Yes, in that case S is {}. {} ∈ P(A) but is not in the image of e. | Hacker News

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		danharaj on March 11, 2017 \| parent \| context \| favorite \| on: Cantor diagonalisation Yes, in that case S is {}. {} ∈ P(A) but is not in the image of e.

danbruc on March 11, 2017 [–]

Thanks, now it clicked. I misinterpreted the image as the image of x instead of as the image of e. And then did something weired that now makes no longer any sense at all.

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