2 ms·
> since there are 2^{\aleph_0} functions from \mathbb{N} to \mathbb{N} I think you mean N->Bool, not N->N
by knaekhoved 4y ago
> since there are 2^{\aleph_0} functions from \mathbb{N} to \mathbb{N}
I think you mean N->Bool, not N->N
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