MATH 409 Lecture 7

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Lecture Slides

Review

• Limits of Sequences
• Properties of Convergent sequences
  • Limit is unique
  • Convergent sequence is bounded
  • Subsequence of convergent sequence converges to same ilmit
  • Modifying a finite number of elements can't affect convergence or change its limit
  • Rearranging elements of a sequence can't affect convergence or change its limit
• Limit Theorems

Examples

Theorem. $\lim _{n\to \infty }{\frac {\sin {\mathrm {e} ^{n}}}{n}}=0$

Proof. $-{\frac {1}{n}}\leq {\frac {\sin {\mathrm {e} ^{n}}}{n}}\leq {\frac {1}{n}}$ for all $n\in \mathbb {N}$ since $-1\leq \sin {x}\leq 1$ for all $x\in \mathbb {R}$. As shown last time, ${\frac {1}{n}}\to 0$ as $n\to \infty$. Then $-{\frac {1}{n}}\to -1\cdot 0=0$ as $n\to \infty$. By the Squeeze theorem, ${\frac {\sin {\mathrm {e} ^{n}}}{n}}=0$ as $n\to \infty$.

quod erat demonstrandum

Theorem. $\lim _{n\to \infty }{\frac {1}{2^{n}}}=0$

Proof. The sequence $\left\{{\frac {1}{2^{n}}}\right\}$ is a subsequence of $\left\{{\frac {1}{n}}\right\}$. Hence it converges to the same limit.

quod erat demonstrandum

Theoreom. $\lim _{n\to \infty }{\frac {(1+2n)^{2}}{1+2n^{2}}}=2$

Proof. Divide both numerator and denominator by $n^{2}$:

${\begin{aligned}{\frac {(1+2n)^{2}}{1+2n^{2}}}&={\frac {\frac {(1+2n)^{2}}{n^{2}}}{\frac {1+2n^{2}}{n^{2}}}}={\frac {\left({\frac {1}{n}}+2\right)^{2}}{\left({\frac {1}{n}}\right)^{2}+2}}\end{aligned}}$

Since ${\frac {1}{n}}\to 0$ as $n\to \infty$, it follows that

• ${\frac {1}{n}}+2\to 0+2=2$ as $n\to \infty$,
• $\left({\frac {1}{n}}+2\right)^{2}\to 2^{2}=4$ as $n\to \infty$
• $\left({\frac {1}{n}}\right)^{2}\to 0^{2}=0$ as $n\to \infty$
• $\left({\frac {1}{n}}\right)^{2}+2\to 0+2=2$ as $n\to \infty$

Finally, ${\frac {\left({\frac {1}{n}}+2\right)^{2}}{\left({\frac {1}{n}}\right)^{2}+2}}\to {\frac {4}{2}}=2$ as $n\to \infty$.

quod erat demonstrandum

Monotone Sequences

A sequence $\left\{x_{n}\right\}$ is called increasing (or nondecreasing) if $x_{1}\leq x_{2}\leq x_{3}\leq \dots$, or, to be precise, $x_{n}\leq x_{n+1}$ for all $n\in \mathbb {N}$.

It is called strictly increasing if $x_{n}<x_{n+1}$ for all $n\in \mathbb {N}$.

Note: These definitions are according to the textbooks. Other people prefer to use nondecreasing for ≤ and increasing for strictly increasing.

Similarly, $\left\{x_{n}\right\}$ is called decreasing (or nonincreasing) if $x_{n}\geq x_{n+1}$ for all $n\in \mathbb {N}$.

It is called strictly decreasing if $x_{n}>x_{n+1}$ for all $n\in \mathbb {N}$.

Examples

• $\left\{{\frac {1}{n}}\right\}$ is strictly decreasing
• The sequence 1, 1, 2, 2, 3, 3, … is increasing, but not strictly increasing
• The sequence -1, 1, -1, 1, … is neither increasing nor decreasing
• A constant sequence $a$, $a$, … is both increasing and decreasing

Theorems

Theorem. Any increasing sequence converges to a limit if it is bounded, and diverges to $+\infty$ otherwise.

