Understanding the Infinite Set of Reals in Cauchy Convergence Proof

  • Thread starter Thread starter sonofagun
  • Start date Start date
  • Tags Tags
    Set
sonofagun
Messages
16
Reaction score
0
Member warned that the homework template is required
I'm reading the proof that a cauchy sequence is convergent.

Let an be a cauchy sequence and let ε=1. Then ∃N∈ℕ such that for all m, n≥N we have
an-am<1. Hence, for all n≥N we have an-aN<1 which implies an<aN+1. Therefore, the set {n∈ℕ: an≤aN+1} is infinite and thus {x∈ℝ : {n∈ℕ: an≤x} is infinite} ≠ ∅.

I can't make sense of the last set. What does it represent and why is it not empty?
 
Physics news on Phys.org
Do you buy the statement that \{n \in \mathbb{N} \colon a_n \le a_N + 1 \} is infinite? If so, what does that tell you about the a_n themselves?
 
  • Like
Likes sonofagun
statdad said:
Do you buy the statement that \{n \in \mathbb{N} \colon a_n \le a_N + 1 \} is infinite? If so, what does that tell you about the a_n themselves?

Okay, I get it now. It's the set of all reals that are ≥ an for infinitely many n's. aN+1 is an element in this set, thus it's not empty.
I'm studying independently so I occasionally get stuck trying to figure out easy concepts like this. Thank you!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top