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This question already has an answer here:

I need to prove the equivalence of sets [-1,4) ~(2,11) using:

a) Cantor-Bernstein Theorem

b) Constructing a bijection between those two sets.

What I have done:

a) I understand that in-order to prove this using Cantor-Bernstein you need to show injection from $f(x): [-1,4)\rightarrow (2,11) $ and $g(x): (2,11) \rightarrow [-1,4)$

I tried to find corresponding $x$'s so that $f(x)$ would always be in range of $(2,11)$ but I couldn't set up a proper ruleset.

b) Contructing a bijection:

If we have $A =$ {-1, ... 4} We would have to construct a $B = $ {2 +- $rule$ .... 11} The result would be a function $f(x) = $ { a set of rules}.

For example $2-x, x < 0$ and $2 \frac{3}4 * x , x > 0$.

Here also I couldn't understand how do I know to look for these rules, if anyone could provide any insight.

Best regards.

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marked as duplicate by Asaf Karagila elementary-set-theory Dec 12 '17 at 17:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Hint: For $(a)$, all you need to do is map $[-1,4)$ inside of $(2,11)$, so adding $4$ would be enough. Don't try to make the map surjective. $\endgroup$ – Michael Burr Dec 9 '17 at 14:11
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Bijectively map [-1,4) to (-4,1].
Bijectively map (-4,1] to (-4,1) by
. . 1/n to 1/(n+1), n in N and all other points to themselves.
Bijectively map (-4,1) to (2,11).
The composition of those maps is a bijection from [-1,4) to (2,11).

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