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Prove that if $\left\{ x_n \right\}$ is an infinite sequence of real numbers, $x \in \mathbb{R}$, and every subsequence $\left\{ x_{n_k} \right\}$ has a subsequence $\left\{ x_{n_{k_j}} \right\}$ with $x_{n_{k_j}} \rightarrow x$, then $x_n \rightarrow x$.

I know that if every subsequence of a sequence converges to the same number, then the sequence converges to that same number. But I don't know if the same can be applied to subsubsequences. So for this problem, can I safely state that because $x_{n_{k_j}} \rightarrow x$, it is also true that $x_{n_k} \rightarrow x$? If this is true, then does that mean every subsequence $\left\{ x_{n_k} \right\}$ also converges to $x$?

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HINT: Suppose that $x_n\not\to x$, and show that $\langle x_n:n\in\Bbb N\rangle$ has a subsequence that is bounded away from $x$.

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    $\begingroup$ @user130018: It means that there is an $\epsilon>0$ such that $|x_{n_k}-x|\ge\epsilon$ for all $k\in\Bbb N$, where $\langle x_{n_k}:k\in\Bbb N\rangle$ is the subsequence in question. The distance of the terms from $x$ is bounded away from $0$, meaning that it has a positive lower bound. $\endgroup$ – Brian M. Scott Dec 8 '14 at 0:24
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    $\begingroup$ @Silencer: I prefer angle brackets to parentheses or curly braces for ordered tuples and sequences; parentheses already have more than enough different uses, and curly braces already have a very different meaning. In particular, it’s important to distinguish the sequence $\langle x_n:n\in\Bbb N\rangle$ from the set $\{x_n:n\in\Bbb N\}$. This angle bracket notation is one of the standard notations, though it’s probably most common among those with a lot of set-theoretic background. $\endgroup$ – Brian M. Scott Dec 8 '14 at 0:41
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    $\begingroup$ @user130018: Your first sentence is correct. it shows that if $M=\{n\in\Bbb N:|x_n-x|\ge\epsilon\}$, then $M$ is infinite, so we can enumerate it as $M=\{m_k:k\in\Bbb N\}$, with $m_k<m_{k+1}$ for each $k\in\Bbb N$. Then $\langle X_{m_k}:k\in\Bbb N\rangle$ is a subsequence of the original sequence, and it has the property that $|x_{m_k}-x|\ge\epsilon$ for all $k\in\Bbb N$. Now use your hypothesis to get a contradiction. $\endgroup$ – Brian M. Scott Dec 8 '14 at 1:49
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    $\begingroup$ @user130018: Yes, it means that the set $M$ has an infinite number of elements. What you’re missing, I think, is that while a sequence $\langle y_n:n\in\Bbb N\rangle$ has an infinite number of terms, some (or even all) of those terms could be the same number, so the set $\{y_n:n\in\Bbb N\}$ might be finite. Consider, for instance, the sequence $\langle (-1)^n:n\in\Bbb N\rangle$; it has a term for each natural number $n$, but the set $\{(-1)^n:n\in\Bbb N\}=\{-1,1\}$ has only two elements. $\endgroup$ – Brian M. Scott Dec 8 '14 at 2:00
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    $\begingroup$ @user130018: That’s why I prefer to use the word each where you have all: for each natural number $N$ there is an $n_N\ge N$ such that ... . The numbers $n_N$ might all be different, or some of them might be the same, but there are certain to be infinitely many different ones. It definitely does not mean that there is a single $n$ that is greater than or equal to all natural numbers: that is indeed impossible. \\ It’s often less confusing to read $\forall x$ as ‘for each $x$’ rather than as ‘for all $x$’. $\endgroup$ – Brian M. Scott Dec 8 '14 at 2:12
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If $x_n$ doesn't converge to $x$, no tail of it converges. But then we can construct a subsequence of elements that are at least $\varepsilon$ from $x$, but that subsequence has to have a subsequence that converges to $x$. By contradiction $x_n$ converges to $x$.

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Suppose that $\{X_n\}$ does not converge to $x$. Then, there is $\varepsilon_0>0$ such that $$\forall N\in\mathbb N,\exists \hspace{.2cm}n=n(N) : n>N~~~and ~~~ |X_n -x|>\varepsilon_0 $$

For $N_1=1$ there exists $n_1$ such that $$n_1>N_1 ~~~and ~~~ |X_{n_1} -x|>\varepsilon_0 $$ Taking successively $N_{k+1}> \max\{N_k, n_k,k+1\}$ there exists $n_{k+1>N_{k+1}}$ such that,

$$ |X_{ n_{k+1}} -x|>\varepsilon_0 $$

It is easy to see that, $\{X_{ n_k}\}_k$ is a subsequence of $\{X_{ n}\}_n$ since $$ n_k< n_{k+1} \quad i.e ~~\text{the map }~~k\mapsto n_k~~~\text{Is one-to-one}$$

However, $$\forall k,~~ |X_{ n_{k}} -x|>\varepsilon_0 \qquad \text{and}~~~\{X_{ n_{k}} \}~~~\text{is bounded} $$

Therefore By Bolzano-Weierstrass Theorem's there exists $\{X_{ n_{k_p} }\}_p$ subsequence of $\{X_{ n_{k} }\}_k$ which converges to some limit $\ell_1 $ but $\{X_{ n_{k_p} }\}_p\to \ell_1$ is also a congering subsequence of $\{X_n\}_n$

By assumption, $\ell=\ell_1$ that is together with the fact $\{X_{ n_{k_p} }\}_p$ is a subsequence of $\{X_{ n_{k} }\}_k$ we have

$$0=\lim_{p\to\infty } |X_{ n_{k_p} }-\ell|>\varepsilon_0>0~~~\text{CONTRADICTION}$$

Note that $$\forall p,~~|X_{ n_{k_p} }-\ell|>\varepsilon_0~~~ Since ~~~\forall k,~~|X_{ n_{k}} -\ell|>\varepsilon_0$$

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