Proof. Let $\left\{x_{n}\right\}$ be an increasing sequence.

First consider the case when $\left\{x_{n}\right\}$ is bounded. In this case, the set $E$ of all elements occurring in the sequence is bounded. By the completeless axiom, any bounded set has a supremum; thus $\sup {E}$ exists.

We claim that $x_{n}\to \sup {E}$ as $n\to \infty$. Take any $\epsilon >0$. Then $\sup {E}-\epsilon$ is not an upper bound of $E$. Hence there exists $n_{0}\in \mathbb {N}$ such that $x_{n_{0}}>\sup {E}-\epsilon$. Since the sequence is increasing, it follows that $x_{n}\geq x_{n_{0}}>\sup {E}-\epsilon$ for $n\geq n_{0}$. At the same time, $x_{n}\leq \sup {E}$ for all $n\in \mathbb {N}$. Therefore $\left|x_{n}-\sup {E}\right|<\epsilon$ for all $n\geq n_{0}$, which proves the claim.

Now consider the case when $\left\{x_{n}\right\}$ is not bounded. Note that the set $E$ is bounded below (as $x_{1}$ is a lower bound). Hence $E$ is not bounded above. Then for any $C\in \mathbb {R}$, there exists $n_{0}\in \mathbb {N}$ such that $x_{n_{0}}>C$. It follows that $x_{n}\geq x_{n_{0}}>C$ for all $n\geq n_{0}$. Thus $\left\{x_{n}\right\}$ diverges to $+\infty$.

quod erat demonstrandum

Theorem. Any decreasing sequence converges to a limit if it is bounded, and diverges to $-\infty$ otherwise.

Proof. Let $\left\{x_{n}\right\}$ be a decreasing sequence. Then the sequence $\left\{-x_{n}\right\}$ is increasing since $a\geq b$ is equivalent to $-a\leq -b$ for all $a,b\in \mathbb {R}$. By the previous theorem, either $-x_{n}\to c$ for some $c\in \mathbb {R}$ as $n\to \infty$, or else $-x_{n}$ diverges to $+\infty$. In the former case, $x_{n}\to -c$ as $n\to \infty$. In the latter case, $x_{n}$ diverges to $-\infty$.

quod erat demonstrandum

Corollary. Any monotone sequence converges to a limit if it is bounded, and diverges to infinity otherwise.

Nested Intervals Property

A sequence of sets $I_{1},I_{2},\ldots$ is called nested if $I_{1}\supset I_{2}\supset \ldots$. That is, $I_{n}\supset I_{n+1}$ 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 n \in \mathbb{N}} .

Theorem. If $\left\{I_{n}\right\}$ is a nested sequence of nonempty closed bounded intervals, then the intersection 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 \bigcap_{n \in \mathbb{N}} I_n} is nonempty. Moreover, if lengths 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 \left| I_n \right|} of the intervals satisfy 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 \left| I_n \right| \to 0} as 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 \to \infty} , then the intersetion consists of a single point.

Proof. 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 I_n = \left[ a_n, b_n \right]} for 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 \in \mathbb{N}} . Since the sequence 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 \left\{ I_n \right\}} is nested, it follows that the sequence 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 \left\{ a_n \right\}} is increasing while the sequence 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 \left\{ b_n \right\}} is decreasing. Besides, both sequences are bonded (since both are contained in the bounded interval 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 I_1} ). Hence both are convergent: 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 \to a} and 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 b_n \to b} as 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 \to \infty} . 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_n \le b_n} 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 n \in \mathbb{N}} , the Comparison Theorem implies 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 \le b} .

We claim 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 \bigcap_{n \in \mathbb{N}} I_n = \left[ a,b \right]} . Indeed, we have 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 \le a} 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 n \in \mathbb{N}} (by Comparison Theorem applied 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 a_1, a_2, \ldots} and constant sequence 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, a_n, \ldots} ). Similarly, 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 b \le b_n} 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 n \in \mathbb{N}} . 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 \left[ a,b \right]} is contained in the intersection.

On the other hand, 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 x < a} , 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 x < a_n} for some 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} so 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 x \not\in I_n} . Similarly, 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 x > b} 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 x > b_m} for some 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 m} so 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 x \not\in I_m} . This proves the claim.

