The Bolzano Weierstrass Theorem

1. Audio
   Observation

   The speaker says, "So let's let our sequence a sub n, n goes from one to infinity, be such a bounded sequence."

2. Formula
   Observation

   He writes the sequence notation on the board.

\{$a_n$\}_{$n=1$}^{$\infty$}

A sequence of real numbers indexed by natural numbers starting at 1.

$n \in \mathbb{N}$, $n \ge 1$

1. Audio
   Observation

   The speaker says, "let's call this point over here c zero, d zero, some closed c zero, d zero, where all of the sequence resides."

2. Diagram
   Observation

   He draws a horizontal line representing the real line and marks two points labeled $c_0$ and $d_0$ with a bracketed interval between them.

$c_0$, $d_0$

Real numbers defining the endpoints of a closed interval that contains all terms of the bounded sequence.

$c_0$, $d_0 \in \mathbb{R}$

1. Audio
   Observation

   The speaker says, "since the sequence is bounded, we have c zero less than or equal to a sub n less than or equal to d zero for all n."

2. Formula
   Observation

   He writes the inequality on the board.

$c_0 \le a_n \le d_0 \text{ for all } n$

The mathematical definition of the sequence being bounded within the interval [$c_0$, $d_0$].

$n \in \mathbb{N}$, $n \ge 1$

1. Formula
   Observation

   Top board text reads "Let {$a_n$}_{$n=1$}^$\infty$ be such a bounded sequence."

{$a_n$}_{$n=1$}^$\infty$

A bounded sequence of real numbers used as the object in the Bolzano-Weierstrass argument.

Sequence indexed by positive integers n.

1. Formula
   Observation

   Top board text reads "We have $c_0 \le a_n \le d_0$ for all n."

2. Diagram
   Observation

   A number line shows endpoints labeled $c_0$ and $d_0$ with the sequence points inside that interval.

$c_0$, $d_0$

Lower and upper bounds of the initial closed interval [$c_0$,$d_0$] containing all terms of the bounded sequence.

Real numbers with $c_0 \le d_0$.

1. Formula
   Observation

   The midpoint label on the number line is written as ($c_0+d_0$)/2.

($c_0+d_0$)/2

Midpoint of the initial interval [$c_0$,$d_0$], used to split it into two halves.

Real number between $c_0$ and $d_0$.

1. Formula
   Observation

   The instructor writes [$c_0$,$c_1$] ∪ [$c_1$,$d_0$].

2. Audio
   Observation

   He says he will call the middle point $c_1$.

$c_1$

The midpoint of [$c_0$,$d_0$], introduced as the dividing point for the first bisection step.

Real number with $c_1=(c_0+d_0)/2$.

1. Formula
   Observation

   The board shows $I_{1,1}$ ∪ $I_{1,2}$.

2. Audio
   Observation

   He names the two subintervals $I_{1,1}$ and $I_{1,2}$.

$I_{1,1}$, $I_{1,2}$

The two half-intervals obtained from splitting [$c_0$,$d_0$] at $c_1$.

Closed subintervals of [$c_0$,$d_0$].

1. Formula
   Observation

   The board adds "call this $\widetilde{I}_1$".

2. Audio
   Observation

   He says one of the two intervals contains infinitely many $a_n$ terms and calls it $I_1$ tilde.

$\widetilde{I}_1$

The chosen one of $I_{1,1}$ or $I_{1,2}$ that contains infinitely many terms of the sequence.

A closed interval selected from the first bisection.

1. Formula
   Observation

   Green writing shows $\widetilde{I}_1$ ⊇ $\widetilde{I}_2$ ⊇ $\widetilde{I}_3$ ⊇ ... ⊇ $\widetilde{I}_k$.

2. Audio
   Observation

   He says they have constructed this nested collection.

$\widetilde{I}_1$,$\widetilde{I}_2$,$\widetilde{I}_3$,$\ldots$,$\widetilde{I}_k$

A finite displayed stage of a nested family of chosen intervals in the induction construction.

Closed intervals with reverse inclusion ordering.

1. Formula
   Observation

   Green writing shows |$\widetilde{I}_k$| = ($d_0-c_0$)/2^k.

2. Audio
   Observation

   He states the length of $I_k$ tilde is half the length of the overall thing, giving ($d_0-c_0$)/2^k.

|$\widetilde{I}_k$|

Length (measure) of the kth chosen nested interval.

