Prove that the following set $A=\{2n−1∥ n∈N\} \subset \mathbb{R}$ is complete.

Question

A subset $S \subset \mathbb{R}$ is complete if every Cauchy sequence consisting of elements of S converges to an element of S

Prove that the following set $A=\{2n−1∥ n∈N\} \subset \mathbb{R}$ is complete.

Note: I don't want the solution (at least not eight now) but I want to run past people my idea of how to prove this.

IDEA:

So we stated in class as an example without proof that $\mathbb{N}$ is complete by way of it being closed. But the professor stressed to us to try and show this set is complete without using the fact about $\mathbb{N}$.

So after playing around with this set I've noticed that it "appears" that the only convergent sequences are sequences of a specific constant term. That being the case, since $A \subset \mathbb{R}$ this would also mean that these sequences are Cauchy. Now if these are the ONLY convergent sequences of this then I could show that these sequences converge to an element of my set $A=\{2n−1∥ n∈N\}$

Here is my issue with this: Saying that this is the only convergent sequence AND the only Cauchy sequence for a countably infinite set is a pretty damn bold statement. Even though it appears to be the case from playing around how could I show that to be true and also is my idea towards showing the set is complete correct?

Answer 1

Hint: A sequence $(x_n)_{n \in \mathbb N}$ in $A$ is a Cauchy sequence if and only if $$ \forall \epsilon > 0 \, \exists n_0 \in \mathbb N \, \forall n, m \geq n_0: \vert x_n - x_m \vert < \epsilon. $$ So take $\epsilon = 1$ for example. What can you say about a sequence the that suffices the assertion above for that explicit $\epsilon$? How does it relate to your observation? :)

Answer 2

Just apply the definition of Cauchy sequence to $\epsilon=2$, for example.

Answer 3

Let $(x_n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $A$. Hence, there exists $n_0\in\mathbb{N}$ such that $|x_n-x_m|<1$, whenever $m,n\geq n_0$. In particular, $|x_n-x_{n_0}|<1$ for all $n\geq n_0$. This means that $x_n=x_{n_0}$ for all $n\geq n_0$. Therefore, $(x_n)_{n\in\mathbb{N}}$ converges to $x_{n_0} \in A$ and, consequently, $A$ is complete.

Answer 4

Since you have a cauchy sequence of points from $S$, and $S \subset \mathbb{R}$ and $\mathbb{R}$ is complete, there is no question as to whether the sequence converges, it certainly converges, the question is whether it converges to something in $S$.

What the other answers are hinting at, is that if you satisfy the cauchy property, then this means that at some $N$, you have $|s_n - s_m| < \epsilon$ for $n,m >N$. But since $s_n \in S$ and $s_m \in S$, they are of the form $$s_n = 2p+1$$ $$s_m = 2q+1 $$

for some $p,q$, but $|s_n - s_m| < \epsilon$ impies $p = q$ for many of the epsilons in the other answers. So for all but finitely many $n$'s, specifically for all but those $n$ with $n<N$, $$s_n = s_m \to s = 2p+1$$ for some $p$, and $s\in A$