Re: cardinals of finite sets

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.

Re: cardinals of finite sets

Quote:

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.

Re: cardinals of finite sets

Quote:

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 $\displaystyle \|A\|$ I mean the cardinality of the set $\displaystyle A.$ .

Then $\displaystyle \|A\|=\|B\|\text{ if and only if }A\overset{f} \leftrightarrow B$, i.e. $\displaystyle f$ is a **bijection**.

To start suppose that there is an injection $\displaystyle \phi : A\to B$ then clearly $\displaystyle \phi(A)\subseteq B.$

Prove that $\displaystyle A\overset{\phi} \leftrightarrow \phi(A)$.

For the converse, assume that $\displaystyle (\exists C\subseteq B)[A\overset{\rho} \leftrightarrow C]$ where $\displaystyle \rho$ is a bijection.

What is your injection?

Re: cardinals of finite sets

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