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).
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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 $n_k \to \infty$.
$\blacksquare$