Suppose that $(x^{(n)})^\infty_{ n =1} $ is a sequence of points in a metric space $(X, d)$. Suppose that $n_1, n_2, … $ is an increasing sequence of integers which are at least as large as $m$, thus
$$1 \leq n_1 < n_2 < n_3 < … $$
Then we call the sequence $(x^{(n_j)})^\infty_{ j =1} $ a subsequence of the original sequence $(x^{(n)})^\infty_{ n =1} $.
Let $(x^{(n)})^\infty_{ n =1}$ be a sequence in $(X, d)$ which converges to some limit $x_0$. Then every subsequence $(x^{(n_j)})^\infty_{ j=1}$ of that sequence also converges to $x_0$.
Suppose that $(x^{(n)})^\infty_{ n =1} $ is a sequence of points in a metric space $(X, d)$, and let $L \in X$.
We say that $L$ is a limit point of $(x^{(n)})^\infty_{n=1}$ iff for every $N \geq 1$ and $\epsilon > 0 $ there exists an $n \geq N $ such that $d(x^{(n)}, L) \leq \epsilon$.
Let $(x^{(n)})^\infty_{ n =1} $ be a sequence of points in a metric space $(X,d) $, and let $L \in X $. Then the following are equivalent:
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$L $ is a limit point of $(x^{(n)})^\infty_{ n =1} $.
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There exists a subsequence $(x^{(n_j)})^\infty_{j=1}$ of the original sequence $(x^{(n)})^\infty_{ n =1}$ which converges to $L$.
Let $(x^{(n)})^\infty_{ n =1} $ be as sequence of points in a metric space $(X, d)$.
$(x^{(n)})^\infty_{ n =1} $ is a Cauchy sequence if
$$\forall \epsilon >0, \exists N \geq 1, \forall j, k \geq N, d(x^{(j)}, x^{(k)}) < \epsilon $$
Let $(x^{(n)})^\infty_{ n =1} $ be a sequence in $(X, d) $ which converges to some limit $x_0 $. Then $(x^{(n)})^\infty_{ n =1} $ is also a Cauchy sequence.
Every subsequence of a Cauchy sequence is also a Cauchy sequence.
Let $(x^{(n)})^\infty_{ n =1} $ be a Cauchy sequence in $(X,d) $. Suppose that there is some subsequence $(x^{(n_j)})^\infty_{ j =1} $ of this sequence which converges to a limit $x_0 $ in $X$. Then the original sequence $(x^{(n)})^\infty_{ n =1} $ also converges to $x_0 $.
Let $(x^{(n)})^\infty_{ n =1} $ be a Cauchy sequence in $(X,d)$.
Suppose that there is some subsequences $(x^{(n_j)})^\infty_{ j =1}$ of this sequence which converges to a limit $x_0$ in $X$. Then the original sequence $(x^{(n)})^\infty_{ n =1} $ also converges to $x_0$.
A metric space $(X,d)$ is said to be complete iff every Cauchy sequence in $(X,d)$ is in fact convergent in $(X,d)$.
Let $(X, d) $ be a metric space, and let $(Y, d|_{ Y \times Y }) $ be a subspace of $(X,d) $. If $(Y, d|_{ Y \times Y }) $ is complete, then $Y $ must be closed in $X$.
Suppose that $(X, d) $ is a complete metric space, and $Y$ is a closed subset of $X$. Then the subspace $(Y, d|_{ Y \times Y }) $ is also complete.
An incomplete metric space such as $(\Q,d) $ may be considered closed in some spaces ($\Q$ is closed in $\Q$) but not in others. (for instance, $\Q $ is not closed in $\R $).
It turns out that given any incomplete metric space $(X,d)$, there exists a completion $(\overline{X}, \overline{d})$, which is a larger metric space containing $(X,d)$ which is complete, and such that $X $ is not closed in $\overline{X} $(indeed, the closure of $X$ in $(\overline{X}, \overline{d})$ will be all of $\overline{X}$).