On 04 Jul 2009, at 04:31, m.a. wrote:
> New comments in italics.
> 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.
> Not OK. You
> previously defined UNION as one OR the other. Now you seem to be
> giving me the same definition for INTERSECTION.
Let us take the set a = {1, 2} and the set b = {2, 3}.
Let us recall the definition (I abandon the capital letters because
they are ugly, and I feel talking louder!)
1) intersection
(a intersection b) = {x such-that (x belongs-to a) AND (x belongs-to
b) }.
So, some x, to belong to the intersection, has to belong
simultaneously to the two sets involved. Only when x is equal to 2, is
that condition verified.
2 belongs to (a intersection b) because 2 belongs to a, AND, 2 belongs
to b.
That condition is not verify for x = 1, nor for x = 3. 3 belongs to b,
but not to a. So 3 is not in the intersection. Nor 1, because it does
not belong to a.
So (a intersection b) = ( {1, 2} intersection {2, 3} ) = {2}.
2) Union
(a union b) = {x such-that (x belongs-to a) OR (x belongs-to b).
Does 1 belong to the union of a and b? That is do we have that 1
belongs-to (a union b)? With same a and b as above.
Let us see.
Does 1 belongs to a union b? Does 1 verify the condition written in
the definition? Do we have that (1 belongs-to a) OR (1 belongs to b))?
A proposition shaped like P OR Q is true in the case one or both of P
and Q is true. It is true that 1 belongs to {1, 2} OR to {2, 3}. A bit
like "any number is odd or is not odd" is always true. So 1 is in the
union.
2 is in the union, because it is true that 2 belongs to a or 2 belongs
to b. Indeed 2 belongs to both of them. And 3 is in the union too,
because iit belongs to one of them again, actually {2, 3}.
So (union b) = {1, 2, 3}.
OK?
Don't hesitate to tell me if it is not OK.
>
>>
>> 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?
>
> Is the missing subset { } ?
Correct.
So the subsets of {1, 2} are { }, {1}, {2}, {1, 2}.
Could you find all subsets of {1, 2, 3}?
And now I give you an exercise which is so much easy that you could
panic, and so I will provide the solution. I have seen often that too
much easy question can make a student panic, and then the prey of out-
of place mockery, and useless loss of confidence.
The easy exercise: could you give me the set of subsets of {1, 2} ?
Solution: You already told me that the subsets of {1, 2} are { }, {1},
{2}, {1, 2}. So, the set of subsets of {1, 2} is
{ { }, {1}, {2}, {1, 2} }
OK? It is just the solutions you give me, enclosed by braces
(accolades) "{", "}". Look at the expression with a spectacle.
If we except the set of books on Brent Meeker's shell, up to now, we
have met mainly set of numbers, like
{1}
{1, 2}
{0, 2, 4, 6, ...}
{0, 1, 2, 3, ...}
Here we met a set of sets.
The set of subsets of a set, can only be, of course, a set of sets.
The set {2, 21, 14} is a set of numbers. The set { { }, {4, 78, 56} }
is a set of sets. It has two elements: the empty set {}, and the set
of numbers {4, 78, 56}. Do not confuse a number, like 24, and a set,
like {24}, which is a set having a number has elements. In particular
it is the case that {4, 78, 56} belongs to { { }, {4, 78, 56} }. Take
it easy, and meditate on the following exercise:
Which of the following are true
{3, 5} included-in {3, 5}
{3, 5} belongs-to {3, 5}
{3, 5} included-in { {3, 5} }
{3, 5} belongs-to { {3, 5} }
Take your time,
Bruno
http://iridia.ulb.ac.be/~marchal/
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