The Bolzano-Weierstrass Theorem

# The Bolzano-Weierstrass Theorem

We will now look at a rather technical theorem known as the Bolzano Weierstrass Theorem which provides a very important result regarding bounded sequences and convergent subsequences.

Theorem 1 (Bolzano-Weierstrass): Let $(a_n)$ be a bounded sequence. Then there exists a subsequence of $(a_n)$, call it $(a_{n_k})$ that is convergent. |

**Proof 1:**Let $(a_n)$ be a bounded sequence, that is the set $\{ a_n : n \in \mathbb{N} \}$ is bounded. Suppose that $a$ is a lower bound for this set, and $b$ is an upper bound for this set, and so $a ≤ a_n ≤ b_n$ for all $n \in \mathbb{N}$. Therefore the set $\{ a_n : n \in \mathbb{N} \}$ is contained in the interval $I_1 = [a, b]$.

- Let's take $n_1 = 1$, i.e., let the first term of the subsequence $(a_{n_k})$ be equal to the first term of the parent sequence $(a_n)$.

- We will now take the interval $I_1$ and divide into two subintervals of equal length. We will call these intervals $I_1'$ and $I_1''$. We will also divide the set of indices into two sets. Let $A_1 := \{ n \in \mathbb{N} : n > n_1, a_n \in I_1' \}$ and let $B_1 = \{ n \in \mathbb{N} : n > n_1, a_N \in I_1'' \}$. We note that $A_1$ is the set of indices $n$ greater than the first index in our subsequence, $n_1$, such that $a_n$ is in the interval $I_1'$. Similarly, $B_1$ is the set of indices $n$ greater than the first index in our subsequence, $n_1$ such that $a_N$ is in the interval $I_1''$.

- We note that both of the sets $A_1$ and $B_1$ cannot be finite since $A_1$ and $B_1$ contain all of the indices of the sequence $(a_n)$ which has infinitely many indices since it is an infinite sequence. If $A_1$ is infinite, then we let $I_2 = I_1'$, and we will let $n_2$ be the smallest natural number index in the set $A_1$, which we are ensured to have by the Well Ordering Principle. If $A_1$ is finite, then $B_1$ is infinite and we let $I_2 = I_1''$ and we let $n_2$ be the smallest natural number index in the set $B_1$, which once again, we are ensured to have by the Well Ordering Principle.

- We will now take the interval $I_2$ and subdivide it into two subintervals of equal length, call them $I_2'$ and $I_2''$. Once again, we will subdivide the set of indices into two sets. Let $A_2 = \{ n \in \mathbb{N} : n > n_2, a_n \in I_2' \}$ and let $B_2 = \{ n \in \mathbb{N} : n > n_2, a_n \in I_2'' \}$.

- Once again, if $A_2$ is infinite, then we will let $I_3 = I_2'$, and let $n_3$ be the smallest natural number index of $A_2$. If $A_2$ is finite, then $B_2$ is infinite and let $n_3$ be the smallest natural number index of $B_2$.

- We will then take the interval $I_3$ and subdivide it into two subintervals of equal length, call them $I_3'$ and $I_3''$. We will also subdivide the set of indices into two sets. $A_3 = \{ n \in \mathbb{N} : n > n_3, a_n \in I_3' \}$ and $B_3 = \{ n \in \mathbb{N} : n > n_3, a_n \in I_3'' \}$

- We will continue this pattern to obtain a sequence of nested intervals $I_1 \supseteq I_2 \supseteq ... \supseteq I_k \supseteq ...$. Furthermore, we will also have a subsequence $(a_{n_k})$ of $(a_n)$ where $a_{n_k} \in I_k$ for every $k \in \mathbb{N}$.

- We also note that the length of any interval $I_k$ is precisely equal to $\frac{b - a}{2^{k-1}}$. By the nested intervals theorem there exists a common point $\xi \in I_k$ for all $k \in \mathbb{N}$, and so we have that:

\begin{align} \mid a_{n_k} - \xi \mid ≤ \frac{b - a}{2^{k-1}} \end{align}

- Therefore the subsequence $(a_{n_k})$ converges to $\xi$. $\blacksquare$

**Proof 2:**Let $(a_n)$ be a bounded sequence. By the The Monotone Subsequence Theorem, every sequence of real numbers has a monotonic subsequence. Let $(a_{n_k})$ be this monotonic subsequence. Since $(a_n)$ is bounded, it follows that $(a_{n_k})$ is also bounded since the values $a_{n_k}$ are contained in the sequence $(a_n)$. Since $(a_{n_k})$ is both bounded and monotonic, then by The Monotone Convergence Theorem, $(a_{n_k})$ is a convergent subsequence. $\blacksquare$

Theorem 2: If $(a_n)$ is a bounded sequence such that every convergent subsequence $(a_{n_k})$ of $(a_n)$ converges to $L$ then $(a_n)$ converges to $L$. |

**Proof:**Suppose that $A = (a_n)$ does not converge to $L$. Then there exists a subsequence $A' = (a_{n_k})$ and an $\epsilon_0 > 0$ such that $\mid a_{n_k} - L \mid ≥ \epsilon_0$ for all $k \in \mathbb{N}$.

- Since $A$ is bounded, any subsequence of $A$ is also bounded, and so the subsequence $A'$ is bounded. Since $A'$ is bounded, then by the Bolzano-Weierstrass theorem there exists a convergent subsequence $A''$ of $A'$. Since $A''$ is a convergent subsequence of $A'$, then it is also a convergent sequence of $A$ by transitivity, and by the hypothesis, $A''$ converges to $L$.

- But then convergence of $A''$ to $L$ contradicts the assumption that $\mid a_{n_k} - L \mid ≥ \epsilon_0$, and so the assumption that $A$ did not converge to $L$ was false. Therefore $(a_n)$ converges to $L$. $\blacksquare$