To show equal cardinality

Question

To show two sets have equal cardinality is it always necessary to show bijection between the sets?

suppose I have two sets A and B isn't it enough to show that there exist one onto mapping from A to B and some another onto mapping form B to A. The reason why I am asking this question is that it sometimes becomes very difficult to construct such a function which gives bijection.

For example, when I wanted to show bijection between rationals and naturals. It was difficult to construct such a function and later found out that
$f(m,n)=2^{m-1}(2n-1)$

will do the work.

Now,recently I came across another problem that show that [0,1] and (0,1) have equal cardinality. I was able to construct onto mapping from (0,1) to [0,1] and also another onto mapping from [0,1] to (0,1) but still now I couldn't construct such a function which will biject the two sets.

Answer 1

It isn't necessary to construct an explicit bijection if you have a result at hand you can apply. For instance, the Schroeder-Bernstein Theorem states that if there are injections $f : A \to B$ and $g : B \to A$, then there exists a bijection $h : A \to B$.

In your last example, it isn't hard to find injections $f : [0,1] \to (0,1)$ and $g : (0,1) \to [0,1]$, so there must exist a bijection between the sets, although we didn't demonstrate a specific example of one.

Answer 2

To biject $[0,1]$ to $(0,1)$ let $A$ be the set of numbers $\frac12 \pm \frac{1}{2n}$ for $n=1,2,...$ Then for each $x$ in A map it to the next (in the sense of closer to $1/2$) element of $A,$ and on the complement map $x$ to itself. For the inverse map from $(0,1)$ back to $[0,1]$ use the set $B=A - \{0,1\}$ and this time map the elements of $B$ "away" from $1/2$ to the next element, except that e.g $1/4$ would map back to $0,$ since in the forward map $0=1/2-1/(2*1)$ went to the right to the point $1/2-1/(2*2)=1/4.$ Similarly $3/4$ maps to $1.$ under the inverse map.

Answer 3

Yes. To show that two sets have the same cardinality you have to prove there exists a bijection between the two sets.

One way is to find an explicit, preferably human readable, formula that defines such bijection.

Another way is to break the equicardinality into steps, and in each intermediate finding a simple bijection between the relevant sets; then use the fact that the composition of bijections is a bijection (for example when you prove that $\Bbb{|R|=|R^N|}$ by passing through $2^\Bbb N$ and $2^{\Bbb{N\times N}}$).

Another way is to use the Cantor-Bernstein theorem which tells you that if there are injections $f\colon A\to B$ and $g\colon B\to A$, then there is a bijection $h\colon A\to B$. This is very useful since more often than not it is easier to come up with injections, than with bijections. (You can use the version that replaces injections by surjections, but in the general case there is an essential use of the axiom of choice here; that being said, in some cases surjections do suffice, e.g. when one of the sets involved is countable.)

Finally, you can use cardinal arithmetic. Cardinal arithmetic is the abstraction of all the previous methods, and it often assumes that you're sufficiently familiar with them to write the details if necessary. For example, $\Bbb {|N|\leq|Z|\leq|N\times N|\leq|N|}$ to establish that $\Bbb Z$ is countable, this implicitly assumes that you're able to generate injections witnessing each $\leq$, and with Cantor-Bernstein's theorem that gives us the wanted equality.

Answer 4

Existence of onto map from $A$ to $B$ gives card$(A)\leq$ card$(B)$ and existence of onto map from $B$ to $A$ gives card$(B)\leq$ card$(A)$ . In either way we can say that card$(A)=$ card$(B)$ i.e. there must be a bijection between $A$ and $B.$ But generally we have to show that two sets are of equal cardinality or not, we are not intersting in bijective map and thats not so simple to find such a bijection.