I considered deleting the post, but this seems more cowardly than just admitting I was wrong. But TIL something!
this post was submitted on 06 Jan 2024
274 points (86.1% liked)
memes
10392 readers
1923 users here now
Community rules
1. Be civil
No trolling, bigotry or other insulting / annoying behaviour
2. No politics
This is non-politics community. For political memes please go to !politicalmemes@lemmy.world
3. No recent reposts
Check for reposts when posting a meme, you can only repost after 1 month
4. No bots
No bots without the express approval of the mods or the admins
5. No Spam/Ads
No advertisements or spam. This is an instance rule and the only way to live.
Sister communities
- !tenforward@lemmy.world : Star Trek memes, chat and shitposts
- !lemmyshitpost@lemmy.world : Lemmy Shitposts, anything and everything goes.
- !linuxmemes@lemmy.world : Linux themed memes
- !comicstrips@lemmy.world : for those who love comic stories.
founded 1 year ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
There is a function which, for each real number, gives you a unique number between 0 and 1. For example,
1/(1+e^x). This shows that there are no more numbers between 0 and 1 than there are real numbers. The formalisation of this fact is contained in the Cantor-Schröder-Bernstein theorem.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