Nonnegative real number.

1. Formula
   Observation

   The board states "Let {$a_n$}_{$n=1$}^$\infty$ be such a bounded sequence" and repeatedly refers to the terms $a_n$.

{$a_n$}_{$n=1$}^$\infty$

A bounded sequence of real numbers whose convergent subsequence is being constructed.

Sequence indexed by positive integers n ∈ ℕ; each term $a_n$ ∈ ℝ.

1. Audio
   Observation

   The speaker states the condition for a bounded sequence.

2. Formula
   Observation

   The inequality $c_0 \le a_n \le d_0$ is written on the board.

A sequence of real numbers is bounded if there exist real numbers $c_0$ and $d_0$ such that every term of the sequence lies within the closed interval [$c_0$, $d_0$].

1. The sequence consists of real numbers.

2. $c_0$ and $d_0$ are fixed real numbers.

1. Audio
   Observation

   The speaker explains the idea of building a nested collection of closed intervals by cutting the initial interval into two parts.

2. Formula
   Observation

   He writes the union of the two sub-intervals on the board.

To extract a convergent subsequence from a bounded sequence, one constructs a nested collection of closed intervals. This is initiated by dividing the initial bounding interval [$c_0$, $d_0$] into two halves at its midpoint.

1. An initial closed interval containing all terms of the sequence must exist.

1. Definition of a Bounded Sequence

1. Formula
   Observation

   Title text on the board reads "Bolzano-Weierstrass: Any bounded sequence of real numbers contains a convergent subsequence."

The clip opens with the theorem statement written across the top: every bounded sequence of real numbers has a convergent subsequence. The rest of the segment develops the bisection/nested-interval construction used to prove it.

1. The sequence consists of real numbers.

2. The sequence is bounded.

1. Formula
   Observation

   Board text: "Let {$a_n$}_{$n=1$}^$\infty$ be such a bounded sequence. We have $c_0 \le a_n \le d_0$ for all n."

2. Diagram
   Observation

   A number line marks $c_0$, the midpoint, and $d_0$, with the sequence contained in that interval.

Because the sequence is bounded, all its terms lie in some closed interval [$c_0$,$d_0$]. This interval is the starting object for the proof, and the midpoint ($c_0+d_0$)/2 is drawn explicitly on the number line.

1. {$a_n$} is bounded.

2. $c_0$ and $d_0$ are chosen as lower and upper bounds.

1. Bolzano-Weierstrass theorem statement shown on board

1. Formula
   Observation

   The instructor writes [$c_0$,($c_0+d_0$)/2] ∪ [($c_0+d_0$)/2,$d_0$].

2. Formula
   Observation

   He then rewrites the same split as [$c_0$,$c_1$] ∪ [$c_1$,$d_0$].

3. Audio
   Observation

   He says he will call the middle point $c_1$.

The proof begins by cutting the initial interval at its midpoint. The two resulting closed subintervals are named $I_{1,1}$ and $I_{1,2}$. This is the first step of the nested-interval construction.

1. Use the midpoint of the current interval.

2. Both subintervals remain closed.

1. Initial enclosing interval for the bounded sequence

1. Formula
   Observation

   Blue writing states "One of these intervals contains infinitely many of the $a_n$".

2. Audio
   Observation

   He explains that if both contained only finitely many terms, the whole sequence would be finite, contradicting that it is infinite.

After splitting [$c_0$,$d_0$], the instructor argues by contradiction that at least one of the two halves must contain infinitely many terms of the sequence. If both halves had only finitely many terms, their union would contain only finitely many terms, which is impossible because the sequence has infinitely many terms.

1. The original interval contains all terms of an infinite sequence.

2. The interval is split into two pieces.

1. Splitting [$c_0$,$d_0$] into two half-intervals

1. Formula
   Observation

   Board text adds "call this $\widetilde{I}_1$".

2. Audio
   Observation

   He says $I_1$ tilde is either $I_{1,1}$ or $I_{1,2}$.

Once one of the two half-intervals is identified as containing infinitely many terms, it is renamed $\widetilde{I}_1$. This notation records the selection step needed to build the nested family.

