Question

_{Show that} (0,1)_{and R have the same cardinality. [Hint: Use the Schroder-Bernstein theorem].}

Short answer

$\left|\left(0,1\right)\left|=\right|R\right|$

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 from A to B.

Step 2:

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

PROOF:

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

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

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, we have then shown that f is a one-to-one function.

By definition (2):

data-custom-editor="chemistry" $\left|(0,1)\right|\left|\le \mathrm{R}\right|$

Step 3:

SECOND PART- Let us definite the function g as:

$g:R\to (0,1),g\left(n\right)=\left\{\begin{array}{l}\frac{1}{2}+2^{-n-1}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}\mathrm{n}\text{non-negative}\\ 2^{n-1}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}\mathrm{n}\text{negative}\end{array}\right.$

Note: Since the powers are always negative and since 2 and 3 are always positive, the fractions are always between 0 and 1 . Moreover, $g\left(n\right)<\frac{1}{2}$when n is negative and $\frac{1}{2}\le g\left(n\right)<1$when n is positive.

Check that g is one-to-one:

If $g\left(m\right)=g\left(n\right)$and m and n are both non-negative: $\frac{1}{2}+2^{n-1}=\frac{1}{2}+2^{-m-1}$implies $-n-1=-m-1$and this also then implies n = m.

If $g\left(m\right)=g\left(n\right)$and m and n are both negative: $2^{n-1}=2^{m-1}$ implies role="math" localid="1668428475343" $-n-1=-m-1$and this also then implies n = m .

If $g\left(m\right)=g\left(n\right)$and m is non-negative and n is negative: $2^{n-1}=\frac{1}{2}+2^{-m-1}$which is impossible as $2^{n-1}$is less than $\frac{1}{2}$and $\frac{1}{2}+2^{-m-1}$is greater than $\frac{1}{2}$.

If $g\left(m\right)=g\left(n\right)$and n is non-negative and m is negative: role="math" localid="1668428623684" $2^{m-1}=\frac{1}{2}+2^{-n-1}$which is impossible as $2^{n-1}$is less than $\frac{1}{2}$and $\frac{1}{2}+2^{-m-1}$is greater than $\frac{1}{2}$.

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

By definition (2):

role="math" localid="1668428704452" $\left|R\right|\le \left|\right(0,1\left)\right|$

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

$\left|\right(0,1\left)\right|=\left|R\right|$.