Nested intervals and the Bolzano-Weierstrass Theorem

Let $\mathbb{X} = \{x_n: n \in \mathbb{N}\}$ be a bounded sequence, so $\vert x\vert \le K$ for some positive real $K$ and for all $x \in \mathbb{X}$. Then any subsequence $\mathbb{Y}$ of $\mathbb{X}$ is bounded by the same number $K$, since $\mathbb{Y} \subset \mathbb{X}$. For each $k \in \mathbb{N}$, define the subsequence $\mathbb{X}_k$ of $\mathbb{X}$ by the formula:

$$\mathbb{X}_k = \{x_{k + n -1 }: n \in \mathbb{N}\}.$$

Note that $\mathbb{X}_1 = \mathbb{X}$. Also for each $n \in \mathbb{N}$, we have $x_n\in\mathbb{X}_n$: in fact $x_n$ is the first term of the subsequence $\mathbb{X}_n$. Note that each set $\mathbb{X}_k$ is bounded by $K$, so each has a supremum and an infimum. Also from the definition of $\mathbb{X}_k$, we see that $\mathbb{X}_j \subset \mathbb{X}_k$, whenever $j \ge k$. For each $k \in \mathbb{N}$ define:

$$b_k = \sup(\mathbb{X}_k), \hspace{10pt} a_k = \inf(\mathbb{X}_k),$$

$$\mathbb{B}= \{b_k: k \in \mathbb{N}\}, \hspace{10pt} \mathbb{A}= \{a_k: k \in \mathbb{N}\}.$$

Then by definition of $a_k$ and $b_k$ and by the fact that $x_k \in \mathbb{X}_k$, we have the inequality, valid for each $k \in \mathbb{N}$:

$$a_k \le x_k \le b_k.$$

Also, when $j \ge k$, since $\mathbb{X}_j \subset \mathbb{X}_k$, we have $a_k \le a_j$ and $b_j \le b_k$.
So the sequence $\mathbb{A}$ is monotonic increasing and the sequence $\mathbb{B}$ is monotonic decreasing.
Define $\mathbb{J}_k = [a_k, b_k]$, for each $k \in \mathbb{N}$.
Then we have just shown that $\{ \mathbb{J}_k: k \in \mathbb{N}\}$ is a nested sequence of closed bounded intervals, since the sequence $\mathbb{A}$ is increasing, the sequence $\mathbb{B}$ is decreasing and we have $a_k \le b_k$, for all $k \in \mathbb{N}$. Put $\mathbb{J} = \bigcap_{k = 1}^\infty \mathbb{J}_k$.
Then by the theory of nested intervals, we have $\mathbb{J} = [\alpha, \beta]$, where $\alpha = \sup(\mathbb{A})$ and $\beta = \inf(\mathbb{B})$ and $\alpha \le \beta$, with $\beta - \alpha = \inf(\{b_k - a_k: k \in \mathbb{N}\}$.
By definition, we write:

$$\alpha = \lim\inf(\mathbb{X}) = \sup(\mathbb{A}), \hspace{10pt} \beta = \lim\sup(\mathbb{X}) = \inf(\mathbb{B}).$$

The first observation is that if $\alpha = \beta$, so the sequences $\mathbb{A}$ and $\mathbb{B}$ have the same limit, then by squeeze applied to the inequality $a_k \le x_k \le b_k$, valid for all $k \in \mathbb{N}$, we get that the sequence $\mathbb{X}$ converges with limit $\alpha$.

Next we construct a subsequence of $\mathbb{X}$ with limit $\beta$.
Let $\epsilon = \{\epsilon_j: j \in \mathbb{N}\}$ be a sequence of positive real numbers, with limit zero. For example we could take $\epsilon_j = j^{-1}$, for each $j \in \mathbb{N}$.

