# Convergent Sequence with Finite Number of Terms Deleted is Convergent

## Theorem

Let $\left({X, d}\right)$ be a metric space.

Let $\left\langle{x_k}\right\rangle$ be a sequence in $X$.

Let $\left\langle{x_k}\right\rangle$ be convergent.

Let a finite number of terms be deleted from $\left \langle {x_k} \right \rangle$.

Then the resulting subsequence is convergent.

## Proof

Suppose the sequence $\left \langle {x_k} \right \rangle$ converges to $x \in X$.

That is for every $\epsilon > 0$ there is some index $N$ such that $d \left({x_n, x}\right) < \epsilon$ for all $n \ge N$.

The same $N$ will work for the new sequence with finitely many terms removed, so the new sequence converges to the same point $x$ as the original sequence.

For the second part note that also adding finitely many terms to a convergent sequence will still result in a convergent sequence.

This implies that removing finitely many terms from a divergent sequence will still result in a divergent sequence (if it were convergent then the original sequence must already have been convergent).

## Lemma 1

Let $\left({X, d}\right)$ be a metric space.

Let $\left\langle{x_k}\right\rangle$ be a sequence in $X$.

Let $\left\langle{x_{n_k} }\right\rangle$ be a subsequence of $\left\langle{x_k}\right\rangle$.

Then:

$\forall k \in \N: k \le n_k$

### Proof of Lemma 1

Note that $1 \le n_1 < n_2 < \ldots < n_k < \ldots$ by definition of a subsequence.

We will proceed with induction on $k$ to prove our claim.

By definition of a subsequence:

$1 \le n_1$

This is the basis for the induction

Suppose $k \le n_k$.

Then:

$k + 1 \le n_k + 1$

By definition of a subsequence:

$n_k < n_{k+1}$

Therefore:

$n_k + 1 \le n_{k+1}$

Thus:

$k + 1 \le n_{k+1}$

$\Box$

## Lemma 2

Let $\left({X, d}\right)$ be a metric space.

Let $\left\langle{x_k}\right\rangle$ be a convergent sequence in $X$.

Then every subsequence of $\left\langle{x_k}\right\rangle$ is convergent.

### Proof of Lemma 2

Let $\left\langle{x_k}\right\rangle$ be a convergent sequence in $X$.

Then $\exists L \in X$ such that $x_k \to L$ as $k \to \infty$.

Let $\left\langle{x_{n_k} }\right\rangle$ be a subsequence of $\left\langle{x_k}\right\rangle$.

We will show that $x_{n_k} \to L$ as $n_k \to \infty$.

Let $\epsilon >0$.

Then $\exists N \in \N$ such that:

$\forall k \ge N: d \left({x_k, L}\right) < \epsilon$

From Lemma 1:

$n_k \ge k$

Therefore:

$\forall n_k \ge k \ge N: d \left({x_{n_k}, L}\right) < \epsilon$

Hence $x_{n_k} \to L$ as $n_k \to \infty$.

$\blacksquare$