Hi Marty,

On 03 Jul 2009, at 00:18, m.a. wrote:

> Bruno,
>               Comments and questions are interspersed below.
>                         marty

> Just tell me if you agree.     I agree and can't understand how I  
> could have been so careless.

So you have to introspect a little bit.

>> Do you remember the empty set? Can you compute:
>> {1, 2} UNION { } = ?  1,2
> OK, but don't forget the accolades.    Are accolades brackets?

Due to Dirac, in Quantum Mechanics, I tend to believe that brackets  
are "<" and ">". parentheses are "(" and ")". I call "{" and "}"  
accolades, but perhaps they are called bracket. The terms are not  
important as far as we understand each other. How would you call "["  
and "]" ?

> You are too quick here, you forget to type the 1.
> {1} UNION { } =  {1 }             Yes, I mistook the {1} for the  
> number of the question...not part of the equation. I tend to  
> overlook the fine points.

It is good that you are aware of that facts. It is what matter. It is  
what will make it possible to develop the familiarity with the  
important fine points which can arise from time to time.

>> Now, an important distinction which will follow us through ...  
>> forever.  I suggest you read attentively the next two paragraphs  
>> two times before breakfast, every day for one week. :), Really take  
>> all your time. It concerns the notion of operation, and relation.
>> INTERSECTION and UNION, are operations on sets, like addition (+,  
>> or PLUS) and multiplication (*, or TIMES) are operation on numbers.  
>> This means, typically, that, if x and y denote numbers, then x + y,  
>> and x * y, will denote, or are equal to, numbers. For example 3 + 4  
>> is equal to 7.
>> Similarly, if x and y denotes, or are equal, to sets, then x  
>> INTERSECTION y denotes, or is equal to, some set. For example {1,2}  
>> INTERSECTION {2, 7} is equal to some set, actually the set {2}.  
>> OK?......No!
>>                                                                 Why 
>>  not the sets {1,2,7} if INTERSECTION means BOTH?

Ah, but the word "both" alone is ambiguous. You could say that the  
UNION of two sets is the merging of BOTH set, and the intersection is  
the given of the elements which are in both set. So the union of {1,  
2} and {2, 7} is {1, 2, 7}, which indeed merges BOTH sets. But for  
computing the intersection, you must ask yourself, does this *element*  
belongs to BOTH set? So, for the intersection of {1, 2} and {2, 7},  
you have to ask yourself the following question: does 1 belong to both  
set? well, the answer is NO. the 1 belongs to the first set but not to  
the second, and so 1 does not belong to the intersection. Does 2  
belongs to both sets? The answer is yes. 2 belongs to {1, 2} and 2  
belongs to {2, 7}. Does 7 belongs to both sets, the answer is no, 7  
belongs to the second set, but does not belong  to the first set, so 7  
is not in the intersection.
Tell me if you are OK with this.

> I'm back!  I give you two last exercises to ponder about, just  in  
> case of insomnia. Again, take your time. I hope Kim follows, and  
> does not look at the solution !
> 1°) In the two relational formula below, one is true, the other is  
> false. Which one are what?
> a)    { } INCLUDED-IN { }True

Very good. All elements of { } are among the elements of { }. This is  
sometimes said to be "trivially" true, because { } has no elements at  
This is an example of an important "fine point".  Examples:

To verify if the set {1, 2, 3} is included in {34, 56, 7, 2, 100, 1,  
45, 3, 4}, you have to check THREE things: does 1 belongs to the  
second set, does 2 belongs to the second set, does 3 belongs to the  
second set.

To verify if the set {1, 2} is included in {34, 56, 7, 2, 100, 1, 45,  
3, 4}, you have to check TWO things: does 1 belongs to the second set,  
does 2 belongs to the second set.

To verify if {1} is included in {34, 56, 7, 2, 100, 1, 45, 3, 4}, you  
have to check ONE things: does 1 belongs to the second set.

To verify if { } is included in {34, 56, 7, 2, 100, 1, 45, 3, 4}, you  
have to verify ZERO thing! So it is automatically true. That is why  
logicians say it is trivially true.

 From this you should understand that the empty set, { }, is included  
to any set.

So { } is included in all the sets:  { }, {1}, {1, 2}, ... {0, 1,  
2, ...}, ....

In particular, as you said correctly, { } is included in { }. Put in  
another way, ({ } INCLUDED-IN { }) = true.

> b)    { } BELONGS-TO { } True

NOT correct. Remember that the empty set is empty, so nothing belongs  
to it. All formula like (x belongs to { }) will be false. You can  
conceive a set as an empty box { }, in which you can fill elements. So  
the set {a, b} is the empty set in which you put the elements a, and  
then, the element b. The accolades "{" and "}" represents the box  
itself, and what is in between the accolades represents the elements  
of the set. You could have guessed the solution because I was helping  
you when saying that one of the proposition is true and the other is  
false, and this means that, like many beginners, you read the  
enunciation of the problem too much quickly. That is why I suggest you  
take your time, and read often, at different time, the enunciation of  
the problems, and actually all explanations as well.

The moral is:

"x belongs to { }"     is never true, or is always false, whatever x  
"{ } included in x"    is always true, or never false, whatever x  

> 2°) And I give you a slightly longer exercise. Can you give me all  
> the subsets of the set {1, 2} ?. That is, can you give me all the  
> sets which are included in the set {1, 2} ? In case of doubt, reread  
> the definitions, reread the examples, and never panic! I give you a  
> hint: the set {1, 2} has four subsets. Can you find them?
>                                                                          {1 
> } {2} {1,2} {2,1}     why not {3} ?

Not too bad. 3/4 correct:

{1} is included in {1, 2}.  Indeed.
{2} is included in {1, 2}. Indeed.
{1, 2} is included in {1, 2}. Indeed.

{2, 1} is included in {1, 2}. Indeed, that is true, but you have to  
remember what you have already agree on: the set {1, 2} is equal to  
the set {2, 1}, so this is not a new solution. It is the preceding one  
in disguised!

Why not {3}? {3} is not included in {1, 2} just because 3 does not  
belong to {1, 2}. Reread the definition of inclusion. A is included in  
B if all the elements of A belongs to B. OK?

So you have found three subsets, among the four. Reading today's  
explanations I think you could find the missing subset. I let you  
search a little bit.

So just one exercise: what is the missing subset?

I apologize if all of this is a bit boring, but soon enough it will be  
highly rewarding. You will see.


You received this message because you are subscribed to the Google Groups 
"Everything List" group.
To post to this group, send email to everything-list@googlegroups.com
To unsubscribe from this group, send email to 
For more options, visit this group at 

Reply via email to