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submitted 2 years ago* (last edited 2 years ago) by balderdash9@lemmy.zip to c/memes@lemmy.world

I considered deleting the post, but this seems more cowardly than just admitting I was wrong. But TIL something!

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[-] lemmington_steele@lemmy.world 1 points 2 years ago

ah, but don't forget to prove that the cardinality of [0,1] is that same as that of (0,1) on the way!

[-] FishFace@lemmy.world 3 points 2 years ago

This is pretty trivial if you know that the cardinality of (0, 1) is the same as that of R ;)

[-] sukhmel@programming.dev 1 points 2 years ago

Isn't cardinality of [0, 1] = cardinality of {0, 1} + cardinality of (0, 1)? One part of the sum is finite thus doesn't contribute to the result

[-] lemmington_steele@lemmy.world 2 points 2 years ago* (last edited 2 years ago)

technically yes, but the proof would usually show that this works by constructing the bijection of [0,1] and (0,1) and then you'd say the cardinalities are the same by the Schröder-Berstein theorem, because the proof of the latter is likely not something you want to demonstrate every day

this post was submitted on 06 Jan 2024
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