Display a bijection between the set [-1,1]
and the set [-1,1] \ [0]
#Countable Sets
edgy crescent
fast joltBOT
Display a bijection between the set [-1,1]
and the set [-1,1] \ [0]
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edgy crescent
The set should be contable because is the intersection of 2 countable sets but idk how to prove it
dapper charm
The set should be contable because is the intersection of 2 countable sets but i...
why'd that be??!
intersections don't conserve any form of uncountablity
they do conserve countablity
for example : $[0,1]\cap [1,2] = \qty{1}$
frank pineBOT
Coffey
dapper charm
in that case they are both uncountable
thats factual but what about the bijection
it'll suffice to frame a bijection $(0,1] \to [0,1]$ as the other part can just be $[-1,0)\to [-1,0)$
frank pineBOT
Coffey
dapper charm
i have an idea
patent trout
yeah that (0,1] -> [0,1] thing is technical
involves picking sequences in a clever way
it's a standard exercise when dealing with cardinality stuff, I have it written somewhere
dapper charm
okay so I think we can do it in this way :
pick out a sequences from (0,1] :
1/1, 1/2, 1/3,...... 1/n,....
and map it to
0, 1/1, 1/2, 1/3,.... 1/n, ....
and everything else to itself
shifted the sequence that it
patent trout
yup, that'll do it
dapper charm
this sort of "removing finite elements" works for any pairs of infinite sets
patent trout
but it definitely has to involve some infinite process
dapper charm
because for them to be even considered infinite they must have a subset such that $A \subset B$ and $\abs{A} =\abs{B}$
frank pineBOT
Coffey
dapper charm
but it definitely has to involve some infinite process
yeah we just picked out a sequence and shifted it
primal fiber
Display a bijection between the set [-1,1]
and the set [-1,1] \ [0]
Oh, I've heard of this before. I think the following works:
f(0) = 1
f(2^(-n)) = 2^(-n - 1), n ≥ 0
Otherwise f(x) = x. This should be a bijection between [-1, 1] and [-1, 0)⋃(0, 1].
The variant I've heard of is a bijection between [0, 1] and (0, 1), I think.
edgy crescent
The set should be contable because is the intersection of 2 countable sets but i...
nvm I was thinking that finite sets where sets that didn't went to + or - infinite
edgy crescent
Oh, I've heard of this before. I think the following works:
f(0) = 1
f(2^(-n)) =...
can you explain where does f(2^n)) = 2(-n.1), n>= 0 came from?
dapper charm
okay so I think we can do it in this way :
pick out a sequences from (0,1] :
1...
@edgy crescent check this too :3