1. Choose one of the two subintervals that contains infinitely many $a_n$ terms.

1. At least one half-interval contains infinitely many sequence terms

1. Formula
   Observation

   Green writing shows $\widetilde{I}_1$ ⊇ $\widetilde{I}_2$ ⊇ $\widetilde{I}_3$ ⊇ ... ⊇ $\widetilde{I}_k$.

2. Audio
   Observation

   He says there is at least one step in the induction and assumes we have constructed this nested collection.

The proof proceeds by induction: after choosing $\widetilde{I}_1$, repeat the same bisection-and-selection process to obtain $\widetilde{I}_2$, $\widetilde{I}_3$, and so on, each contained in the previous one. The displayed chain records the nested structure being built.

1. Each new interval is chosen from a bisection of the previous one.

2. Each chosen interval still contains infinitely many terms of the sequence.

1. Naming the chosen infinite-content interval as $\widetilde{I}_1$

1. Formula
   Observation

   Green writing shows |$\widetilde{I}_k$| = ($d_0-c_0$)/2^k.

2. Audio
   Observation

   He says the length is half the length of the overall thing, giving ($d_0-c_0$)/2^k.

Each bisection halves the interval length, so after k steps the chosen interval has length ($d_0-c_0$)/2^k. This formula is central to the later convergence argument because the lengths tend to 0.

1. Start from interval length $d_0-c_0$.

2. Each step bisects the current interval.

1. Inductive construction of a nested interval family

1. Formula
   Observation

   Top of board: "Bolzano-Weierstrass: Any bounded sequence of real numbers contains a convergent subsequence."

The theorem asserts that every bounded sequence of real numbers has at least one convergent subsequence. In this clip, the proof begins from a bounded sequence {$a_n$}_{$n=1$}^$\infty$ with $c_0 \le a_n \le d_0$ for all n.

1. The sequence consists of real numbers.

2. The sequence is bounded.

1. Audio
   Observation

   The instructor says they will construct the next interval by splitting Ĩ_k in half so that infinitely many $a_n$ are in Ĩ_{$k+1$}.

2. Formula
   Observation

   Board text: "Split Ĩ_k in half s.t. infinitely many $a_n$ are in Ĩ_{$k+1$}."

Given a nested interval Ĩ_k containing infinitely many terms of the sequence, split it into two halves. At least one half must contain infinitely many terms; choose that half as Ĩ_{$k+1$}. This produces a nested sequence of closed intervals whose lengths shrink by a factor of 2 each step.

1. Each Ĩ_k is a closed interval.

2. Each Ĩ_k contains infinitely many terms of {$a_n$}.

3. The construction proceeds inductively over k.

1. Bolzano-Weierstrass theorem statement on the board

1. Formula
   Observation

   Board writes |Ĩ_k| = ($d_0 - c_0$)/2^k.

2. Audio
   Observation

   The instructor says the measure result is exactly the same when splitting in half.

The initial bounding interval has length $d_0 - c_0$. Each bisection halves the previous interval length, so after k steps the selected interval has length ($d_0 - c_0$)/2^k.

1. $c_0 \le d_0$.

2. Each new interval is obtained by halving the previous interval.

1. Inductive construction of nested intervals

1. Audio
   Observation

   The speaker states the theorem verbally.

2. Formula
   Observation

   The speaker writes the theorem statement on the board.

Any bounded sequence of real numbers contains a convergent subsequence.

1. The sequence consists of real numbers.

2. The sequence is bounded.

For any bounded sequence of real numbers, there exists a convergent subsequence.

1. Formula
   Observation

   The theorem statement is written at the top of the board.

Any bounded sequence of real numbers contains a convergent subsequence.

1. The sequence is a sequence of real numbers.

2. The sequence is bounded.

For every bounded real sequence, there exists a convergent subsequence.

1. Formula
   Observation

   Written statement: "One of these intervals contains infinitely many of the $a_n$".

2. Audio
   Observation

   He justifies it by saying that if both had finitely many terms, the sequence would have finitely many terms.

If an interval containing all terms of an infinite sequence is split into two subintervals, then at least one of the two subintervals contains infinitely many terms of the sequence.

1. The original interval contains all terms of the sequence.

2. The sequence has infinitely many terms.

3. The interval is divided into two subintervals whose union is the original interval.

There exists at least one of the two subintervals containing infinitely many sequence terms.

