Prove $x_n$ converges to $a$ iff every subsequence of $x_n$ also converges to $a$.

Prove $x_n$ converges to $a$ iff every subsequence of $x_n$ also converges to $a$. Suppose that $\{x_n\}$ is a sequence in $\mathbb R$.

Definitions available:

(1) A sequence of real numbers $\{x_n\}$ is said to converge to a real number $a\in \mathbb R$ if and only if for $\quad$ every $\epsilon\gt 0$ there is an $N\in \mathbb N$ such that $n\ge N \to |x_n-a|\lt \epsilon$.

(2) By a subsequence of a sequence $\{x_n\}_{n \in \mathbb N}$, we mean a sequence of the form $\{x_{n_k}\}_{k \in \mathbb N}$, where $\quad$ each $n_k\in \mathbb N$ and $n_1\lt n_2\lt \cdots$.

(3) If $\{x_n\}_{n \in \mathbb N}$ converges to $a$ and $\{x_{n_k}\}_{k \in \mathbb N}$ is any subsequence of $\{x_n\}_{n \in \mathbb N}$ then $x_{n_k}$ converges $\quad$ to $a$ as $k\to \infty$.

My attempt:

Let $x_n$ converge to $a$ and $\{x_{n_k}\}$ be a subsequence of $\{x_n\}$ where $n_k\in \mathbb N$. Since we know for every $\epsilon \gt 0$ there is an $N\in\mathbb N$ such that $n\ge N \to |x_n-a|\lt \epsilon$, if we minimize $b$ such that $n_b\ge N$ , then $|x_n-a|\lt \epsilon$ for any $k\ge b$. Therefore, $\{x_{n_k}\}$ converges to $a$.

Does this work for the first direction? For the second direction, is it acceptable to say that if every subsequence converges to $a$, it is trivial that $\{x_n\}$ converges to $a$, because $\{x_n\}$ is a subsequence of itself?

• Does not (3) prove the first direction? – shooting-squirrel Sep 12 '14 at 5:28
• For the second direction, i'll state it informally but, you'll know how to write it in epsilon-delta form afterwards. If the main sequence does not converge to a than that epsilon-delta inequality is not satisfied(write it down!). Immediately, it follows that you have at least 2 infinite groups of numbers in the sequence clumped up in two different intervals.(Here you can see a contradiction.) – shooting-squirrel Sep 12 '14 at 5:31
• Your argument for the second direction is absolutely correct. $(x_n)$ is a perfectly respectable subsequence of itself. – Bungo Sep 12 '14 at 5:34
• Informally, when you say a sequence converges to a number, you are saying that except for some numbers in the sequence(finite number of numbers) all other lye in any desired interval(as small as you like) centered at the number the sequence converges. – shooting-squirrel Sep 12 '14 at 5:34
• @Bungo The theorem is also true if you would leave that out. – shooting-squirrel Sep 12 '14 at 5:36