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/sci/ - Science & Math

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Anonymous Sun Jan 13 08:37:03 2019 No.10292690 [Reply] [Original]

Does \mathbb{R}^2=\mathcal{P}(\mathbb{R}^2)

>>	Anonymous Sun Jan 13 08:41:15 2019 No.10292701 >>10292690 Apt image for this thread

>>	Anonymous Sun Jan 13 08:43:45 2019 No.10292707 >>10292690 Is the P for projective space or the set of subsets? No either way.

>>	Anonymous Sun Jan 13 08:45:30 2019 No.10292709 >>10292690 [math] \mathbb{R}^2=\mathcal{P}(\mathbb{R}^2) [/math] how about you tell us what P is you retarded nigger

>>	Anonymous Sun Jan 13 09:00:29 2019 No.10292745 >>10292707 >>10292709 power set/set of all subsets I'm kinda confused why R^2 != P(R^2)

>>	Anonymous Sun Jan 13 09:05:30 2019 No.10292759 >>10292745 wait nvm i understand P(R^2) = { empty set, ... the set of every single point on R^2 ... the set of all the combinations of every single point on R^2, and R^2} r-right?

>>	Anonymous Sun Jan 13 09:16:15 2019 No.10292788 File: 18 KB, 326x294, 1546906176924.jpg [View same] [iqdb] [saucenao] [google] >>10292690 >[math](0, 0) \in \mathbb{R}^2[/math] >[math]{(0, 0)} \in \mathcal{P}(\mathbb{R}^2)[/math] >what is set equality

>>	Anonymous Sun Jan 13 09:17:24 2019 No.10292790 File: 92 KB, 638x1000, 1547149430881.jpg [View same] [iqdb] [saucenao] [google] >>10292690 correction [math]\{(0, 0)\} \in \mathcal{P}(\mathbb{R}^2)[/math]

>>	Anonymous Sun Jan 13 09:19:57 2019 No.10292795 >>10292690 not equal, not in bijection