Theorem:Bolzano-Weierstrass

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 $a\leq a_{n}\leq b_{n}$ for all $n\in \mathbb {N}$. Therefore the set $\{a_{n}:n\in \mathbb {N} \}$ is contained in the interval $I_{1}=[a,b]$.

<img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-22%20at%2011.36.10%20PM.png" alt="Screen%20Shot%202014-10-22%20at%2011.36.10%20PM.png" class="image" />

• 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 $I_{1}''$.

<img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-23%20at%2012.00.55%20AM.png" alt="Screen%20Shot%202014-10-23%20at%2012.00.55%20AM.png" class="image" />

• 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 $B_{2}=\{n\in \mathbb {N} :n>n_{2},a_{n}\in I_{2}''\}$.

<img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-23%20at%2012.14.38%20AM.png" alt="Screen%20Shot%202014-10-23%20at%2012.14.38%20AM.png" class="image" />

• Once again, if $A_{2}$ is infinite, then we will let $I_{3}=I_{2}'$, and let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 $B_{3}=\{n\in \mathbb {N} :n>n_{3},a_{n}\in I_{3}''\}$

<img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-23%20at%2012.23.20%20AM.png" alt="Screen%20Shot%202014-10-23%20at%2012.23.20%20AM.png" class="image" />

• 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 $k\in \mathbb {N}$.

<img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-23%20at%2012.30.02%20AM.png" alt="Screen%20Shot%202014-10-23%20at%2012.30.02%20AM.png" class="image" />

• 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:

${\begin{aligned}\mid a_{n_{k}}-\xi \mid \leq {\frac {b-a}{2^{k-1}}}\end{aligned}}$

• Therefore the subsequence $(a_{n_{k}})$ converges to $\xi$. $\blacksquare$

• Proof 2: Let $(a_{n})$ be a bounded sequence. By the <a href="/the-monotone-subsequence-theorem">The Monotone Subsequence Theorem</a>, 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 <a href="/the-monotone-convergence-theorem">The Monotone Convergence Theorem</a>, $(a_{n_{k}})$ is a convergent subsequence. $\blacksquare$

Theorem 2: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} is a bounded sequence such that every convergent subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{n_k})} of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} .

• Proof: Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A = (a_n)} does not converge to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} . Then there exists a subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A' = (a_{n_k})} and an Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_0 > 0} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid a_{n_k} - L \mid \geq \epsilon_0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in \mathbb{N}} .

• Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} is bounded, any subsequence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} is also bounded, and so the subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A'} is bounded. Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A'} is bounded, then by the Bolzano-Weierstrass theorem there exists a convergent subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A''} of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A'} . Since $A''$ is a convergent subsequence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A'} , then it is also a convergent sequence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} by transitivity, and by the hypothesis, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A''} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} .

• But then convergence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A''} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} contradicts the assumption that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid a_{n_k} - L \mid \geq \epsilon_0} , and so the assumption that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} did not converge to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} was false. Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}