Do sets $A = \{0, 1, 2, 3\} \cup [\pi, 4)$ and $B = (-1, 0)\cup(1, 2)$ have same cardinality? [closed]

I guess I should try to make some bijective function that maps elements from A to B if I want to prove they have same cardinality, but I don't know how to do it. I know hot to prove that sets $\mathbb{Q}$ and $\mathbb{N}$ have same cardinality, but this seems different.

Solution 1:

We know that $|A\cup B|=\kappa+\lambda$ where $|A|=\kappa$, $|B|=\lambda$, and $A\cap B=\emptyset.$ Also, the following is correct $$\aleph_0+2^{\aleph_{0}}=2^{\aleph_{0}}$$ Indeed, $$2^{\aleph_{0}}\leq \aleph_0+2^{\aleph_{0}}\leq 2^{\aleph_{0}}+2^{\aleph_{0}}\leq 2\times 2^{\aleph_{0}}=2^{\aleph_{0}}$$ Now, Clearly, $$n+2^{\aleph_{0}}=2^{\aleph_{0}}$$ for every $n\in\mathbb N.$ So, since $|[\pi,4)|=2^{\aleph_{0}}$ so $|A|=2^{\aleph_{0}}$. Also, $|B|=2^{\aleph_{0}}.$ This finishes the proof.

Solution 2:

I will present a proof without cardinal arithmetic, only by building bijections (almost) explicitely.

First, it is not hard to see that :

$$A \simeq \{-1,-2,-3,-4,-5,-6,-7\} \cup (-1,0) \cup (1,2) $$ (where $\simeq$ means that two sets are in bijection)

Therefore, the only thing we have to show is that $\{-1\} \cup (-1,0) \simeq (-1,0)$ (then we can iterate this seven times to fit all the stray points into $(-1,0)$).

To do this, let us define : $$f \left\{ \begin{array}{ccc} \{-1\} \cup (-1,0) & \longrightarrow & (-1,0)\\ x &\longmapsto & \left\{\begin{array}{cl} x/2 & \text{if } \quad \exists n\in\mathbb N, x = -2^{-n}\\ x & \text{else}\end{array}\right. \end{array}\right.$$ You can check that this is a bijection, which ends the proof.