i dont understand what does this means:
If A and B are any two non empty sets, then a function f from A to B is a subset of A x B, with two important restrictions:
(i) ∀a ∈ A, (a,b) ∈ f for some b ∈ B.
(ii) If (a,b) ∈ f and (a,c) ∈ f then b=c.
please help me,,,,,
If A and B are any two non empty sets, then a function f from A to B is a subset of A x B, with two important restrictions:
(i) ∀a ∈ A, (a,b) ∈ f for some b ∈ B.
(ii) If (a,b) ∈ f and (a,c) ∈ f then b=c.
please help me,,,,,
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"If A and B are non-empty sets"
Pretty straightforward. A and B must contain at least one element.
"a subset of A x B"
A x B is the cartesian product of A and B. That is, all ordered pairs (a, b) so that a is in A and b is in B. For example:
{1, 2, 3} x {x, y} = {(1, x), (2, x), (3, x), (1, y), (2, y), (3, y)}
A subset of A x B is exactly that: a subset of this product set. A subset of A x B contains some (not necessarily all) pairs of the form (a, b).
"∀a ∈ A, (a,b) ∈ f for some b ∈ B"
Translates to: "For each a in A, there is at least one b in B so that (a, b) is in f". Take, for example:
A = {1, 2, 3}
B = {x, y}
f = {(1, x), (1, y), (2, x)}
If we consider a = 2, then there is some b = x such that (a, b) = (2, x) is in f. If we consider a = 1, there is at least one b so that (a, b) is in f. In fact, both (1, x) and (1, y) are in f, so b could be x or y. However, if a = 3, then there is no such b so that (a, b) is in f. So, for this set, condition (i) is not upheld, and so it is not a function.
"If (a,b) ∈ f and (a,c) ∈ f then b=c."
This says that (a, b) and (a, c) are both in f, then we must have b = c. This means that, if (a, b) is in f, it is the *only* ordered pair in f starting with a. Note, this does not guarantee that there is a pair in f beginning with a given a, just that if there is one, there must be unique. Take the previous example:
A = {1, 2, 3}
B = {x, y}
f = {(1, x), (1, y), (2, x)}
Does this satisfy (ii)? If we took any two pairs with the same first entry, would the second entries be necessarily equal? That is, for any given first coordinate of a pair in f, is there exactly one pair with that first coordinate? No, since (1, x) and (1, y) are in f, but x does not equal y. So, this set does not satisfy (ii) either. If we had instead:
f = {(1, y), (2, x)}
then we would have (ii) satisfied, but not (i), since there is no b such that (3, b) is in f. If we add a different pair:
f = {(1, y), (2, x), (3, x)}
then we have a function. For each a, a pair beginning with a exists in f (so it satisfies (i)), and that pair is unique (so it satisfies (ii)). Basically (i) and (ii) can be simplified into the following statement:
"If a is in A, then there exists a *unique* (a, b) in f"
This allows us to use function notation. We say that f(a) = b, if (a, b) is in f. Condition (i) means that f(a) is defined to be at least one number for anything in the domain A. Condition (ii) means that f(a) cannot be defined to be multiple values for the same a.
Hope that helps!
Pretty straightforward. A and B must contain at least one element.
"a subset of A x B"
A x B is the cartesian product of A and B. That is, all ordered pairs (a, b) so that a is in A and b is in B. For example:
{1, 2, 3} x {x, y} = {(1, x), (2, x), (3, x), (1, y), (2, y), (3, y)}
A subset of A x B is exactly that: a subset of this product set. A subset of A x B contains some (not necessarily all) pairs of the form (a, b).
"∀a ∈ A, (a,b) ∈ f for some b ∈ B"
Translates to: "For each a in A, there is at least one b in B so that (a, b) is in f". Take, for example:
A = {1, 2, 3}
B = {x, y}
f = {(1, x), (1, y), (2, x)}
If we consider a = 2, then there is some b = x such that (a, b) = (2, x) is in f. If we consider a = 1, there is at least one b so that (a, b) is in f. In fact, both (1, x) and (1, y) are in f, so b could be x or y. However, if a = 3, then there is no such b so that (a, b) is in f. So, for this set, condition (i) is not upheld, and so it is not a function.
"If (a,b) ∈ f and (a,c) ∈ f then b=c."
This says that (a, b) and (a, c) are both in f, then we must have b = c. This means that, if (a, b) is in f, it is the *only* ordered pair in f starting with a. Note, this does not guarantee that there is a pair in f beginning with a given a, just that if there is one, there must be unique. Take the previous example:
A = {1, 2, 3}
B = {x, y}
f = {(1, x), (1, y), (2, x)}
Does this satisfy (ii)? If we took any two pairs with the same first entry, would the second entries be necessarily equal? That is, for any given first coordinate of a pair in f, is there exactly one pair with that first coordinate? No, since (1, x) and (1, y) are in f, but x does not equal y. So, this set does not satisfy (ii) either. If we had instead:
f = {(1, y), (2, x)}
then we would have (ii) satisfied, but not (i), since there is no b such that (3, b) is in f. If we add a different pair:
f = {(1, y), (2, x), (3, x)}
then we have a function. For each a, a pair beginning with a exists in f (so it satisfies (i)), and that pair is unique (so it satisfies (ii)). Basically (i) and (ii) can be simplified into the following statement:
"If a is in A, then there exists a *unique* (a, b) in f"
This allows us to use function notation. We say that f(a) = b, if (a, b) is in f. Condition (i) means that f(a) is defined to be at least one number for anything in the domain A. Condition (ii) means that f(a) cannot be defined to be multiple values for the same a.
Hope that helps!