1. Audio
   Observation

   The instructor says, "by the nested interval property, there is a point a which is in the intersection n goes from 1 to infinity of $I_n$ tilde."

2. Formula
   Observation

   Board writes: "By the nested interval property there is a point a ∈ ⋂_{$n=1$}^$\infty$ Ĩ_n."

1. The video names the nested interval property but does not separately state its full hypotheses on screen in this clip.

For the constructed nested sequence of closed intervals, there exists a point a belonging to every interval, i.e. a ∈ ⋂_{$n=1$}^$\infty$ Ĩ_n.

1. The intervals form a nested sequence.

2. The intervals are closed.

3. The construction has been carried out inductively for all positive integer indices.

There exists a real number a such that for every $n \ge 1$, a ∈ Ĩ_n.

1. Formula
   Observation

   Written explicitly on the top left of the blackboard.

Any bounded sequence of real numbers contains a convergent subsequence.

1. Let \{$a_n$\}_{$n=1$}^{$\infty$} be a bounded sequence of real numbers.

Universal quantification over all bounded sequences of real numbers; existential quantification over convergent subsequences.

1. Audio
   Observation

   The speaker outlines the initial steps of the proof, including defining the sequence, establishing its bounds, and proposing the method of nested intervals.

2. Formula
   Observation

   The speaker writes the sequence definition, the bounding inequality, and the split intervals on the board.

3. Diagram
   Observation

   The speaker draws the real line and the initial bounding interval.

1. Expression

   $$\{a_n\}_{n=1}^{\infty}$$
   Explanation

   Let the sequence be denoted as {$a_n$} from $n=1$ to infinity.

   Justification

   Definition of the object being analyzed in the theorem.

   Shown in the video
2. Expression

   $$c_0 \le a_n \le d_0 \text{ for all } n$$
   Explanation

   Since the sequence is bounded, there exist real numbers $c_0$ and $d_0$ such that all terms lie in the closed interval [$c_0$, $d_0$].

   Justification

   Definition of a bounded sequence.

   Shown in the video
3. Expression

   $$\left[c_0, \frac{c_0+d_0}{2}\right] \cup \left[\frac{c_0+d_0}{2}, d_0\right]$$
   Explanation

   The initial interval [$c_0$, $d_0$] is split into two closed sub-intervals at its midpoint.

   Justification

   Strategy to build a nested collection of closed intervals to extract a convergent subsequence.

   Shown in the video

The proof begins by establishing a bounded interval containing the sequence and splitting it into two halves to initiate the construction of nested intervals.

1. Formula
   Observation

   The board shows the split [$c_0$,($c_0+d_0$)/2] ∪ [($c_0+d_0$)/2,$d_0$], then [$c_0$,$c_1$] ∪ [$c_1$,$d_0$], then $I_{1,1}$ ∪ $I_{1,2}$.

2. Audio
   Observation

   He narrates naming the midpoint $c_1$ and the two subintervals.

1. Expression

   $$c_0 \le a_n \le d_0 \quad \text{for all } n$$
   Explanation

   Start from the boundedness assumption, so all sequence terms lie in [$c_0$,$d_0$].

   Justification

   Given directly on the board as the setup for the proof.

   Shown in the video
2. Expression

   $$c_1=\frac{c_0+d_0}{2}$$
   Explanation

   Introduce the midpoint of the initial interval.

   Justification

   Written on the number line and then named $c_1$ in the algebraic decomposition.

   Shown in the video
3. Expression

   $$[c_0,d_0]=[c_0,c_1]\cup[c_1,d_0]$$
   Explanation

   Split the original interval into two closed halves meeting at $c_1$.

   Justification

   Direct rewriting shown on the board.

   Shown in the video
4. Expression

   $$[c_0,d_0]=I_{1,1}\cup I_{1,2}$$
   Explanation

   Rename the two halves as $I_{1,1}$ and $I_{1,2}$.

   Justification

   Explicit notation added on the board.

   Shown in the video
5. Expression

   $$\text{One of } I_{1,1},I_{1,2} \text{ contains infinitely many } a_n$$
   Explanation

   At least one half must contain infinitely many terms of the sequence.

   Justification

   Stated on the board and justified verbally by contradiction: if both were finite, the whole sequence would be finite.

