Thread: cardinals of finite sets

Hello,


I have the following question:
Prove that a set A has the same cardinality of a subset of a Set B, if and only if exists an injective function A to B.


I find it hard to prove it because I can easily find a set A which is:


A={1,2}
B={1,2,3,4}
C={1,2,3}


C is a subset of B. C's cardinality is bigger than A's.


There is an injective function from A to B, of course.


where am I wrong?

Thank you!

A and B have the same cardinality if there is a 1-1 mapping of A onto B (bijective).

If there is a 1-1 mapping of A onto a subset of B (the same cardinality), there exists a 1-1 mapping into B (injective).
If there is an injective mapping into a subset of B, The mapping is onto that subset from A, and A and that subset have the same cardinaity.

Originally Posted by nerazzurri10

Prove that a set A has the same cardinality of a subset of a Set B, if and only if exists an injective function A to B.


I find it hard to prove it because I can easily find a set A which is:


A={1,2}
B={1,2,3,4}
C={1,2,3}


C is a subset of B. C's cardinality is bigger than A's.


There is an injective function from A to B, of course.


where am I wrong?

If there is an injective function from A to B, it means that A has the same cardinality as some subset of B, not every subset of B. So, C may have a greater cardinality than A's, but there is another subset of B that has the same cardinality as A's.

Originally Posted by nerazzurri10

I have the following question:
Prove that a set A has the same cardinality of a subset of a Set B, if and only if exists an injective function A to B.

Let's first agree on notation and definitions. By I mean the cardinality of the set .
Then , i.e. is a bijection.

To start suppose that there is an injection then clearly
Prove that .

For the converse, assume that where is a bijection.
What is your injection?

Nice example of using symbology to reword an existing proof (#2)