Suppose that $(x_n)_{n \geq 1}$ is a convergent sequence. Denote $(x_{n_j})_{j \geq 1}$ as a subsequence of $(x_n)_{n \geq 1}$.

Question: What is the precise definition of $\lim_{j \rightarrow \infty}{x_{n_j}}=x$?

Based on the definition of limit, we have

$$ \lim_{j \rightarrow \infty}{x_{n_j}}=x \Longleftrightarrow \forall \epsilon>0, \exists J \in \mathbb{N}, \forall j (j \geq J \Rightarrow | x_{n_j} - x | < \epsilon)$$

But I think something is strange here. In the definition, we do not specify what is our $n_j$, but it appears in $|x_{n_j}-x|<\epsilon$. Is the definition above correct?


Saying $(x_{n_j})$ is a subsequence means that $(n_j)$ is a strictly increasing sequence of natural numbers, so there is nothing strange. $n_j$ is defined for every $j\in\mathbb N$.

So, when you say

In the definition, we do not specify what is our $n_j$,

That's not true. The value of $n_j$ is defined when you decide which subsequence you are looking at.

For example, if $x_n = (-1)^n \frac{n+1}{n}$, then we can examine the subsequence $(x_{2j})$ (defined by $n_j=2j$). This is a convergent subsequence of $(x_n)$ because $$\lim_{j\to\infty} x_{2j} = 1$$

  • $\begingroup$ But in my definition, there is no quantifier attached to $n_j$, which I feel worried. $\endgroup$ – Idonknow Jan 27 '17 at 14:43
  • $\begingroup$ @Idonknow Yes there is. The $\forall j$. Remember, $n_j$ is simply some number dependent on $j$, just like $a_n$ is a number dependent on $n$. So, just like you don't need a quantifier over $a_n$, but one over $n$, you don't need a quantifier over $n_j$, but one over $j$. $\endgroup$ – 5xum Jan 27 '17 at 14:47
  • $\begingroup$ The set $\{n_j\}_{j\in\mathbb{N}}$ is a fixed subset of $\mathbb{N}$, indexed by the subscript $j$. Basically you don't have a quantifier because it's fixed. $\endgroup$ – mathematician Jan 27 '17 at 14:47

Instead of giving the definition of convergence, which you already seem to know, let's prove that the subsequence converges to the same limit at the original sequence.

First thing to note is that $n_j$ must be an increasing sequence in terms of the index $j \in \mathbb{N}$. We know for instance that $1 \le n_1$, and $n_1 < n_2$. This means that $2 \le n_2$. In particular $j < n_j$ for all $j \in \mathbb{N}$.

Now suppose that $\epsilon > 0$. There is an $N\in \mathbb{N}$ for which given any $n > N$ we have $$| x_n - x| < \epsilon.$$

Now consider $n_N$. For any $m > n_N$ we know that $m > n_N \ge N$. Thus $$|x_m - x | < \epsilon$$ for all $m > n_N$. Since for any $j > N$ we have $n_j > n_N$ this means $$|x_{n_j} - x| < \epsilon$$ for all $j > N$.

Thus we have for any $\epsilon > 0$, there is an $N \in \mathbb{N}$ such that for any $j > N$ we have $$|x_{n_j} - x| < \epsilon$$ and that proves the convergence of the subsequence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.