I am trying to show that $|(0,1)| = $$|(0,1) \cup ${$2,3,4,5$}$|$ using the Schroder-Bernstein Theorem.

Therefore, I need to find injections

  1. $f:(0,1) \rightarrow (0,1) \cup ${$2,3,4,5$} and

  2. $g:(0,1) \cup ${$2,3,4,5$} $\rightarrow (0,1)$

(1) We can define $f(x)=x$ since we do not have to consider 2,3,4,5.

(2) We can define $g(x)= \frac{x}{6}$ in order to direct 2,3,4,5 to a number $\in (0,1)$

Therefore, by the Schroder-Bernstein Theorem, $|(0,1)| = $$|(0,1) \cup ${$2,3,4,5$}$|$

Is this valid?

  • 1
    $\begingroup$ Yes, this is good. $\endgroup$ Apr 25 at 21:05
  • 1
    $\begingroup$ Alternatively, consider the numbers $a_n=\frac{1}{n}$, $n=2,3,\ldots,$, and define $g$ so that it sends $a_n$ to $a_{n+4}$, $2$ to $a_2$, $3$ to $a_3$, $4$ to $a_4$, and $5$ to $a_5$, and sends every element of $(0,1)$ other than the $a_n$s to themselves. This gives a bijection directly. $\endgroup$ Apr 25 at 21:08


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