# Proving One-to-One Function f: X -> Y

• mathboy
In summary, the conversation is about proving that a function is one-to-one given the condition that f(X-A) = Y-f(A) for any subset A of the domain X. The participants discuss different approaches and suggest trying to prove the contrapositive or assuming that a \neq b by contradiction.
mathboy
[SOLVED] One-to-one function

Let f: X -> Y. Suppose f(X-A) = Y-f(A) for any A in X. Prove that f is one-to-one.

Suppose f(a)=f(b). I tried letting A = {a,b}, A= X-{a,b}, but I can't manage to prove that a=b. Can someone help?

mathboy said:
Let f: X -> Y. Suppose f(X-A) = Y-f(A) for any A in X. Prove that f is one-to-one.

Suppose f(a)=f(b). I tried letting A = {a,b}, A= X-{a,b}, but I can't manage to prove that a=b. Can someone help?

You need to clean up your notation. What are $X$ and $Y$? Sets?

X is the domain of f and Y is the range of f, so they are sets. A is any subset of X. So the question is asking to prove that if f(complement of A)= complement of f(A), then f is one-to-one.

Let a be in X and set A=X-{a}. What's f(X-A) now?

You might want to try proving the contrapositive: if f is not one to one, then there exist a set A such that f(complement of A) is not complement of f(A).

If f is not one to one, then there exist x1 and x2 such that f(x1)= f(x2). What if A= {x1}?

mathboy said:
Let f: X -> Y. Suppose f(X-A) = Y-f(A) for any A in X. Prove that f is one-to-one.

Suppose f(a)=f(b). I tried letting A = {a,b}, A= X-{a,b}, but I can't manage to prove that a=b. Can someone help?
In situations like this, you can always try assuming that $a \neq b$ by contradiction. The proof might not require it, but you can sometimes find direction that way.

## 1. What does it mean for a function to be one-to-one?

A one-to-one function, also known as an injective function, is a function in which each element of the domain (X) is mapped to a unique element in the codomain (Y). This means that no two elements in the domain can have the same output in the codomain.

## 2. How can we prove that a function is one-to-one?

To prove that a function f: X -> Y is one-to-one, we can use the horizontal line test. This test involves drawing a horizontal line on the graph of the function. If the line intersects the graph at more than one point, then the function is not one-to-one. If the line only intersects the graph at one point, then the function is one-to-one.

## 3. Can a function be both one-to-one and onto?

Yes, a function can be both one-to-one and onto. A function that is both one-to-one and onto is called a bijection. This means that every element in the codomain (Y) is mapped to by exactly one element in the domain (X), and every element in the codomain has a preimage in the domain.

## 4. What is the difference between a one-to-one function and a one-to-one correspondence?

A one-to-one function is a function in which each element in the domain is mapped to a unique element in the codomain. A one-to-one correspondence, also known as a bijection, is a function that is both one-to-one and onto. In other words, a one-to-one correspondence is a function in which each element in the domain is mapped to a unique element in the codomain and each element in the codomain has a preimage in the domain.

## 5. Can a one-to-one function have the same output for different inputs?

No, a one-to-one function cannot have the same output for different inputs. This is because in a one-to-one function, each input has a unique output. If two different inputs had the same output, then the function would not be one-to-one.

• Calculus and Beyond Homework Help
Replies
11
Views
336
• Calculus and Beyond Homework Help
Replies
5
Views
420
• Calculus and Beyond Homework Help
Replies
8
Views
290
• Calculus and Beyond Homework Help
Replies
3
Views
226
• Calculus and Beyond Homework Help
Replies
1
Views
595
• Calculus and Beyond Homework Help
Replies
20
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
672
• Calculus and Beyond Homework Help
Replies
2
Views
531
• Calculus and Beyond Homework Help
Replies
3
Views
433
• Calculus and Beyond Homework Help
Replies
3
Views
799