# Equal sets and bijective correspondence

1. Dec 7, 2011

### mizunoami

1. The problem statement, all variables and given/known data

If [n] and [m] are equal, then they are bijective correspondent.

I define $f \subset\{(n,m)\mid n \in [n], m\in [m]\}$. Suppose [n]=[m]. Let$(n,m_1),(n,m_2)\in f.$ Because [n]=[m], then $m_1=m_2$. So for all $n \in [n]$, there exists a unique $m\in [m]$ such that f(n)=m. So f is a function.

Next I want to prove f is surjective and injective. But I'm stuck. How can I make use of the supposition [n]=[m] to prove surjectivity and injectivity?

Thanks.

2. Dec 8, 2011

### HallsofIvy

Staff Emeritus
Do you mean "equal" or "equivalent"? In order to be equal, sets A and B must be identical- they contain exactly the same elements. In order to be equivalent, they must have the same cardinality- contain exactly the same number of elements. In either case, there is an obvious bijection.

3. Dec 8, 2011

### mizunoami

I mean equal - so both sets contain the same elements.

Yeah, there is obviously a bijection. But how can I show that using a rigorous mathematical proof?

4. Dec 8, 2011

### HallsofIvy

Staff Emeritus
The obvious way- show that f, from [m] to [n], defined by f(x)= x, is both "one-to-one" and "onto". And that should be easy- if f(a)= f(b), what must be true of a and b? Let c be a member of [n]. What member of [m] is mapped to c?