Build a bijection between $mathcal{P}(mathbb{Z}^+)$ and $mathcal{P} (mathbb{Z})$











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I tried building a function out of the cardinality of the two, but I'm not sure if you can do that. Is there a better way of approaching the problem?










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    I would try and establish a bijection between $mathbb{Z}^+$ and $mathbb{Z}$. This would induce a bijection between their power sets. Suppose $f:mathbb{Z}^+tomathbb{Z}$ is bijective. Then we define $f^*:mathcal{P}(mathbb{Z}^+)tomathcal{P}(mathbb{Z})$ by $$f^*({z_1,z_2,...})={f(z_1),f(z_2),...}.$$
    – Melody
    Nov 15 at 5:17















up vote
0
down vote

favorite
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I tried building a function out of the cardinality of the two, but I'm not sure if you can do that. Is there a better way of approaching the problem?










share|cite|improve this question


















  • 2




    I would try and establish a bijection between $mathbb{Z}^+$ and $mathbb{Z}$. This would induce a bijection between their power sets. Suppose $f:mathbb{Z}^+tomathbb{Z}$ is bijective. Then we define $f^*:mathcal{P}(mathbb{Z}^+)tomathcal{P}(mathbb{Z})$ by $$f^*({z_1,z_2,...})={f(z_1),f(z_2),...}.$$
    – Melody
    Nov 15 at 5:17













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up vote
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I tried building a function out of the cardinality of the two, but I'm not sure if you can do that. Is there a better way of approaching the problem?










share|cite|improve this question













I tried building a function out of the cardinality of the two, but I'm not sure if you can do that. Is there a better way of approaching the problem?







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asked Nov 15 at 5:13









Umuko

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  • 2




    I would try and establish a bijection between $mathbb{Z}^+$ and $mathbb{Z}$. This would induce a bijection between their power sets. Suppose $f:mathbb{Z}^+tomathbb{Z}$ is bijective. Then we define $f^*:mathcal{P}(mathbb{Z}^+)tomathcal{P}(mathbb{Z})$ by $$f^*({z_1,z_2,...})={f(z_1),f(z_2),...}.$$
    – Melody
    Nov 15 at 5:17














  • 2




    I would try and establish a bijection between $mathbb{Z}^+$ and $mathbb{Z}$. This would induce a bijection between their power sets. Suppose $f:mathbb{Z}^+tomathbb{Z}$ is bijective. Then we define $f^*:mathcal{P}(mathbb{Z}^+)tomathcal{P}(mathbb{Z})$ by $$f^*({z_1,z_2,...})={f(z_1),f(z_2),...}.$$
    – Melody
    Nov 15 at 5:17








2




2




I would try and establish a bijection between $mathbb{Z}^+$ and $mathbb{Z}$. This would induce a bijection between their power sets. Suppose $f:mathbb{Z}^+tomathbb{Z}$ is bijective. Then we define $f^*:mathcal{P}(mathbb{Z}^+)tomathcal{P}(mathbb{Z})$ by $$f^*({z_1,z_2,...})={f(z_1),f(z_2),...}.$$
– Melody
Nov 15 at 5:17




I would try and establish a bijection between $mathbb{Z}^+$ and $mathbb{Z}$. This would induce a bijection between their power sets. Suppose $f:mathbb{Z}^+tomathbb{Z}$ is bijective. Then we define $f^*:mathcal{P}(mathbb{Z}^+)tomathcal{P}(mathbb{Z})$ by $$f^*({z_1,z_2,...})={f(z_1),f(z_2),...}.$$
– Melody
Nov 15 at 5:17















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