Convex Sets - Introduction

1. Audio
   Observation

   The speaker says the topic is special sets in $R^n$ called convex sets.

2. Formula
   Observation

   Later written expressions use x,$y \in \mathbb{R}^n$ and $C \subseteq \mathbb{R}^n$.

$\mathbb{R}^n$

The ambient Euclidean space in which the sets are considered.

All n-dimensional real vectors.

1. Formula
   Observation

   Written line: Let x,$y \in \mathbb{R}^n$.

2. Audio
   Observation

   Speaker says let x and y be two elements in $R^n$.

x,y

Two arbitrary points/vectors used to define a line segment and test convexity.

x,$y \in \mathbb{R}^n$; later in the convexity condition they are taken from C.

1. Formula
   Observation

   Written set-builder uses $\lambda \in [0,1]$.

2. Audio
   Observation

   Speaker says lambda is anywhere from zero to one inclusive.

$\lambda$

Scalar interpolation parameter between two points.

$\lambda \in [0,1]$.

1. Formula
   Observation

   Written line: A set $C \subseteq \mathbb{R}^n$ is convex if ...

2. Audio
   Observation

   Speaker says a subset of $R^n$ is convex if for all x and y in the set, C contains the line segment between x and y.

C

A candidate convex set, i.e. a subset of $\mathbb{R}^n$.

$C \subseteq \mathbb{R}^n$.

1. Formula
   Observation

   Written line: H := { $x \in \mathbb{R}^n$ : $a^T x \geq \beta$ }.

2. Audio
   Observation

   Speaker says consider the set H given by the set of x in $R^n$ such that a transpose x is greater than or equal to beta.

H

A specific example set introduced at the end of the clip.

$H \subseteq \mathbb{R}^n$.

1. Formula
   Observation

   Written expression contains $a^T x$.

2. Audio
   Observation

   Speaker says a transpose x.

1. The video does not explicitly state that a is nonzero.

a

Vector defining the linear inequality for H.

Implicitly a vector in $\mathbb{R}^n$ so that $a^T x$ is defined.

1. Formula
   Observation

   Written inequality ends with $\geq \beta$.

2. Audio
   Observation

   Speaker says greater than or equal to beta.

$\beta$

Scalar threshold in the halfspace inequality.

Implicitly a real scalar.

1. Formula
   Observation

   The board writes H := { x ∈ ℝ^n : $a^T x \ge β$ }.

2. Audio
   Observation

   The speaker refers to the set H and later says points must satisfy the condition defining the set.

H

The halfspace consisting of all vectors x in ℝ^n whose inner product with a is at least β.

subset of ℝ^n

1. Formula
   Observation

   x appears in the definition H := { x ∈ ℝ^n : $a^T x \ge β$ } and later in Take x, $y \in H$.

2. Audio
   Observation

   The speaker says to take x and y from H.

x

An arbitrary vector in ℝ^n; in the proof it is one chosen point of H.

ℝ^n

1. Formula
   Observation

   The board writes where a ∈ ℝ^n, $a \ne 0$.

2. Audio
   Observation

   The speaker says a is an n-vector and that we require a to be non-zero.

a

A fixed nonzero coefficient vector defining the halfspace inequality.

ℝ^n \ {0}

1. Formula
   Observation

   The board writes β ∈ ℝ.

2. Audio
   Observation

   The speaker says beta is just a real constant.

β

A fixed real threshold in the halfspace inequality.

ℝ

1. Formula
   Observation

   The board writes Take x, $y \in H$.

2. Audio
   Observation

   The speaker says let us take x and y from H.

y

A second chosen point of H used to form a line segment with x.

ℝ^n

1. Audio
   Observation

   Speaker introduces convex sets as special sets in $R^n$ and says that if you pick any two points in the set, then every point on the line segment between those two points is also in the set.

2. Diagram
   Observation

   Title 'Convex Sets' is visible while the verbal definition is given.

The clip first gives the geometric idea before formal notation: a set is convex when any two chosen points from the set have their entire connecting line segment contained in the set.

