Exercise 1.5.4

Question

\begin{enumerate}[label=(\alph*)] 
  \item Show $(a, b) \sim \mathbf{R}$ for any interval $(a, b)$.
  \item Show that an unbounded interval like $(a, \infty)=\{x: x>a\}$ has the same cardinality as $\mathbf{R}$ as well.
  \item Using open intervals makes it more convenient to produce the required 1-1, onto functions, but it is not really necessary. Show that $[0,1) \sim(0,1)$ by exhibiting a 1-1 onto function between the two sets.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We will start by finding $f : (-1, 1) 	o \mathbf{R}$ and then transform it to $(a, b)$. Example 1.5.4 gives a suitable $f$
    $$
    f(x) = \frac{x}{x^2 - 1}
    $$
    The book says to use calculus to show $f$ is bijective, first we will examine the derivative
    $$
    f'(x) = \frac{x^2 - 1 - 2x^2}{(x^2 - 1)^2} = - \frac{x^2 + 1}{(x^2 - 1)^2}
    $$
    The denominator and numerator are positive, so $f'(x) < 0$ for all $x \in (0,1)$.
    This means no two inputs will be mapped to the same output, meaning $f$ is one to one (a rigorous proof is beyond our current ability)

To show that $f$ is onto, we examine the limits
    $$
    \begin{aligned}
      \lim_{x 	o 1^{-}} \frac{x}{x^2 - 1} &= - \infty \
      \lim_{x 	o {-1}^{+}} \frac{x}{x^2 - 1} &= + \infty
    \end{aligned}
    $$
    Then use the intermediate value theorem to conclude $f$ is onto.

Now we shift $f$ to the interval $(a, b)$
    $$
    g(x) = f\left(\frac{2x - 1}{b - a} - aight)
    $$
    Proving $g(x)$ is also bijective is a straightforward application of the chain rule.
  \item We want a bijective $h(x)$ such that $h(x) : (a, \infty) 	o (-1, 1)$ because then we could compose them to get a new bijective function $f(h(x)) : (a, \infty) 	o \mathbf{R}$.

Let
    $$
    h(x) = \frac{2}{x - a + 1} - 1
    $$
    We have $h : (a, \infty) 	o (1, -1)$ since $h(a) = 1$ and $\lim_{x 	o \infty} h(x) = 1$.

Meaning that $f(h(x)) : (a, \infty) 	o \mathbf{R}$ is our bijective map.
  \item With countable sets adding a single element doesn't change cardinality since we can just shift by one to get a bijective map. we'll use a similar technique here to essentially outrun our problems. Define $f : [0,1) 	o (0,1)$ as
    $$
    f(x) = \begin{cases}
      1/2 &	ext{if } {x=0} \
      1/4 &	ext{if } {x=1/2} \
      1/8 &	ext{if } {x=1/4} \
      \vdots \
      x &	ext{otherwise}
    \end{cases}
    $$
    Now we prove $f$ is bijective by showing $y = f(x)$ has exactly one solution for all $y \in (0,1)$.

If $y = 1/2^n$ then the only solution is $y = f(1/2^{n-1})$ (or $x=0$ in the special case $n=1$),
    If $y 
e 1/2^n$ then the only solution is $y = f(y)$.
   \end{enumerate}

Exercise 1.5.4

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Exercise 1.5.4

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