# Union of two sets

• Jan 28th 2010, 05:36 AM
novice
Union of two sets
$\{\emptyset\}\cup A$=?

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 $\{\emptyset\}\cup A$ ?
• Jan 28th 2010, 06:06 AM
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 $A\subset (A\cup \left\{\emptyset\right\})$.

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 $A\cap \left\{\emptyset\right\} = \left\{\emptyset\right\}$
• Jan 28th 2010, 08:17 AM
novice
Quote:

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 $A$.

Quote:

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 $\{\emptyset, \{\emptyset\},\{\{\emptyset\}\}\}\cap A=\emptyset.$
• Jan 28th 2010, 08:44 AM
Dinkydoe
I can better paint the picture by making it more concrete: Let $A = \left\{a_1,a_2,\cdots \right\}$. Then $A\cup \left\{\emptyset\right\}=\left\{\emptyset, a_1,a_2,\cdots \right\}$

Quote:
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.
• Jan 28th 2010, 08:54 AM
novice
Quote:

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.
• Jan 28th 2010, 03:30 PM
Drexel28
Quote:

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...
• Jan 28th 2010, 09:30 PM
novice
Quote:

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.
• Jan 28th 2010, 09:35 PM
Drexel28
Quote:

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."
• Jan 28th 2010, 10:02 PM
VonNemo19
Quote:

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?
• Jan 28th 2010, 10:03 PM
Drexel28
Quote:

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.
• Jan 29th 2010, 03:02 AM
canyiah
set identities
[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).
• Jan 29th 2010, 04:06 AM
HallsofIvy
Quote:

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.
• Jan 29th 2010, 07:41 AM
novice
Quote:

Originally Posted by Drexel28
You are doing what is called Naive Set Theory. In other words, it's naive because... "

I sounds very funny.(Rofl)

Quote:

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.
• Jan 29th 2010, 02:23 PM
canyiah