1. Applies to subsets of $\mathbb{R}^n$.

2. The statement is about every pair of points in the set, not just some pair.

1. Formula
   Observation

   Written: Let x,$y \in \mathbb{R}^n$. The line segment between x,y is the set { ($1-\lambda$)$x + \lambda y$ : $\lambda \in [0,1]$ }.

2. Audio
   Observation

   Speaker says this set is the set of points on the line segment between x and y.

For two points x and y in $\mathbb{R}^n$, the line segment joining them is represented algebraically by all affine combinations ($1-\lambda$)$x+\lambda y$ with $\lambda$ ranging from 0 to 1 inclusive.

1. x,$y \in \mathbb{R}^n$.

2. $\lambda \in [0,1]$.

1. x,y
2. $\lambda$

1. Formula
   Observation

   Written: A set $C \subseteq \mathbb{R}^n$ is convex if $\forall x$,$y \in C$, C contains the line segment between x and y.

2. Audio
   Observation

   Speaker states the same condition verbally and adds that this lets one talk about convex sets in n dimensions, not just in one, two, or three dimensions.

A subset C of $\mathbb{R}^n$ is convex precisely when, for every pair of points x and y in C, the whole line segment between x and y is also contained in C.

1. $C \subseteq \mathbb{R}^n$.

2. The condition must hold for all x,$y \in C$.

3. It is equivalent to requiring ($1-\lambda$)$x+\lambda y \in C$ for all $\lambda \in [0,1]$.

1. Definition of the line segment between two points in $R^n$
2. C

1. Formula
   Observation

   Written: Consider the set H := { $x \in \mathbb{R}^n$ : $a^T x \geq \beta$ }.

2. Audio
   Observation

   Speaker says let's look at one special case and considers this set.

1. The clip does not yet prove that H is convex.

2. The video does not explicitly name H a halfspace.

At the end of the clip, the lecturer introduces a concrete set H consisting of all points x in $\mathbb{R}^n$ satisfying the linear inequality $a^T x \geq \beta$, setting up a standard example for checking convexity.

1. $x \in \mathbb{R}^n$.

2. The defining condition is $a^T x \geq \beta$.

1. H
2. a
3. $\beta$

1. Formula
   Observation

   The board shows H := { x ∈ ℝ^n : $a^T x \ge β$ } with where a ∈ ℝ^n, $a \ne 0$ and β ∈ ℝ.

2. Audio
   Observation

   The speaker states that a is an n-vector, beta is a real constant, and a is required to be non-zero.

The video introduces the set H as the collection of all vectors x in ℝ^n satisfying the linear inequality $a^T x \ge β$, with a fixed nonzero vector a and fixed real constant β. This is the object whose convexity will be proved.

1. a is a nonzero vector in ℝ^n

2. β is a real constant

3. membership in H is tested by the inequality $a^T x \ge β$

1. Audio
   Observation

   The speaker says, 'We will show that this is convex' and explains that if we pick any two points from H, H contains the line segment between those two points.

2. Formula
   Observation

   No separate formula is written for the criterion, but the next lines instantiate it with x, $y \in H$ and $z = (1-λ)x + λy$.

The proof strategy stated in the video is the standard convex-set criterion: take any two points x and y in H and show that every point on the line segment joining them also lies in H. The subsequent construction of z implements exactly this criterion.

1. x and y are arbitrary points of H

2. the goal is to prove the whole segment between x and y is contained in H

1. Definition of the halfspace H

1. Audio
   Observation

   The speaker asks how to test if something is an element of H and answers that it has to satisfy the condition defining the set, namely a transpose times the point is greater than or equal to beta.

2. Formula
   Observation

   The board already displays $a^T x \ge β$ in the definition of H.

To verify that a candidate point belongs to H, one checks whether it satisfies the defining inequality. In this proof, showing $z \in H$ is therefore reduced to proving $a^T z \ge β$.

1. The candidate point must lie in ℝ^n

2. Membership is equivalent to satisfying the defining inequality

1. Definition of the halfspace H

1. Audio
   Observation

   The lecturer says a halfspace is a set of points in $R^n$ satisfying a single linear inequality where the coefficients of x are not all zero.

