# Proof that sequence not convergent to $x$ contain subsequence completely not contained in neighborhood of x

Can anyone verify if the following proof is correct and/or suggest any improvement to be made?

Let $$x \in \mathbb{R}$$ and let $$\{x_n\}$$ be sequence of real numbers not convergent to $$x$$. Then there is $$\varepsilon > 0$$ and subsequence $$\{x_{n_k}\}$$ such that for all $$k \in \mathbb{N}$$ we have $$|x_{n_k}-x| \geq \varepsilon$$.

\textbf{Proof:} Assume that for all $$\varepsilon > 0$$ and for all subsequences $$\{x_{n_k}\}$$ there exists $$K \in \mathbb{N}$$ such that $$|x_{n_k}-x| < \varepsilon$$.

Now let $$\varepsilon$$ be given. Let $$N_{1}$$ be minimal natural number such that $$|x_{N_1}-x| < \varepsilon$$. Now let $$\{x_{n_k}\}$$ be subsequence such that $$n_k = N_1+1, N_1+2, \ldots$$. There is still some minimal number $$N_{2} = n_k$$ such that $$|x_{N_2}-x| < \varepsilon$$. Proceeding in the same manner, we find number $$N_3$$ and, in general, we obtain sequence of natural numbers $$\{N_j\}$$, such that subsequence $$\{x_{N_j}\}$$ consists, by construction, of all elements of $$\{x_n\}$$ such that $$|x_n-x| < \varepsilon$$. But this leaves two possibilities for us.

First one is that $$\mathbb{N} \setminus \{N_j\}$$ is finite, from what it follows that $$\{x_n\}$$ converges to $$x$$. Indeed, it is enough to take $$N' = \max (\mathbb{N} \setminus \{N_j\})$$ and for $$n > N'$$ then $$|x_n-x| < \varepsilon$$, completing proof by contrapositive.

Another case is that, assuming that $$\{x_n\}$$ does not converge to $$x$$, we see that for subsequence $$\{x_m\}$$ where $$m \in \mathbb{N} \setminus \{N_j\}$$, there is no such $$M$$ such that $$|x_{n_k}-x| < \varepsilon$$, which contradicts to assumption made.

• How do you know that outside $(x_{N_j})$, no element is near $x$? – Tito Eliatron Sep 15 '20 at 17:53
• because for given $\varepsilon$ , i first choose $N_1$ to be minimal number for which $|x_n-x|<\varepsilon$ holds, then consider subsequence consisting of all terms with indices $> N_1$, obtain from this sequence next $N_2$ and by this process i collect all such $N$'s into one sequence of naturals. – Commander Vimes Sep 15 '20 at 17:58
• Upd: i found some unclearness in the considering case when I get contrapositive, which I will clean up right now – Commander Vimes Sep 15 '20 at 17:59
• I fixed subtlety in the part with contrapositive making it clearer. – Commander Vimes Sep 15 '20 at 18:09

## 1 Answer

Proving this by contraposition seems (in my opinion) a bit clumsy.

Here is a way to prove directly the implication. If $$(x_n)$$ does not converge to $$x$$, by definition, that means that there exists $$\varepsilon > 0$$ such that $$\forall N \in \mathbb{N}$$, $$\exists n > N$$ with $$|x_n - x| \geq \varepsilon$$ (let's call this property $$(*)$$).

Take this $$\varepsilon >0$$, and let's construct recursively the subsequence $$(x_{n_k})$$ using the property $$(*)$$. For $$N=0$$, you get that there exists $$n_0 > 0$$ such that $$|x_{n_0}-x|\geq \varepsilon$$. Now, let's suppose that you constructed $$n_k$$. There applying the property $$(*)$$ to $$N=n_k$$, you deduce that there exists an integer $$n_{k+1} > n_k$$ such that $$|x_{n_{k+1}}-x|\geq \varepsilon$$.

By induction, you constructed a strictly increasing sequence $$(n_k)$$ such that for all $$k$$, $$|x_{n_k}-x| \geq \varepsilon$$.

• Yes, this one is much more direct approach than mine, thank you! But still, can you find whether is there any error in the proof? Or it is correct? – Commander Vimes Sep 15 '20 at 18:30