# Prove: If a sequence converges, then every subsequence converges to the same limit.

I need some help understanding this proof:

Prove: If a sequence converges, then every subsequence converges to the same limit.

Proof:

Let $s_{n_k}$ denote a subsequence of $s_n$. Note that $n_k \geq k$ for all $k$. This easy to prove by induction: in fact, $n_1 \geq 1$ and $n_k \geq k$ implies $n_{k+1} > n_k \geq k$ and hence $n_{k+1} \geq k+1$.

Let $\lim s_n = s$ and let $\epsilon > 0$. There exists $N$ so that $n>N$ implies $|s_n - s| < \epsilon$. Now $k > N \implies n_k > N \implies |s_{n_k} - s| < \epsilon$.

Therefore: $\lim_{k \to \infty} s_{n_k} = s$.

1. What is the intuition that each subsequence will converge to the same limit
2. I do not understand the induction that claims $n_k \geq k$

1. A sequence converges to a limit $L$ provided that, eventually, the entire tail of the sequence is very close to $L$. If you restrict your view to a subset of that tail, it will also be very close to $L$.

2. An example might help. Suppose your subsequence is to take every other index: $n_1 = 2$, $n_2 = 4$, etc. In general, $n_k = 2k$. Notice $n_k \geq k$, since each step forward in the sequence makes $n_k$ increase by $2$, but $k$ increases only by $1$. The same will be true for other kinds of subsequences (i.e. $n_k$ increases by at least $1$, while $k$ increases by exactly $1$).

• Sorry Professor Mohr, I meant, what does $k$ represent. Sorry, that was the question Mar 23, 2021 at 7:16
• $k$ runs from 1 to infinity and allows me to name the members of the subsequence: $n_1$, $n_2$, etc. Mar 23, 2021 at 7:19
• So $k$ itself is not the index right? Mar 23, 2021 at 7:29
• @Negrawh The various values of $k$ serve to index the terms of the subsequence. Using the terminology in the problem, the original sequence would be $(s_1, s_2, s_3, \dots)$, while the subsequence would be $(s_{n_1}, s_{n_2}, s_{n_3}, \dots)$. It's merely a convenient way to remind the reader "I am choosing some subset of values that $n$ ranges over to form my subsequence". Mar 24, 2021 at 0:28

For part 1, if there were a subsequence that didn't converge to the same limit, then we could find a neighborhood around the original point such that infinitely-many subsequence terms were outside the neighborhood. Then infinitely many sequence terms are outside the neighborhood, and so the sequence can't converge to the original point.

As for part 2, let me put an alternative proof forward. Suppose by way of contradiction that there is some $k$ such that $n_k<k$. We know that $n_1\geq 1$ since $1$ is the least positive integer, so we necessarily have $k>1.$ Now, $n_j<n_{j+1}$ for all $j$, so we know that the $n_j$ are all distinct, and in particular, for all $j\leq k$ we have $n_j\leq k-1$ (since $n_j\leq n_k<k$). But then $\{n_1,...,n_k\}$ is $k$-element subset of the set $\{1,...,k-1\}$, which is impossible, since a subset can't be strictly larger than a set. There's the desired contradiction.

$$n_1\geq 1$$, because $$(n_k)$$ is defined to be a strictly increasing sequence of indices.
Assume $$n_k\geq k$$ for some $$k\in\mathbb{N}$$, what follows by definition is that $$n_{k+1} > n_k\geq k$$.
If $$m,n\in\mathbb{N}$$ such that $$m>n$$, then $$m\geq n+1$$. (property of natural numbers)
We now conclude that $$n_{k+1}\geq n_k +1\geq k+1$$
• I know this is a very old answer, but I wanted to ask: couldn't we just say that $n_{k+1} > n_k \ge k \implies n_{k+1}>k$ and, using the property of natural numbers, say that $n_{k+1}>k \implies n_{k+1} \ge k+1$ without using the inequality $n_{k+1} \ge n_k+1$? Feb 4 at 3:43