1 20. What is the cardinality of each of these sets? c) {,{}} d) {,{},{,{}}} 2 L
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Question
1 20. What is the cardinality of each of these sets?
c) {,{}}
d) {,{},{,{}}}
2 Let A, B, and C be sets. Show that
d) (A C) (C B) = .
e) (B A) (C A) = (B C) A.
3 Find f g and g f , where f (x) = x2 + 1 and g(x) = x + 2, are functions from R to R.
4 Suppose L = {a, b, c, d} and a function f : L L is defined as follows:
f(a) = d, f(b) = b, f(c) = c, f(d) = d
Now, answer the following questions: [2 + 2 = 4 points]
a. Is f one-to-one? Explain your answer. [Do not just write Yes/No. Explanation required.]
b. Is f onto? Justify your answer. [Do not just write Yes/No. Explanation required.]
Explanation / Answer
Cardinality of the sets: Simply speaking cardinality of a set is the number of objects in the set. An object can be an individual element, or a set, or even a set of subsets. Please do keep in mind that is an empty set which is exactly equal to {}. Therefore the cardinality of is 0, where as the cardinality of {} is one. Why because this is equal to {{}} which means its a set of one element, where that element is an empty set. Therefore the cardinality of superset is 1, where as the cardinality of subset is 0.
1. c. {,{}} In this set, it is holding 2 elements, therefore the cardinality of this superset is 2. Please do keep in mind that, it is having 2 elements, where first element is an empty set, and second element is a set which inturn is holding an empty set.So, irrespective of internal contents, in abstract its having 2 elements.
d. {,{},{,{}}}. Speaking on the same lines above, this set is having a cardinality of 3. The outermost set is holding a collection of 3 elements:
1. An empty set.
2. A set which inturn holds an empty set.
3. A set which holds 2 elements, where the first element is an empty set, and the second element is a set which inturn holds an empty set.
Totally, its a collection of 3 objects/elements, therefore the cardinality of the outermost set is 3.
2. d. (A-C) is a set of elements that are left in A other than C.
(C-B) is a set of elements that are left in C other than B.
The common part between both these sets is nothing ().
e. (B-A) is a set of elements that are left in B other than A.
(C-A) is a set of elements that are left in C other than A.
A union of both these sets: (B-A)U(C-A) is the leftovers of B and C together that are not shared by A.
(BUC) is a set of elements that are part of B together with C, irrespective of A.
The part that is being shared by A in the above set is:
(BUC) - A, is the removal of the part from the above set that is being shared by A which will lead to the same
(B-A)U(C-A.)
3. f g where f(x) = x2+1. and g(x) = x+2.
g(x) = x+2.
f(g(x)) = f(x+2) = (x+2)2+1 = x2+4+4x+1 = x2+4x+5.
g f: g(f(x)) = g(x2+1) =x2+1+2 = x2+3.
4. f(a) = d, f(b) = b, f(c) = c, f(d) = d.
Is f one-to-one? No. f is not one-to-one. One-to-one function is a function where every element in the first set when applied to a function, should lead to a unique element in the other set. Here in this example when you applied the function f on a and d, both will lead to the same result, d. Therefore the function f is not said to be one-to-one. Exactly speaking f is a many-to-one function.
Is f onto? A function f is said to be onto, if every element in the second set has a corresponding element in the first set.
In the above example f is not onto, as both sets (first set and second set) are of elements L(a,b,c,d), and in the second set, the element a is not having the corresponding element in the first set. In other words, the element a in the second set can not be reached by applying the function f on any element in the first set.
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