Let , let , let . Then,
I. The main properties of the first set mapping are:
1a: obvious by definition of f(A) or f.
1b: obvious by definition of f
1c: Let so that for some one i we have . That is, . Reversing the arguments, we get the other subset relation. So that . The image of the union is the union of images.
1d: Let so that belongs to each . So that LHS is clearly a subset of .
II) Now, we want to verify the following useful/famous relations of the second set mapping:
Answer 2a: obvious from definition.
Answer 2b: obvious from definition.
Answer 2c: Let belong to at least one so that for some i. In other words,
Hence, , where U is the universal set
Hence, . Hence,
Reversing the arguments proves the other subset inequality.
Hence, . QED.
More Problems :
A) Two mappings and are said to be equal (and we write them as f=g) if for every x in X. Let f, g, and h be any three mappings of a non-empty set X into itself, and now prove that the multiplication of mappings is associative in the sense that .
Let , , and be any three mappings of X into itself. Let and let . Now, we know that . So that now we want to find assuming .
Now, on the other hand, means just as in LHS. QED.
B) Let X be a non-empty set. The identity mapping on X is the mapping of X onto itlsef defined by for every x. Then, sends each element of the set X to itself; that is, leaves fixed each element of X. Prove that for any mapping f of X into itself. If f is one-to-one onto, so that its inverse exists, show that . Prove further that is the only mapping of X into itself which has this property; that is, show that if g is a mapping of X into itself such that , then . Hint: , or
B(i) TPT: for any mapping f of X into itself. Proof: Consider and such that . Hence, . Now, on the other also. QED.
B(ii): Let f be one-to-one onto such that . Hence, exists. Also, . Let . Hence, by definition and because is also one-to-one and onto, . Hence, clearly, . QED.
B(iii) : Prove that such a is unique.
Solution to B(iii): Let there be another function g such that . Also, we have . Use the hint ! 🙂 QED.
Let X and Y be non-empty sets and f a mapping of X into Y. Prove the following: (i) f is one-to-one if and only if there exists a mapping g of Y into X such that (ii) f is onto if and only if there exists a mapping h of Y into X such that .
Solution C(i) : Given such that if , then . Now, we want a mapping such that . Let us construct a g such that and . Then, . Similarly, . Here, we have assumed that and . So, knowing f to be one-to-one, we can explicitly construct a mapping g such that and . QED. On the other hand, let it be given that X and Y are two non-empty sets, that and such that . That is, if some , then we have . That is, . This forces, if , then . Now, if there be an element , it will force . Hence, different inputs give rise to different outputs and so f is one-to-one. QED.
Solution C(ii) : Given and f is onto, that is, . We want to show a function such that . Since , with X as domain, Y as codomain equal to range, f is one-to-one, onto. So . QED.
Let X be a non-empty set and f a mapping of X into itself. Show f is one-to-one onto if and only there exists a mapping g of X into itself such that . If there exists a mapping g with the property, then there is one and only one such mapping. Why?
Solution D: From previous question, C we are done.
Let X be a non-empty set, and let f and g be one-to-one mappings of X onto itself. Show that fg is also a one-to-one mapping of X onto itself and that .
HW. Hint: consider the action of the functions on each element.
Let X and Y be non-empty sets and f a mapping of X into Y. If A and B are, respectively, subsets of X and Y, prove the following:
(a) , and is true for all B if and only if f is onto.
(b) , and is true for all A if and only if f is one-to-one.
(c) is true for all and if and only if f is one-to-one.
(d) is true for all A if and only if f is onto.
(e) If f is onto — so that, is true for all A — then, is true for all A if and only if f is also one-to-one.
Solution Fa: Put , so that f is onto. To prove the other subset relationship, simply reverse the arguments.
We need to apply the definition that different inputs give rise to different outputs.
😦 I hope to blog later)
It certainly implies that .
(I hope to blog later).
Given that and and . Then, is , that is, , that is , that is, . Put , then f is onto. Now, , that is, . This implies . To prove the other subset relationship, simply reverse the arguments. QED.