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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.

Bob Parker
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3 Answers3

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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.

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

zwim
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  • Can you explain your answer for (a) please? – Bob Parker Oct 09 '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 09 '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 09 '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 09 '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 09 '17 at 19:58
  • Yes, you've got it. – zwim Oct 09 '17 at 20:04
  • Sorry, but how did you determine the Arg(z) and figure out the bijections for (c)? – Bob Parker Oct 09 '17 at 20:39
  • $\Re(z)<0,\Im(z)>0$ is the second quadrant, so $\theta\in]\frac{\pi}2,\pi[$. Multiplying by $i$ is a rotation of $\perp$ angle and $z^2$ on the unit circle double the angle $(e^{i\theta})^2=e^{i2\theta}$. Finally $2z$ takes care of the amplitude. Draw a picture of the two angular sectors, you'll understand how to convert one in the other. – zwim Oct 09 '17 at 21:07
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(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

M.ihsan
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