# Show that the following pairs of sets have the same cardinality by giving explicit bijections between them

Show that the following pairs of sets have the same cardinality by giving explicit bijections between them:

(a) $M_{2×2}(\mathbb C)$ and $\mathbb R$8

(b) $(0,1)$ and $(−1,+\infty)$

(c) The sets $\{z\in\mathbb C \mid 0<|z|<1,\ 0<\arg(z)<\frac{\pi}4\}\quad\text{ and }\\\{z\in\mathbb C \mid 0<|z|<2,\ \Re(z)<0,\ \Im(z)>0\}$

How would I find the bijections between these sets that have to same cardinality? Thanks.

• for b you can use something like an $\tan$ – user392395 Oct 9 '17 at 19:13

(a) $M_{2x2}(\mathbb C)\equiv \mathbb C^4\equiv\mathbb R^8$ and $\mathbb R^n$ equipotent to $\mathbb R$ but explicit bijection is not straightforward.

(b) $(0,1)\overset{\frac 1x}{\longmapsto}(1,+\infty)\overset{x-2}{\longmapsto}(-1,+\infty)$

(c) $\{0<|z|<1,0<\theta<\frac{\pi}4\}\overset{z^2}{\longmapsto}\{0<|z|<1,0<\theta<\frac{\pi}2\}\overset{iz}{\longmapsto}\{0<|z|<1,\frac{\pi}2<\theta<\pi\}\overset{2z}{\longmapsto}\{0<|z|<2,\Re(z)<0,\Im(z)>0\}$

• Can you explain your answer for (a) please? – Bob Parker Oct 9 '17 at 19:31
• @BobParker A complex $2\times 2$ matrix is basically $4$ complex numbers. And a complex number is basically $2$ reals. So a complex $2\times 2$ matrix can be represented by $8$ reals coefficients. – zwim Oct 9 '17 at 19:34
• Okay that makes sense. But in order to show an explicit bijection, that explanation wouldn't suffice, right? – Bob Parker Oct 9 '17 at 19:39
• just show that $(a,b,c,d,e,f,g,h)\mapsto\begin{bmatrix}a+ib & c+id \\ e+if & g+ih\end{bmatrix}$ is a bijection, should not be too hard. i.e. given $z=a+ib$ then $a=\frac{z+\bar z}2$ and $b=\frac{z-\bar z}2$ – zwim Oct 9 '17 at 19:40
• Thanks so much for the help. So there are essentially two primary bijections that apply to all 8 real numbers? – Bob Parker Oct 9 '17 at 19:58

For $(a)$, I reckon what's fundamental is that you find a bijection between $\mathbb{R}$ and $\mathbb{R}^2$. Hint: try to use digits of the decimal expansion!

For $(b)$, you can try to modify functions like $\tan(x)$ or $\frac1x$.

For $(c)$, try to draw the sets. The bijection should be pretty obvious once you've pictured them.

(B) set (0'1) have cardinality C and Also (1, infinite) cardinality C Using result same cardinality of the set equivalent Impiles (0'1)~(1'infinite) Implis that is bijective Define map: F(x)=1/x