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(a) Integers divisible by 3, and the even positive integers.

(b) $\Bbb {R}$, and the interval $(0,\infty)$.

(c) The interval $[0, 2)$, and the set [5,6)∪[7,8).

(d) The intervals $(-\infty, -1)$ and $(-1, 0)$.

I know that cardinality means that there is a bijection between the two sets, and that means there is a surjection and injection. For the first one I think you can simply make a function that is a bijection and prove that it is an injection and surjection. I am not sure how to do the last three.

Sorry for the poor formatting in advance and if you need more information about the question I am more than happy to provide it.

Thank you.

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  • $\begingroup$ ok thank you, fixed it $\endgroup$ – smith Nov 27 '18 at 2:47
  • $\begingroup$ Do you mean the set $[5,6)\cup [7,8)$? $\endgroup$ – TonyK Nov 27 '18 at 2:50
  • $\begingroup$ yes, it has been fixed $\endgroup$ – smith Nov 27 '18 at 2:51
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Hints:

(b) Map $x \in \mathbb{R}$ by $x \to e^x$

(c) Map $x \in [0, 2)$ by $x \to x + 5$ if $x \in [0,1)$ and $x \to x + 6$ if $x \in [1,2)$

(d) Map $x \in (-\infty,-1)$ by $x \to \frac{1}{x}$

Show that these are all bijections.

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(b) $\mathbb{R}$, and the interval (0,∞).

Consider $f : \mathbb{R}\rightarrow (0,\infty)$ defined by $f(x)=e^x.$

(c) The interval [0,2), and the set [5,6) or [7,8).

Consider $g: [0,2)\rightarrow [5,6)$ defined by $g(x)=5+\frac{x}{2}$ or $g'(x)=7+\frac{x}{2}$ if codomain is $[7,8)$.

(d) The intervals (−∞,−1) and (−1,0).

Consider $h: (-1,0)\rightarrow (-\infty,-1)$ defined by $h(x)=\tan(\frac{\pi}{2}x)-1.$

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a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:A\to \mathbb{Z}$ defined with $$n\longmapsto {n\over 3}$$

and then $g:\mathbb{Z}\to C$ defined by:

$$ g(n) \begin{cases} = 4n+2 & \text{if } n\geq 0, \\ = 4|n| & \text{if } n < 0. \\ \end{cases} $$

Then let $g\circ f$ will do the job.

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The bijections, reading left to right, can be chosen as:

(a) $|4x/3|+2[x\ge 0]$ (b) $x+\sqrt{1+x^2}$ (c) $x+5+\lfloor x\rfloor$ (d) $1/x$.

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