So the equivalence classes here can be thought of as the limits to which the sequences in them converge? If this is the case I understand how the Reals are built. If not, I still see holes.
From Terence Tao's book Analysis I: Let x ≥ 0 be a non-negative real, and let n ≥ 1 be a positive integer. We define x^(1/n), also known as the nthroot of x, by the formula x^(1/n):= sup{y ∈ R : y ≥ 0 and y^n≤ x}. So 2^(1/2) = sup{y ∈ R : y ≥ 0 and y^2≤ 2}.
In any good real analysis book, this construction should be included. Also set theory books cover this topic. Maybe just look through some of them and take the one you like the most :)
Sorry but around 7:41 the order relation is defined by some delta>0 Is this delta a rational number? It seems like you're using circular logic if delta is real because how to you define being greater than 0
Do we need atleast 2 unique Cauchy sequences to form an equivalence class? for instance if there is a real number which only has 1 Cauchy sequence, would it form an equivalence class?
You say at the end that by construction, every Cauchy sequence in R is convergent. But while we have shown that the elements of R are themselves Cauchy sequences in Q, we have not shown anything about Cauchy sequences in R ( which would be sequences of sequences in Q). I don't see how we have shown the completeness axiom.
@@brightsideofmaths The proof for this appears to be subtle. In this paper ( pi.math.cornell.edu/~kahn/reals07.pdf ), the author takes the same approach of constructing R from Cauchy sequences in Q. To prove convergence, they take three pages (34-37). They construct a new sequence of rationals L that approximate the sequence of reals, and then prove that the sequence of reals converges to L. The proof seems more involved than other proofs in this series. Would you consider doing a part 5 of this series to go over the proof of convergence from this construction?
every cauchy sequence is a convergent sequence. Is it because real number construction now contains caushy sequences themselves, so the limit 'a' is itself a cauchy sequence whereas that in case of rational numbers construction is not being always convergent because numbers like root 2 , e , pi are cauchy sequences whose limit 'a' cant be expressed as m/n hence does not belong to set Q? Also its not clear why 1 /=0 in (M) property.
Great series!
Analysis is so fun
Beauty ❤️ Thank you so much sir.
Awesome, was waiting for this one!
Great going!!
So the equivalence classes here can be thought of as the limits to which the sequences in them converge? If this is the case I understand how the Reals are built. If not, I still see holes.
You understand!
Start learning complexs next? :D
I guess that is a fitting name :)
Thank you! Given a Cauchy sequence that represents 2, how do you calculate a Cauchy sequence that represents the square root of 2?
From Terence Tao's book Analysis I:
Let x ≥ 0 be a non-negative real, and let n ≥ 1 be a positive integer. We define x^(1/n), also known as the nthroot of x, by the formula x^(1/n):= sup{y ∈ R : y ≥ 0 and y^n≤ x}.
So 2^(1/2) = sup{y ∈ R : y ≥ 0 and y^2≤ 2}.
This was a very interesting topic with nice presentation! Does anyone know about a book or other resource that does this construction of real numbers?
In any good real analysis book, this construction should be included. Also set theory books cover this topic. Maybe just look through some of them and take the one you like the most :)
Sorry but around 7:41 the order relation is defined by some delta>0
Is this delta a rational number? It seems like you're using circular logic if delta is real because how to you define being greater than 0
It's a rational number and > 0 for rational numbers is already defined. We don't use circular logic but rather step by step generalisations :)
Do we need atleast 2 unique Cauchy sequences to form an equivalence class? for instance if there is a real number which only has 1 Cauchy sequence, would it form an equivalence class?
Yes, one member is the equivalence class is enough. However, this is not the case here. We have infinitely many members in an equivalence class here.
for the building of the set C why can't we say that the limits of both sequences reach the limit 1/3?
We could do that but we also want to cover all other limits and even the Cauchy sequences that have no limit in Q.
How this equivalence class construction will also include irrational numbers?
Why do you think it's not the case?
Hi ! So , Can we say that some rational number donot converge to their supremum , but real number do ?
Yes, exactly. (You should of monotonically increasing sequences of numbers then)
How do we define convergence here? For every rational epsilon > 0, blablabla?
Exactly!
You say at the end that by construction, every Cauchy sequence in R is convergent. But while we have shown that the elements of R are themselves Cauchy sequences in Q, we have not shown anything about Cauchy sequences in R ( which would be sequences of sequences in Q). I don't see how we have shown the completeness axiom.
You can write down the completeness. Take a Cauchy sequence in R and then it works :)
@@brightsideofmaths The proof for this appears to be subtle. In this paper ( pi.math.cornell.edu/~kahn/reals07.pdf ), the author takes the same approach of constructing R from Cauchy sequences in Q. To prove convergence, they take three pages (34-37). They construct a new sequence of rationals L that approximate the sequence of reals, and then prove that the sequence of reals converges to L. The proof seems more involved than other proofs in this series. Would you consider doing a part 5 of this series to go over the proof of convergence from this construction?
every cauchy sequence is a convergent sequence. Is it because real number construction now contains caushy sequences themselves, so the limit 'a' is itself a cauchy sequence whereas that in case of rational numbers construction is not being always convergent because numbers like root 2 , e , pi are cauchy sequences whose limit 'a' cant be expressed as m/n hence does not belong to set Q? Also its not clear why 1 /=0 in (M) property.