# Cardinality between union of open uncountable set with finite set

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?

• Yes, this is good. Apr 25 at 21:05
• 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. Apr 25 at 21:08