Hi Marty,

On 03 Jul 2009, at 00:18, m.a. wrote: > Bruno, > Comments and questions are interspersed below. > > marty <snip> > 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 all. 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 represents. "{ } included in x" is always true, or never false, whatever x represents. > > 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. Bruno --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. 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