STEP-BY-STEP LESSON · §7.1
Functions Defined on General Sets
Check the definition of a function, its domain, and its codomain.
Before you begin
The Language of Relations and Functions
A relation from A to B is a set of selected pairs in A×B. An object may relate to several objects or none.
R⊆A×B
Every x in A must have exactly one associated y in B. Different inputs may share an output.
f(x)=y
Set Theory: Definitions and the Element Method of Proof
The statement x∈A concerns one object. A⊆B says that every element of A belongs to B.
A⊆B ⇔ ∀x(x∈A ⇒ x∈B)
X=Y means X⊆Y and Y⊆X. One inclusion still allows extra elements in the second set.
X=Y ⇔ (X⊆Y ∧ Y⊆X)
Symbols
- f:X→Y
- function from X to Y
- f(x)
- output for input x
- f⁻¹(T)
- preimage set of T; not necessarily an inverse function
Definitions and notation for this topic
Function
f: A → B
Assigns exactly one element of B to each element of A.
Every allowed input must have exactly one output.
Separate glossary example
Let f:ℤ→ℤ be defined by f(x)=x+1. Then f(2)=3.
f(x)=x+1
Function evaluation
f(x)
For f:A→B and x∈A, f(x) is the unique output in B assigned to input x. It is an element of B, not generally a set.
The declared function determines the operation; f(x) does not mean f multiplied by x. Distinguish an input element x from an input subset S in f(S).
Separate glossary example
For f:ℤ→ℤ, f(x)=x², f(−3)=9; the image of the set {−3} is f({−3})={9}.
Image / range
f(S)={f(x):x∈S}For f:A→B and S⊆A, the image f(S) consists of outputs reached by inputs in S. The range of f is f(A).
Different inputs may produce the same output; the image is a set, so repeated outputs are listed once. This is different from the preimage f⁻¹(T).
Separate glossary example
For f:ℤ→ℤ, f(x)=x² and S={−2,2,3}, f(S)={4,9}.
For f:ℤ→ℤ with f(x)=2x, f(ℤ)=2ℤ.
Preimage of a set
f⁻¹(T)={x∈A:f(x)∈T}For any function f:A→B and T⊆B, f⁻¹(T) is the set of all inputs whose outputs lie in T. No bijectivity is required.
The result is a set and may be empty or have several elements. This use of f⁻¹ does not claim an inverse function exists.
Separate glossary example
For f:ℝ→ℝ, f(x)=x², f⁻¹({4})={−2,2} and f⁻¹({−1})=∅.
Domain / codomain
f:A→B
In f:A→B, A is the domain of allowed inputs and B is the declared codomain. The image is the set of outputs actually attained.
Changing the codomain can change whether a function is onto. A/B is not notation for these two roles.
Separate glossary example
For f:ℤ→ℤ with f(x)=2x, both domain and codomain are ℤ, but f(ℤ) is the set of even integers.
f:ℤ→ℤ, f(x)=2x
Step by step
Step 1 / 6
Specify inputs and allowed outputs
X is the set of allowed inputs; Y is the declared set of possible outputs. Both are part of the function specification.
f:X→Y
Without X, we cannot check that every input has an output.
Worked example
X={1,2}, Y={a,b,c}, f(1)=a, f(2)=a. Is this a function?
- Both inputs, 1 and 2, have assigned outputs.
- Each has exactly one output in Y; matching outputs are allowed. Thus it is a function.
- The range is {a}, while the codomain is {a,b,c}.
Optional self-check
For X={1,2}, only f(1)=a is specified. Is that sufficient?
Show answer
No: no output has been specified for input 2.