Exercise 1.5.7

Question

Consider the open interval $(0,1)$, and let $S$ be the set of points in the open unit square; that is, $S=\{(x, y): 0<x, y<1\}$.
  \begin{enumerate}[label=(\alph*)] 
  \item Find a 1-1 function that maps $(0,1)$ into, but not necessarily onto, $S$. (This is easy.)
  \item Use the fact that every real number has a decimal expansion to produce a $1-1$ function that maps $S$ into $(0,1)$. Discuss whether the formulated function is onto. (Keep in mind that any terminating decimal expansion such as $.235$ represents the same real number as $.234999 \ldots .$)
   \end{enumerate}

The Schröder-Bernstein Theorem discussed in Exercise 1.5.11 can now be applied to conclude that $(0,1) \sim S$.

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We scale and shift up into the square. $f(x) = \frac 12 x + \frac 13$
  \item Let $g : S 	o (0, 1)$ be a function that interleaves decimals in the representation without trailing nines, padding with zeros if necessary. $g(0.32, 0.45) = 0.3425$, $g(0.1\bar 9, 0.2) = g(0.2, 0.2) = 0.22$, $g(0.1, 0.23) = 0.1203$, $g(0.1, 0.\bar 2) = 0.12\overline{02}$, etc.

Every real number can be written with two digit representations, one with trailing 9's and one without.
    However $g(x, y) = 0.d_1d_2\dots\bar9$ is impossible since it would imply $x = 0.d_1\dots\bar9$ and $y = 0.d_2\dots\bar9$ but the definition of $g$ forbids this.
    therefore $g(s)$ is unique, and so $g$ is 1-1.

Is $g$ onto? No since $g(x, y) = 0.1$ has no solutions, since we would want $x = 0.1$ and $y = 0$ but $0 
otin (0, 1)$.
   \end{enumerate}

Exercise 1.5.7

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Exercise 1.5.7

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