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
ah, but don’t forget to prove that the cardinality of [0,1] is that same as that of (0,1) on the way!
This is pretty trivial if you know that the cardinality of (0, 1) is the same as that of R ;)
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
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