   Shown in the video
6. Expression

   $$\widetilde{I}_1 \in \{I_{1,1},I_{1,2}\}$$
   Explanation

   Choose one such half and call it $\widetilde{I}_1$.

   Justification

   Explicitly written as "call this $\widetilde{I}_1$".

   Shown in the video

The first step of the nested-interval construction produces a closed subinterval $\widetilde{I}_1$ that still contains infinitely many terms of the sequence.

1. Formula
   Observation

   Green writing shows $\widetilde{I}_1$ ⊇ $\widetilde{I}_2$ ⊇ $\widetilde{I}_3$ ⊇ ... ⊇ $\widetilde{I}_k$.

2. Audio
   Observation

   He says to assume we have constructed this nested collection and that each interval has length ($d_0-c_0$)/2^k.

1. The clip states the inductive hypothesis and displays the pattern, but the full induction proof for all k is not completed within this excerpt.

1. Expression

   $$\widetilde{I}_1 \supseteq \widetilde{I}_2 \supseteq \widetilde{I}_3 \supseteq \cdots \supseteq \widetilde{I}_k$$
   Explanation

   Assume a finite stage of the construction has produced a nested chain of chosen intervals.

   Justification

   Written explicitly in green on the board as the inductive setup.

   Shown in the video
2. Expression

   $$|\widetilde{I}_k|=\frac{d_0-c_0}{2^k}$$
   Explanation

   Record the length of the kth chosen interval.

   Justification

   Written explicitly on the board and explained verbally as halving at each step.

   Shown in the video

By the end of the clip, the proof has established the nested structure and the shrinking-length formula needed for the later convergence argument.

1. Formula
   Observation

   The board lays out the proof from the bounded sequence assumption through nested intervals, intersection point, subsequence selection, and convergence.

2. Audio
   Observation

   The instructor narrates the inductive split, invokes the nested interval property, picks $a_{n_k}$ ∈ Ĩ_k, bounds |$a_{n_k}-a$|, and concludes $a_{n_k} \to a$.

1. The clip does not show the initial proof of why one half must contain infinitely many terms; it is assumed from earlier context on the board.

1. Expression

   $$\text{Let } \{a_n\}_{n=1}^{\infty} \text{ be a bounded sequence with } c_0 \leq a_n \leq d_0 \text{ for all } n.$$
   Explanation

   Start with a bounded real sequence and enclose all its terms in the initial closed interval [$c_0$,$d_0$].

   Justification

   Stated directly on the board as the setup for the theorem.

   Shown in the video
2. Expression

   $$\widetilde{I}_1 \supseteq \widetilde{I}_2 \supseteq \widetilde{I}_3 \supseteq \cdots \supseteq \widetilde{I}_k, \quad |\widetilde{I}_k| = \frac{d_0-c_0}{2^k}$$
   Explanation

   Assume nested intervals have been constructed up to stage k, each containing infinitely many sequence terms and having length reduced by repeated halving.

   Justification

   Written on the board as the induction hypothesis for the construction.

   Shown in the video
3. Expression

   $$\text{Split } \widetilde{I}_k \text{ in half s.t. infinitely many } a_n \text{ are in } \widetilde{I}_{k+1}.$$
   Explanation

   To continue the construction, divide Ĩ_k into two halves and select a half that still contains infinitely many terms of the sequence.

   Justification

   Audio explanation plus board text; the underlying pigeonhole idea is that finitely many terms in both halves would imply only finitely many in their union.

   Shown in the video
4. Expression

   $$\exists a \in \bigcap_{n=1}^{\infty} \widetilde{I}_n$$
   Explanation

   The nested closed intervals have a common point a.

   Justification

   Explicit invocation of the nested interval property.

   Shown in the video
5. Expression

   $$\text{Pick } a_{n_k} \in \widetilde{I}_k$$
   Explanation

   For each k, choose a sequence term lying in the kth interval; these chosen terms form the candidate subsequence.

   Justification

   Possible because each Ĩ_k was constructed to contain infinitely many $a_n$.

   Shown in the video
6. Expression

   $$|a_{n_k} - a| \leq \frac{d_0-c_0}{2^k}$$
   Explanation

   Since both $a_{n_k}$ and a lie in Ĩ_k, their distance cannot exceed the length of Ĩ_k.

   Justification

   Interval containment plus the formula for |Ĩ_k|.

