#correct me proof of uncountability of real numbers
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finite vortex
so, correct me if i'm wrong, you kind of permute the natural numbers with this beta function then say that there cannot be a number between a(b(i)) and a(b(i+1))?
if that's your proof, i don't think it's valid
beta may not exist
for example N -> Q bijective exists
and for any 2 rationals the rational between them is also found
and the reason that can be is because beta does not exist in the case of bijecting N -> Q
i think
deft bloom
why would beta not exist the only condition for beta's existence would be ordering, no?
finite vortex
there are infinitely many rationals between 0 and 1 and we can get all of them in the image of N in alpha
but then we would have to assign infinity to 1
you can't really order them that way
deft bloom
why could you not order them in that case I don't understand
oh wait
I get why you couldn't
oh okay yeah I see how I assumed the existence of beta nvm thx
finite vortex
deft bloom
because alpha could be a function defined like a+1/n
and in that case there are infinitely many points points in the range of alpha in between [a,a+1]
finite vortex
true
deft bloom
true
by induction if I prove that for all natural numbers n, there exists a (infinite) closed interval of real numbers not mapped by alpha by any of the natural numbers up to n, is that sufficient proof that there exists such c?
finite vortex
well yeah of course up to some finite n alpha is going to leave some closed intervals of real numbers
im not sure how this proves alpha is not surjective though
deft bloom
no each interval
is a sub interval of the last
so the intersection of all of the intervals made by induction isn't empty
therefor c is in that intersection
deft bloom
im not sure how this proves alpha is not surjective though
by induction make an interval for each natural number such that alpha(n) is not in that interval, prove that the intersection of all such intervals isn't empty, and isn't that a proof?
finite vortex
i don't know
deft bloom
oh
deft bloom
this is new proof the question is at the top of this thread please tell me if this works
I proved by induction that up to any n there exists a set containing atleast one element which is not mapped by alpha with the current n or any of the n beforehand.
finite vortex
https://cdn.discordapp.com/attachments/1105092511392870400/1120593941427916851/5...
are you sure this doesn't prove that the rationals are uncountable
because if you only consider Q the "intersection of all sets is nonempty" thing seems still to apply
deft bloom
are you sure this doesn't prove that the rationals are uncountable
yeah you're right but now I'm just confused because the hint in this book literally says use induction by forming sub intervals
it says "by induction define a sequence of closed intervals such that each are a subinterval of ecahother and such that alpha(n) is not in the (n+1)th interval) and the use fact that the intersection of all intervals is nonempty)
deft bloom
would you know how to use induction such that we don't run into that problem
deft bloom
that seems like pretty much telling you to do what you did
yeah that's why I am confused now lol
finite vortex
all i know is cantor's diagonalization
deft bloom
anyway I guess I just wont do this question
finite vortex
are you just looking at this book for fun
or is it for an assignment
out of curiosity
deft bloom
I'm not in uni yet so just for funsies
finite vortex
ah
deft bloom