Clearly, the length 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 \left[ a,b \right]} cannot exceed 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 \left| I_n \right|} for any 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 \in \mathbb{N}} . 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 \left| I_n \right| \to 0} as 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 \to \infty} implies 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 \left[ a,b \right]} is a degenerate interval: 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 = b} .

quod erat demonstrandum

A few remarks:

1. The theorem may not hold if the intervals are open. Counterexample: 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 I_n = \left( 0,\frac{1}{n} \right)} , 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 \in \mathbb{N}} The intervals are nested and bounded, but their intersection is empty 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 \frac{1}{n} \to 0} as 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 \to \infty} .
2. The theorem may not hold if the intervals are not bounded. Counterexample: 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 I_n = \left[ n, \infty \right)} , 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 \in \mathbb{N}} . The intervals are nested and closed, but their intersection is empty since the sequence 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 \left\{ n \right\}} diverges 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 +\infty} .
3. This is another form of the completeness axiom.
4. This theorem can be used to prove the existence of a decimal expansion

Bolzano-Weierstrass Theorem

Theorem. Every bounded sequence of real numbers has a convergent subsequence.

Proof. 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 \left\{ x_n \right\}} be a bounded sequence of real numbers. We are going to build a nested sequence of intervals 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 I_n = \left[ a_n, b_n \right]} for 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 \in \mathbb{N}} , such that each 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 I_n} contains infinitely many elements 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 \left\{ x_n \right\}} and $\left|I_{n+1}\right|={\frac {\left|I_{n}\right|}{2}}$ for all $n\in \mathbb {N}$. The sequence is built inductively.

Basis. First we set $I_{1}$ to be any closed bounded interval that contains all elements of $\left\{x_{n}\right\}$ (such an interval exists because the sequence $\left\{x_{n}\right\}$ is bounded).

Induction. Now assume that for some 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 \in \mathbb{N}} the interval 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 I_n} is already chosen and it contains infinitely many elements 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 \left\{ x_n \right\}} . Then at least one of the subintervals 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 I' = \left[ a_n, \frac{a_n + b_n}{2} \right]} and 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 I'' = \left[ \frac{a_n + b_n}{2}, b_n \right]} also contains infinitely many elements 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 \left\{ x_n \right\}} . We set 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 I_{n+1}} to be such an interval. By construction, 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 I_{n+1} \subset I_n} and 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 \left| I_{n+1} \right| = \frac{\left| I_n \right|}{2}} .

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 \left| I_{n+1} \right| = \frac{\left| I_n \right|}{2}} 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 n \in \mathbb{N}} , it follows by induction 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 \left| I_n \right| = \frac{\left| I_1 \right|}{2^{n-1}}} 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 n \in \mathbb{N}} . As a consequence, 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 \left| I_n \right| \to 0} as 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 \to \infty} . By the Nested Intervals Property, the intersection of the intervals consists of a single number 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} .

Next we are going to build a strictly increasing sequence of natural numbers 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 \left\{ n_k \right\}} 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 x_{n_k} \in I_k} 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}} . The sequence is built inductively:

Basis. First 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_1 = 1} .

Induction. Now assume that for some $k\in \mathbb {N}$ the number 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_k} is already chosen. Since the interval 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 I_{k+1}} contains infinitely many elements of the sequence 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 \left\{ x_n \right\}} , there exists 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 m > n_k} such that $x_{m}\in I_{k+1}$. We set 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_{k+1} = m} .

Now we claim that 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 \left\{ x_{n_k} \right\}_{k \in \mathbb{N}}} of the sequence 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 \left\{ x_n \right\}} 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 a} . Indeed, for any $k\in \mathbb {N}$, the points 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 x_{n_k}} and 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} both belong to the interval 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 I_k} . Hence 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 \left| x_{n_k} - a \right| \le \left| I_k \right|} . Since $\left|I_{k}\right|\to 0$ as 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 \to \infty} , it follows 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 x_{n_k} \to a} as 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 \to \infty} .

quod erat demonstrandum