   Shown in the video
7. Expression

   $$\frac{d_0-c_0}{2^k} \to 0 \quad \text{as } k \to \infty$$
   Explanation

   The upper bound on |$a_{n_k}-a$| tends to zero because repeated division by 2 makes the interval lengths arbitrarily small.

   Justification

   Standard limit of a constant divided by $2^k$.

   Derived from the video
8. Expression

   $$a_{n_k} \to a$$
   Explanation

   Therefore the selected subsequence converges to a.

   Justification

   Comparison/squeeze argument using the bound |$a_{n_k}-a$| ≤ ($d_0-c_0$)/2^k and the fact that the right-hand side tends to 0.

   Shown in the video

A convergent subsequence {$a_{n_k}$} of the original bounded sequence has been constructed, proving the Bolzano-Weierstrass theorem for this case.

1. Formula
   Observation

   The entire right side and bottom left of the board detail the step-by-step construction of the proof.

2. Audio
   Observation

   Speaker concludes: 'And that's the proof of Bolzano-Weierstrass.'

1. Expression

   $$c_0 \le a_n \le d_0 \text{ for all } n$$
   Explanation

   Establish the initial bounds for the bounded sequence \{$a_n$\}.

   Justification

   Definition of a bounded sequence.

   Shown in the video
2. Expression

   $$[c_0, \frac{c_0+d_0}{2}] \cup [\frac{c_0+d_0}{2}, d_0]$$
   Explanation

   Split the initial interval [$c_0$, $d_0$] into two halves.

   Justification

   Bisection method.

   Shown in the video
3. Expression

   $$\text{One of these intervals contains infinitely many of the } a_n$$
   Explanation

   Identify which half-interval contains infinitely many terms of the sequence and label it $\widetilde{I}_1$.

   Justification

   Pigeonhole principle applied to infinite sets.

   Shown in the video
4. Expression

   $$\widetilde{I}_1 \supseteq \widetilde{I}_2 \supseteq \widetilde{I}_3 \supseteq \dots \supseteq \widetilde{I}_k$$
   Explanation

   Iteratively construct a nested sequence of closed intervals, where each $\widetilde{I}_k$ contains infinitely many terms of \{$a_n$\}.

   Justification

   Recursive application of the bisection and infinite subset selection.

   Shown in the video
5. Expression

   $$|\widetilde{I}_k| = \frac{d_0 - c_0}{2^k}$$
   Explanation

   Calculate the length of the k-th interval.

   Justification

   Each bisection halves the length of the previous interval.

   Shown in the video
6. Expression

   $$a \in \bigcap_{n=1}^{\infty} \widetilde{I}_n$$
   Explanation

   Assert the existence of a limit point 'a' common to all nested intervals.

   Justification

   Nested interval property of real numbers.

   Shown in the video
7. Expression

   $$\text{Pick } a_{q_k} \in \widetilde{I}_k$$
   Explanation

   Construct a subsequence by selecting one term from each interval $\widetilde{I}_k$.

   Justification

   Axiom of choice / constructive selection from infinite subsets.

   Shown in the video
8. Expression

   $$|a_{q_k} - a| \le \frac{d_0 - c_0}{2^k}$$
   Explanation

   Bound the distance between the subsequence term and the limit point 'a'.

   Justification

   Both points lie within the same interval $\widetilde{I}_k$ of length $\frac{d_0 - c_0}{2^k}$.

   Shown in the video
9. Expression

   $$a_{q_k} \to a$$
   Explanation

   Conclude that the subsequence converges to 'a'.

   Justification

   Comparison test, since $\frac{d_0 - c_0}{2^k} \to 0$ as $k \to \infty$.

   Shown in the video

The constructed subsequence \{$a_{q_k}$\} converges to the real number a, proving the Bolzano-Weierstrass theorem.

1. Diagram
   Observation

   The speaker draws a horizontal line with an arrow on the right, labeling it 'R' for the real line. He then marks two points on the line, labeling them '$c_0$' and '$d_0$', and draws a bracketed interval between them.

1. Horizontal line representing the real number line ($\mathbb{R}$)

2. Point labeled $c_0$

3. Point labeled $d_0$

4. Bracketed interval between $c_0$ and $d_0$

1. Drawing of the real line

2. Marking of points $c_0$ and $d_0$

3. Drawing of the interval [$c_0$, $d_0$]