28

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.

**Drexel28** · January 28th 2010, 08:35 PM

Originally Posted by novice

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."

**VonNemo19** · January 28th 2010, 09:02 PM

Originally Posted by Dinkydoe

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?

**Drexel28** · January 28th 2010, 09:03 PM

Originally Posted by VonNemo19

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.

**canyiah** · January 29th 2010, 02:02 AM

[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).

**HallsofIvy** · January 29th 2010, 03:06 AM

Originally Posted by novice

$\{\emptyset\}\cup A$ =?

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.

**novice** · January 29th 2010, 06:41 AM

Originally Posted by Drexel28

You are doing what is called Naive Set Theory. In other words, it's naive because... "

I sounds very funny.

Originally Posted by Drexel28

$A\cup B$ 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.

**canyiah** · January 29th 2010, 01:23 PM

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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