properties of a function∗
CWoo†
2013-03-21 21:23:30
Let X, Y be sets and f : X → Y be a function. For any A ⊆ X, define
f (A) := {f (x) ∈ Y | x ∈ A}
and any B ⊆ Y , define
f −1 (B) := {x ∈ X | f (x) ∈ B}.
So f (A) is a subset of Y and f −1 (B) is a subset of X.
Let A, A1 , A2 , Ai be arbitrary subsets of X and B, B1 , B2 , Bj be arbitrary
subsets of Y , where i belongs to the index set I and j to the index set J. We
have the following properties:
1. If A1 ⊂ A2 , then f (A1 ) ⊆ f (A2 ). In particular, f (A) ⊆ f (X).
S
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2. f (A1 ∪ A2 ) = f (A1 ) ∪ f (A2 ). More generally, f ( i Ai ) = i f (Ai ).
3. f (A1 ∩ A2 ) ⊆ f (A1 ) ∩ f (A2 ). The equality fails in the example where f is
a real function defined by f (x) = x2 and A1 = {1}, A2 = {−1}. Equality
occurs iff f is one-to-one:
Suppose f (x) = f (y) = z. Pick A1 = {x} and A2 = {y}. Then
f (A1 ∩ A2 ) = f (A1 ) ∩ f (A2 ) = {z} 6= ∅. This means that
A1 ∩ A2 6= ∅. Since both A1 and A2 are singletons, A1 = A2 ,
or x = y.
Conversely, let’s show that f is one-to-one then f (A1 ∩ A2 ) =
f (A1 ) ∩ f (A2 ). To do this, we only need to show the right
hand side is included in the left, and this follows since if x ∈
f (A1 ) ∩ f (A2 ) then for some a1 ∈ A1 and a2 ∈ A2 we have
x = f (a1 ) = f (a2 ). As f is one-to-one, a1 = a2 and so a1 lies
in A1 ∩ A2 and x is in f (A1 ∩ A2 ).
∗ hPropertiesOfAFunctioni created: h2013-03-21i by: hCWooi version: h38497i Privacy
setting: h1i hDefinitioni h03E20i
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T
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More generally, f ( i Ai ) ⊆ i f (Ai ).
4. f (A1 ) − f (A2 ) ⊆ f (A1 − A2 ): If y ∈ f (A1 ) − f (A2 ), then y = f (x) for
some x ∈ A1 . If x ∈ A2 , then y = f (x) ∈ f (A2 ) as well, a contradiction.
So x ∈ A1 − A2 , and y = f (x) ∈ f (A1 − A2 ). The inequality is strict in
the case when f : Z → Z given by f (x) = 1, and A1 = Z and A2 = {2}.
5. A ⊆ f −1 f (A). Again, one finds that equality fails for the real function
f (x) = x2 by selecting A = {1}. Equality again holds iff f is injective:
Suppose x ∈ f −1 f (A). By definition this means that f (x) =
f (a) for some x ∈ A, and since f is injective we have x = a ∈ A.
It follows that f −1 f (A) ⊆ A. Convserly, if f (x) = f (y) = z,
then {x, y} = f −1 f ({x, y}) = f −1 ({z}). On the other hand
{x} = f −1 f ({x}) = f −1 ({z}). So {x, y} = {x}, x = y.
6. If B1 ⊆ B2 , then f −1 (B1 ) ⊆ f −1 (B2 ). In particular, f −1 (B) ⊆ f −1 (Y ).
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7. f −1 (B1 ∪ B2 ) = f −1 (B1 ) ∪ f −1 (B2 ). More generally, f −1 ( j Bj ) =
S −1
(Bj ).
jf
T
8. f −1 (B1 ∩ B2 ) = f −1 (B1 ) ∩ f −1 (B2 ). More generally, f −1 ( j Bj ) =
T −1
(Bj ).
jf
9. f −1 (Y − B) = X − f −1 (B). As a result, f −1 (B1 − B2 ) = f −1 (B1 ) −
f −1 (B2 ).
10. f f −1 (B) ⊆ B. Yet again, one finds that equality fails for the real function
f (x) = x2 by selecting B = [−1, 1]. Equality holds iff f is surjective:
Suppose f is onto. Pick any y ∈ B ⊂ Y . Then y = f (x) for
some x ∈ X. In other words, x ∈ f −1 (B) and hence y = f (x) ∈
f f −1 (B). Now suppose the convserse, then pick B = Y , and
we have Y = f f −1 (Y ) = f (X).
11. Combining ?? and ??, we have that f f −1 f (A) = f (A) and f −1 f f −1 (B) =
f −1 (B). Let’s show the first equality:
From ??, A ⊆ f −1 f (A), so that f (A) ⊆ f f −1 f (A) (by 1). Set
B = f (A). Then by ??, f f −1 f (A) = f f −1 (B) ⊆ B = f (A).
Remarks.
• f −1 f and f f −1 mean the compositions of the function and its inverse as
defined at the beginning of the entry, so that f −1 f (A) = f −1 (f (A)) and
f f −1 (B) = f (f −1 (B)).
• From the definition above, we see that a function f : X → Y induces two
functions [f ] and [f −1 ] defined by
[f ] : 2X → 2Y such that [f ](A) := f (A) and
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[f −1 ] : 2Y → 2X such that [f −1 ](B) := f −1 (B).
The last property ?? says that [f ] and [f −1 ] are quasi-inverses of each
other.
• f is a bijection iff [f ] and [f −1 ] are inverses of one another.
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