Question

Use the Schroder-Bernstein theorem to show that $\left(0,1\right)$and (0,1)have the same cardinality.

Short answer

Required answer is $\left|\left(0,1\right)\right|=\left|\left[0,1\right]\right|$.

Step-by-step solution

Step 1:

DEFINITION (1):

Shroder - Bernstein theorem:

If$\left|A\right|\le \left|B\right|and\left|B\right|\le \left|A\right|,then\left|A\right|=\left|B\right|$.

DEFINITION (2):

$\left|A\left|\le \right|B\right|$if and only if there is a one-to-one function A to B.

Step 2:

To proof:$\left|\left(0,1\right)\right|=\left|\left[0,1\right]\right|$

PROOF:

FIRST PART- Let us definite the function f as (since all elements of are also elements of [0,1]):

$f:(0,1)\to [0,1],f\left(n\right)=n$

Check that f is one-to-one: if $f\left(m\right)=f\left(n\right)$, then by definition of f

m = n

By the definition of one-to-one, then shown that f is a one-to-one function.

By definition (2):

$\left|\left[0,1\right]\right|\le \left|\left[0,1\right]\right|$

Step 3:

SECOND PART- Let us definite the function g as:

$g:[0,1]\to (0,1),g\left(n\right)=\frac{n+1}{3}$

Note: $\frac{n+1}{3}\in (0,1)whenn\in (-1,2)$.

Since data-custom-editor="chemistry" $[0,1]\subset (-1,2),\frac{\mathrm{n}+1}{3}\in (0,1)$for all $n\in \left[0,1\right]$.

Check that g is one-to-one: If g(m) = g(n) , then by definition of g

$\frac{m+1}{3}=\frac{n+1}{3}$

Multiply each side of the equation by :

m + 1 = n + 1

Subtract from each side of the equation:

m = n

By the definition of one-to-one, we have then shown that g is a one-to-one function.

$\left|\right[0,1\left]\right|\le \left|\right(0,1\left)\right|$

By definition (2):

CONCLUSION- Since $\left|\right[0,1\left]\right|\le \left|\right(0,1\left)\right|and\left|\right(0,1\left)\right|\le \left|\right[0,1\left]\right|$, Schroder-Bernstein theorem tells us then:

$\left|\left(0,1\right)\right|=\left|\left[0,1\right]\right|$