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.

  • $\begingroup$ How do you know that outside $(x_{N_j})$, no element is near $x$? $\endgroup$ Commented Sep 15, 2020 at 17:53
  • $\begingroup$ 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. $\endgroup$ Commented Sep 15, 2020 at 17:58
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    $\begingroup$ Upd: i found some unclearness in the considering case when I get contrapositive, which I will clean up right now $\endgroup$ Commented Sep 15, 2020 at 17:59
  • $\begingroup$ I fixed subtlety in the part with contrapositive making it clearer. $\endgroup$ Commented Sep 15, 2020 at 18:09

1 Answer 1


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 $k$, $|x_{n_k}-x| \geq \varepsilon$.

  • $\begingroup$ 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? $\endgroup$ Commented Sep 15, 2020 at 18:30

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