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Use Schröder-Bernstein Theorem to show construct a bijection between $(0,1)$ and $(0,1]$

Let $f : (0,1) \rightarrow (0,1] $ be defined as $x \mapsto x$

and $g : (0,1] \to (0,1)$ be defined as $x \mapsto \frac{x}{2}$

The theorem states if we have two injective functions $f,g$ between sets then there exists a bijective function $h$ between them.

Define

$A^* \subset (0,1)$ to be $\{x \in (0,1) : x = \frac{p}{q} \,\&\, p = 1 \, \& \, q = 2^k, 1 \le k \} $ and

$ A' \subset (0,1) $ to be equal to $(0,1) \backslash A^*$

and

$B^* \subset (0,1]$ to be $\{x \in (0,1] : x = \frac{p}{q} \,\&\, p = 1 \, \& \, q = 2^j, 0 \le j \}$

$ B' \subset (0,1] $ to be equal to $(0,1] \backslash B^*$

Define $h : A \to B $ to be defined by $ h(x) \displaystyle \Bigg \{ \begin{array} x f(x) = x \,\, \text{if} \, x \in A' \\ g^{-1}(x) = 2x \, \text{if} \, x \in A^*\end{array} \\ $

So $A^* \cup A' = (0,1) \,\&\, B' \cup B^* = (0,1] \,\&\, A' \cap A^* = \emptyset \,\&\, B' \cap B^* = \emptyset $

and

$A^* \mapsto B^* \, \& \, A' \mapsto B'$

oliverjones
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    What's your question? – Rob Arthan Sep 26 '15 at 22:30
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    Parroting @RobArthan's concern. What are you asking? This second half of your post seems to be unrelated to the first half.. Are you trying to find what the bijection is? If so, that's not necessary.. – Cameron Williams Sep 26 '15 at 22:32
  • two sets have the same size if theres a bijection between them, I am supposed to use the above theorem to show this. the theorem says a bijection exists between two sets if there is an injective function for each set to the other. I tagged it solution-verification to have my solution verified. – oliverjones Sep 26 '15 at 22:36
  • changed the title so that its more clear – oliverjones Sep 26 '15 at 22:46

1 Answers1

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All that Schröder-Bernstein requires is injective functions in both directions. You have that in $f$ and $g$. All you needed to do after defining them is show that they are injective, which is trivial. There is no need to define the various sets $A$ and $B$ or the function $h$.

Paul Sinclair
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