subsequence converges to L implies L is a limit point of sequence

Question

Proposition: Let $(a_n)^\infty_0$ be a sequence of real numbers, and let $L$ be a real number. Then the following two statements are logically equivalent:

(a) $L$ is a limit point of $(a_n)^\infty_0$

(b) There exists a subsequence of $(a_n)^\infty_0$ which converges to $L$.

I am trying to show that (b) implies (a). I am confused how to relate the different epsilons and Ns together to serve the proof:

Proof: Let $(b_n)^\infty_0$ be a subsequence that converges to $L$:

$\forall \epsilon> 0$ there exists an $N$ such that $$|b_n − L|\leq\epsilon,\forall n\geq N$$

For $L$ to be a limit point of the sequence $(a_n)^\infty_0$, we need: $$\forall \epsilon >0, \forall N\geq0, \exists n\geq N,\text{ such that }|a_n − L| ≤\epsilon$$

I am confused how to continue. Should I fix an epsilon and relate the Ns?

Aaron Maroja · Answer 1 · 2014-12-28T17:05:25.050

2

Let $X \subset \mathbb{R}$ and $L \in \mathbb{R}$.

$(b)\implies (a)$

Consider $\lim x_n = L$, where $x_n \in X - \{L\}$. For every $n_0 \in \mathbb{N}$ the set $\{x_n ; n> n_0\}$ is infinite, because if it wasn't there would be a $x_{n_1}$ repeating itself infinite times and this would give us a constant sequence such that $\lim x_{n_1} \neq L$. Then by definition of limit we have that $(L-\epsilon, L + \epsilon)$ centered at $L$ has an infinity points of $X$, so $L$ is a limit point of $X$, that is $(L-\epsilon, L + \epsilon) \cap X - \{L\} \neq \emptyset$.

edited Dec 28 '14 at 17:05

answered Dec 28 '14 at 16:44

Aaron Maroja

17,571

That is different. I am asking about limit points not limits. two different concepts. – MAS Dec 28 '14 at 16:46
It's my english not that good. It's accumulation point, right? You are using the definition of limit though. I'll change my answer. – Aaron Maroja Dec 28 '14 at 16:48
yes, it is also called accumulation point. I am using the definition of a limit for branch b. thanks. – MAS Dec 28 '14 at 17:00
Did you get $(a) \implies (b)$? – Aaron Maroja Dec 28 '14 at 17:14

score 0 · Answer 2 · answered Jul 15 '18 at 02:40

For other people interested (and for my future reference), this is proposition 6.6.6 (exercise 6.6.5) of Terence Tao's Analysis 1 book (with plenty of hints given about $(a) \implies (b)$).

By definition, we say $(b_n)_0^\infty$ is a subsequence of $(a_n)_0^\infty$ if there exists a strictly increasing function $f: \mathbb{N} \to \mathbb{N}$ such that $\forall n \in \mathbb{N}, b_n = a_{f(n)}$.

Now on to the proof of $(b) \implies (a)$:

Let any $\epsilon > 0 $ and $N \in \mathbb{N}$ be fixed. Consider the subsequence $(b_n)_{n=N}^\infty$. Since $(b_n)_{n=N}^\infty$ is a subsequence of $(b_n)_{n=0}^\infty$, hence a subsequence of $(a_n)_{n=0}^\infty$, $(b_n)_{n=N}^\infty$ also converges to $L$ (proposition 6.6.5). Therefore $\exists k \geq N$, s.t., $|b_k - L|\leq \epsilon$, which implies $|a_{f(k)} - L| \leq \epsilon$. It's easy to verify by induction that $\forall x\in \mathbb{N}, f(x) \geq x$, so $f(k) \geq k \geq N$. Since we can exhibit a number $f(k) \geq N$ s.t. $|a_{f(k)} - L| \leq \epsilon$ for arbitrary $\epsilon > 0 $ and $N \in \mathbb{N}$, $L$ is a limit point of $(a_n)_{n=0}^\infty$.

subsequence converges to L implies L is a limit point of sequence

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