# Convergent Sequence with Finite Number of Terms Deleted is Convergent

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## Theorem

Let $\struct {X, d}$ be a metric space.

Let $\sequence {x_k}$ be a sequence in $X$.

Let $\sequence {x_k}$ be convergent.

Let a finite number of terms be deleted from $\sequence {x_k}$.

Then the resulting subsequence is convergent.

## Proof

Suppose the sequence $\sequence {x_k}$ converges to $x \in X$.

That is for every $\epsilon > 0$ there is some index $N$ such that $\map d {x_n, x} < \epsilon$ 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).

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## Lemma 1

Let $\struct {X, d}$ be a metric space.

Let $\sequence {x_k}$ be a sequence in $X$.

Let $\sequence {x_{n_k} }$ be a subsequence of $\sequence {x_k}$.

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 $\struct {X, d}$ be a metric space.

Let $\sequence {x_k}$ be a convergent sequence in $X$.

Then every subsequence of $\sequence {x_k}$ is convergent.

### Proof of Lemma 2

Let $\sequence {x_k}$ be a convergent sequence in $X$.

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

Let $\sequence {x_{n_k} }$ be a subsequence of $\sequence {x_k}$.

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

Let $\epsilon >0$.

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

- $\forall k \ge N: \map d {x_k, L} < \epsilon$

From Lemma 1:

- $n_k \ge k$

Therefore:

- $\forall n_k \ge k \ge N: \map d {x_{n_k}, L} < \epsilon$

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

$\blacksquare$