• Put $N_1 = 1$.
  First, since $b_{N_1} - \epsilon_{1} = b_1 - \epsilon_1$ is less than the supremum $b_{N_1}$ of the set $\mathbb{X}_{N_1} = \mathbb{X}_1 = \mathbb{X}$, there is an $n_1 \in \mathbb{N}$, so $n_1 \ge N_1$, such that:
  $$b_{N_1} - \epsilon_1 < x_{n_1} \le b_{N_1}.$$

• Pick an integer $N_2 > n_1$.
  Then since $b_{N_2} - \epsilon_{2}$ is less than the supremum $b_{N_2}$ of $\mathbb{X}_{N_2}$, there is an $n_2 \in \mathbb{N}$, with $n_2 \ge N_2$, such that:
  $$b_{N_2} - \epsilon_2 < x_{n_2} \le b_{N_2}.$$

• Pick an integer $N_3 > n_2$.
  Then since $b_{N_3} - \epsilon_{3}$ is less than the supremum $b_{N_3}$ of $\mathbb{X}_{N_2}$, there is an $n_3 \in \mathbb{N}$, with $n_3 \ge N_3$, such that:
  $$b_{N_3} - \epsilon_3 < x_{n_3} \le b_{N_3}.$$

We continue forever, yielding sequences $\{N_k: k \in \mathbb{N}\}$ and $\{n_k: k \in \mathbb{N}\}$ such that each sequence is strictly increasing and they are interwoven in the sense that $N_k \le n_k < N_{k + 1} \le n_{k + 1}$, for each $k \in \mathbb{N}$ and such that we have, for each $k \in \mathbb{N}$, the inequality:

$$b_{N_k} - \epsilon_k < x_{n_k} \le b_{N_k}.$$

Then the sequence $\{b_{N_k}:k \in \mathbb{N}\}$ is a subsequence of the sequence $\mathbb{B}$, so converges with limit $\beta$. So, by squeeze applied to the inequality $b_{N_k} - \epsilon_k < x_{n_k} \le b_{N_k}$, valid for any $k \in \mathbb{N}$, we have that the subsequence $\{ x_{n_k}: k \in \mathbb{N}\}$ also converges, with the same limit $\beta$. Similarly we construct a subsequence of the sequence $\mathbb{X}$ with limit $\alpha$.
Let $\zeta = \{\zeta_j: j \in \mathbb{N}\}$ be a sequence of positive real numbers, with limit zero.

• Put $M_1 = 1$.
  First, since $a_{M_1} + \zeta_{1} = a_1 + \zeta_1$ is greater than the infimum $a_{M_1}$ of $\mathbb{X}_{M_1} = \mathbb{X}_1 = \mathbb{X}$, there is an $m_1 \in \mathbb{N}$, so $m_1 \ge M_1$, such that:
  $$a_{M_1} \le x_{m_1} < a_{M_1}+ \zeta_1.$$

• Pick $M_2 > m_1$.
  Then since $a_{M_2} + \zeta_{2}$ is greater than the infimum $a_{M_2}$ of $\mathbb{X}_{M_2}$, there is an $m_2 \in \mathbb{N}$, with $m_2 \ge M_2$, such that:
  $$a_{M_2} \le x_{m_2} < a_{M_2}+ \zeta_2.$$

• Pick $M_3 > m_2$.
  Then since $a_{M_3} + \zeta_{3}$ is greater than the infimum $a_{M_3}$ of $\mathbb{X}_{M_3}$, there is an $m_3 \in \mathbb{N}$, with $m_3 \ge M_3$, such that:
  $$a_{M_3} \le x_{m_3} < a_{M_3}+ \zeta_3.$$

