# A question about the proof of Theorem $3.44$ of Bartle.

Specifically, the part where it proves that $$(ii) \implies (iii)$$, since I am not clear about what he means by "Continuing in this way, a sub-succession is obtained."

I have read on some internet pages that this refers to induction, however I am not sure what would be the specific property to be demonstrated by induction, that is, what would be the base case and which series would be the inductive step.

Theorem $$3.44$$

Let $$X = (X_n)$$ be a sequence of real numbers, Then the following are equivalent:

i) The sequence $$X = (X_n)$$ does not converge to $$x \in \mathbb{R}$$.

ii) There exists an $$\epsilon_0 >0$$ such that for any $$k \in \mathbb{N}$$, there exists $$n_k \in \mathbb{N}$$ such that $$n_k \geq k$$ and $$|X_{n_k}-x|\geq \epsilon_o$$

iii) There exists an $$\epsilon_o >0$$ and a subsequence $$X´=(X_{n_k})$$ of $$X$$ such that $$|X_{n_k}-x|\geq \epsilon_o$$ for all $$k \in \mathbb{N}$$

Proof. i) $$\Rightarrow$$ ii) If $$(X_n)$$ does not converge to $$x$$, then for some $$\epsilon_0>0$$ it is impossible to find natural number $$k$$ such that for all $$n\geq k$$ the terms $$X_n$$ satisfy $$|x_n-x|<\epsilon_0$$ holds. In other words, for each $$k \in \mathbb{N}$$ there a exists natural number $$n_k \geq k$$ such that $$|X_{n_k}-x|\geq \epsilon_o$$.

Proof. ii) $$\Rightarrow$$ iii) Let $$\epsilon_0$$ be as in (ii) and let $$n_1 \in \mathbb{N}$$ be such that $$n_1 \geq 1$$ and $$|X_{n_1}-x|\geq \epsilon_o$$.

Now let $$n_2 \in \mathbb{N}$$ be such that $$n_2>n_1$$ and $$|X_{n_2}-x|\geq \epsilon_o$$; let $$n_3 \in \mathbb{N}$$ be such that $$n_3>n_2$$ and $$|X_{n_3}-x|\geq \epsilon_o$$. Continue in this way to obtain a subsequence $$X´=(X_{n_k})$$ of $$X$$ such that $$|X_{n_k}-x|\geq \epsilon_o$$ for all $$k \in \mathbb{N}$$.

• What was said in proof before "Continuing in this way..."? Commented Jul 17, 2021 at 23:05
• My edit was for a typo, 8th line, $n_k\in \Bbb R$ changed to $n_k\in \Bbb N$. Commented Jul 18, 2021 at 8:50

Let $$\epsilon_0$$ be as in (ii) and let $$n_1 \in \mathbb{N}$$ be such that $$n_1 \geq 1$$ and $$|X_{n_1}-x|\geq \epsilon_o$$.
Now let $$n_2 \in \mathbb{N}$$ be such that $$n_2>n_1$$ and $$|X_{n_2}-x|\geq \epsilon_0$$; let $$n_3 \in \mathbb{N}$$ be such that $$n_3>n_2$$ and $$|X_{n_3}-x|\geq \epsilon_0$$.
So, "continuing in this way" basically means that you go on choosing $$n_k>n_{k-1}$$ and $$|X_{n_k}-x|\geq \epsilon_0$$ and define this to be the the sequence $$\{X_{n_k}\}$$.