Hi, I’m reading the notes of my Professor and he says that if we prove that ]0,1[ is an uncountable set then |R is also an uncountable set. Can you explain why?
#Uncountable set
real pendant
nova elmBOT
Hi, I’m reading the notes of my Professor and he says that if we prove that ]0,1...
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real pendant
My doubt comes cause ]0,1[ has also irrational numbers
versed crescent
Consider the function $\tan(\pi\left(x+\frac{1}{2}\right))$. It's a bijection between which two sets?
radiant cloudBOT
mPear Category Theorem
versed crescent
Alternatively, suppose that $\mathbb{R}$ were countable. Then we would have some bijection $\phi:\mathbb{N}\to\mathbb{R}$. But, $]0,1[\subseteq\mathbb{R}$; so there is some subset $S\subseteq\mathbb{N}$ such that $\phi(S)=]0,1[$.
radiant cloudBOT
mPear Category Theorem
odd grove
My doubt comes cause ]0,1[ has also irrational numbers
Well, so does ℝ.
odd grove
Was thinking of R as Q
Oh, I see 😅
Yeah, that wouldn't be possible in that case, of course.
lost temple
$$|(0,1)| \leqslant |\mathbb R| \leqslant |(0,1)| $$
radiant cloudBOT
aL
lost temple
cantor bernstein schroeder shenanigans
real pendant
What?
lost temple
in other words, it's enough to find an injective function R -> (0,1)
real pendant
R<=0,1 ??
lost temple
the first inequality is obvious
real pendant
Yes
lost temple
$$ \mathbb R \to (-\pi/2, \pi/2) \to (0,1) $$
radiant cloudBOT
aL
lost temple
these injections can also be found
real pendant
I can see the injections on the inverse order not in bigger sets going to smaller sets
lost temple
arctan x is injective (bijective even)
and then (-pi/2,pi/2) -> (0,1) injection is easy enough
odd grove
I mean, I just agree with mPear's proposition - just take a bijective function with the given sets as the domain and range. Pretty easy in this case.
real pendant
and then (-pi/2,pi/2) -> (0,1) injection is easy enough
That I can see
lost temple
so you have injections (0,1)->R and R->(0,1)
assuming you have proved R is uncountable, you're done
real pendant
Ok thanks so in general if A,B are uncountable A<= B<=A is always possible?
real pendant
Can you give a simple example?
lost temple
R and P(R)
both are uncountable
but one is strictly bigger in cardinality than the other
real pendant
What is P(R)?
lost temple
$$ \mathcal P(A) = {X \subseteq A} $$
radiant cloudBOT
aL
hasty wraith
Hi, I’m reading the notes of my Professor and he says that if we prove that ]0,1...
Because (0, 1) is a strict subset of R, therefore R must be at least as large as (0, 1).
Just like how the prime numbers are a strict subset of the integers, therefore a proof of infinitely many primes is a proof of infinitely many integers.
inland oliveBOT
@real pendant