# How can I find an injective function for infinite pairwise disjoint sets?

• mathboy
In summary, the function f is not one-to-one because there is no injective function that allows for x to belong to only one A_i.
mathboy
[SOLVED] Find a one-to-one function

Let A1, A2, A3... be infinite pairwise disjoint sets. Find an injective function

f : A1 U A2 U A3 U... -> A1 x A2 x A3 x...

For all i, I choose fixed elements a_i in each A_i. If x is in A_n, I tried

f(x) = (a1, a2,...,a_n-1, x, a_n+1,...).

But this doesn't work because, e.g., f(a1)=f(a2). So I added the extra rule

f(a_n) = (a1, a2,...,a_n-1, b_n, a_n+1,...),

where a_n not= b_n. But this still doesn't work because f(a_n)=f(b_n). Can someone help me find an injective function that works?

Or at least prove that such an injective function exists using the axiom of choice (without simply stating that the cardinality of the domain is smaller than the cardinality of the range, because using this fact is not allowed in my question, unless I prove it using the axiom of choice).

Last edited:
If the cardinality of domain is generally smaller than the cardinality of the range, which is true, there generally is no one-to-one function. Do you just mean an injective function?

Yes, injective function.

You're almost there. You just need to know where to map the a_i's. That the sets A_i are infinite will be useful here.

Use the axiom of choice to pick an element in each A_i. Call that 'a' in the cartesian product. If x is in the union then it belongs to exactly one A_i since they are pairwise disjoint. Define f(x) to be 'a' but only change the ith element.

Dick said:
Use the axiom of choice to pick an element in each A_i. Call that 'a' in the cartesian product. If x is in the union then it belongs to exactly one A_i since they are pairwise disjoint. Define f(x) to be 'a' but only change the ith element.

Isn't that what I did in the first post? But it runs into problems, e.g. f(a1)=f(a2).

How? 'a'=(a1,a2,a3,a4...). f(a1)=(b1,a2,a3,a4,...). f(a2)=(a1,b2,a3,a4...). ... ... ..., I don't know how many ...'s I need here. I'm leaning on the pairwise disjoint condition.

Ok, so for each i, using the axiom of choice, we use a choice function c: P(A_i) -> A_i and define (let's say x is in A_n),

f(x) = (a1, a2,...,a_n-1, c(A_n-{x}), a_n+1,...).

So the nth component has changed. Now suppose x and y are distinct points in A_n (the same A_n). How do we guarantee that f(x) not= f(y)? Perhaps change the domain of the choice function c to subsets of the form A_n-{x} and make the choice function injective somehow?

Last edited:
Right. Now you just just need an injective map from A_i->A_i-{a_i}. A_i is infinite. How do you define infinite? If your definition is that A_i can be placed into 1-1 correspondence with a proper subset, that should be pretty easy.

## What is a one-to-one function?

A one-to-one function is a type of mathematical function where each element of the domain is uniquely mapped to an element in the range. This means that for every input, there is only one output.

## How do I know if a function is one-to-one?

To determine if a function is one-to-one, you can use the horizontal line test. If a horizontal line intersects the graph of the function at more than one point, then the function is not one-to-one. If the line only intersects at one point, then the function is one-to-one.

## Why are one-to-one functions useful?

One-to-one functions are useful in many areas of mathematics and science because they allow us to easily find unique solutions to problems. They are also important in fields such as cryptography and data encryption.

## How can I find a one-to-one function?

To find a one-to-one function, you can start by choosing a simple function such as f(x) = x and then making modifications to ensure that each input has a unique output. For example, you can add or subtract a constant, multiply or divide by a non-zero number, or use trigonometric functions.

## What is the inverse of a one-to-one function?

The inverse of a one-to-one function is a function that "undoes" the original function. It swaps the roles of the domain and range, so that the inputs become the outputs and vice versa. In other words, if f(x) is a one-to-one function, then its inverse is f^-1(x).

• Calculus and Beyond Homework Help
Replies
1
Views
368
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
3K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
13
Views
1K
• Differential Equations
Replies
7
Views
462
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Engineering and Comp Sci Homework Help
Replies
7
Views
929
• Calculus and Beyond Homework Help
Replies
1
Views
953