We continue forever, yielding sequences $\{M_k: k \in \mathbb{N}\}$ and $\{m_k: k \in \mathbb{N}\}$ such that each sequence is strictly increasing and they are interwoven in the sense that $M_k \le m_k < M_{k + 1} \le m_{k + 1}$, for each $k \in \mathbb{N}$ and such that we have, for each $k \in \mathbb{N}$, the inequality:

$$a_{M_k} \le x_{m_k} \le a_{M_k} + \zeta_k.$$

Then the sequence $\{a_{M_k}:k \in \mathbb{N}\}$ is a subsequence of the sequence $\mathbb{A}$, so converges with limit $\alpha$. So, by squeeze applied to the inequality $a_{M_k} \le x_{m_k} \le a_{M_k} + \zeta_k$, valid for any $k \in \mathbb{N}$, we have that the subsequence $\{ x_{m_k}: k \in \mathbb{N}\}$ also converges, with the same limit $\alpha$.

So we proved that the sequence $\mathbb{X}$ has two subsequences, one with limit $\beta$ and another with limit $\alpha$. Consequently, if $\alpha \ne \beta$, the sequence $\mathbb{X}$ cannot be convergent.

So we have proved:

• A bounded sequence $\mathbb{X}$ is convergent if and only if $\lim\inf(\mathbb{X}) = \lim\sup(\mathbb{X})$ and then its limit is $\lim\inf(\mathbb{X})$.

Also we have proved:

• The Bolzano-Weierstrass Theorem: every bounded sequence has a convergent subsequence.

In fact we have just constructed one or two such subsequences, one with limit $\lim\inf(\mathbb{X})$ the other with limit $\lim\sup(\mathbb{X})$.

Next let $\mathbb{Y}$ be any subsequence of $\mathbb{X}$.
Then the sequence $\mathbb{Y}_k= \{ y_j: j \ge k, j \in \mathbb{N}\}$ is a subsequence of the sequence $\mathbb{X}_k = \{x_m: m \ge k\}$, since $y_j = x_{n_j}$, for some $x_{n_j}$ and $n_j \ge j$, for each $j \in \mathbb{N}$.
So $a'_k = \inf(\mathbb{Y}_k)\ge a_k$ and $b'_k = \sup(\mathbb{Y}_k)\ge b_k$, so we have:
$\alpha' = \lim\inf(\mathbb{Y}) = \lim_{k \rightarrow \infty} a_k' \ge \alpha$ and $\beta' = \lim\sup(\mathbb{Y}) = \lim_{k \rightarrow \infty} b'_k \le \beta$.
In particular, if $\mathbb{Y}$ is convergent, its limit is $\alpha' = \beta'$, so we have proved:

• If $\mathbb{Y}$ is a subsequence of a bounded sequence $\mathbb{X}$, then we have the inequality:
  $$\lim\inf(\mathbb{X})) \le \lim\inf(\mathbb{Y})\le \lim\sup(\mathbb{Y}) \le \lim\sup(\mathbb{X}).$$

• If $\mathbb{Y}$ is a convergent subsequence of a bounded sequence $\mathbb{X}$, then its limit $\lim(\mathbb{Y})$ satisfies the inequality:
  $$\lim\inf(\mathbb{X}) \le \lim(\mathbb{Y}) \le \lim\sup(\mathbb{X}).$$

Finally we have the result:

• Let $\mathbb{X}$ be a bounded sequence. Suppose that every convergent subsequence of $\mathbb{X}$ has the same limit $L$. Then $\mathbb{X}$ is also convergent with limit $\mathbb{L}$.

This follows since we have shown that there is a convergent subsequence of $\mathbb{X}$ with limit $\lim\sup(\mathbb{X})$ and another convergent subsequence of $\mathbb{X}$ with limit $\lim\inf(\mathbb{X})$ so these limits must be equal and equal to $L$, so by the above results, since $\lim \inf(\mathbb{X}) = \lim \sup(\mathbb{X}) = L$, we have that $\mathbb{X}$ converges with limit $L$.

The converse of course is true also: if a sequence $\mathbb{X}$ converges with limit $L$ then all subsequences of $\mathbb{X}$ converge with the same limit $L$.