2. Formula
   Observation

   The board shows H := { x ∈ ℝ^n : $a^T x \ge β$ } with a ∈ ℝ^n, $a \ne 0$, β ∈ ℝ.

3. Diagram
   Observation

   The word Halfspace is written above the displayed set definition.

A halfspace is the subset of ℝ^n consisting of all points that satisfy one linear inequality with a nonzero coefficient vector. In this clip the displayed form uses ≥ β.

1. a ∈ ℝ^n

2. $a \ne 0$

3. β ∈ ℝ

1. Convexity test via convex combinations

1. Audio
   Observation

   The lecturer says if we take an intersection of finitely many halfspaces, what we get is something called a polyhedron.

2. Formula
   Observation

   The board writes A polyhedron is the intersection of finitely many halfspaces.

A polyhedron is defined here as the intersection of finitely many halfspaces. This makes it a convex set because each halfspace is convex and intersections preserve convexity.

1. The number of halfspaces is finite.

2. Each $H_i$ is a halfspace as previously defined.

1. Halfspace
2. Intersection of convex sets

1. Formula
   Observation

   The board takes x, $y \in H$, forms $z = (1-λ)x + λy$ with $λ \in [0,1]$, then proves $a^T z \ge β$.

2. Audio
   Observation

   The lecturer says this implies z is in H, so H is convex.

To prove a set is convex, take any two points in the set, form their convex combination, and show that the resulting point still satisfies the defining condition of the set.

1. x and y must be arbitrary points of the set under consideration.

2. The conclusion must hold for every $λ \in [0,1]$.

1. Formula
   Observation

   The board writes Fact: The intersection of two convex sets is convex.

2. Audio
   Observation

   The lecturer states the same fact and then says applying it repeatedly gives the result for any number of convex sets.

The clip presents a basic closure property of convex sets: intersecting convex sets yields another convex set. The written statement is for two sets, and the lecturer verbally extends it to any finite or arbitrary family by repeated application.

1. Each set being intersected must already be convex.

1. Convexity test via convex combinations

1. Audio
   Observation

   A polyhedron is simply the intersection of finitely many halfspaces.

2. Formula
   Observation

   A polyhedron is the intersection of finitely many halfspaces.

A polyhedron is defined as the intersection of finitely many halfspaces.

1. The number of halfspaces must be finite.

1. Audio
   Observation

   Speaker says in two dimensions, disks are convex.

2. Diagram
   Observation

   A filled disk is drawn and labeled 'Convex'.

3. Animation
   Observation

   Two red points inside the disk are joined by a red segment lying entirely within the disk.

In two dimensions, disks are convex sets.

1. The set under discussion is a disk in the plane.

Universal over pairs of points in the disk, as expressed by the visual test.

1. Audio
   Observation

   Speaker says here is an example of a set that is not convex because there are points on the segment between two chosen points that are not in the set.

2. Diagram
   Observation

   A kidney-shaped region is drawn and labeled 'Not convex'.

3. Animation
   Observation

   Two red points in different lobes are connected by a red segment passing outside the shaded region.

The illustrated non-disk region is not convex because some line segments between points of the set leave the set.

1. The set is the drawn kidney-shaped region.

2. The two selected points lie in different lobes.

Existential: there exist two points in the set whose connecting segment is not contained in the set.

1. Audio
   Observation

   Speaker says in one dimension, the convex sets are precisely the intervals.

1. No proof is given in this clip.

2. The exact class of intervals meant is not further specified on screen.

In one dimension, the convex sets are precisely the intervals.

1. The ambient space is one-dimensional.

Characterization over all one-dimensional sets.

1. Audio
   Observation

   Speaker says now we can talk about convex sets and line segments in n dimensions, not just in one, two, or three dimensions, and to check if a set is convex you just need to check that it satisfies this definition.

Once the line segment is defined algebraically in $\mathbb{R}^n$, a set is convex exactly when it satisfies the stated containment condition for all pairs of its points.

1. The line segment between two points is defined as \{($1-\lambda$)$x+\lambda y$:\lambda\in[0,1]\}.

Universal over all x,y in the set and all $\lambda \in [0,1]$.

1. Audio
   Observation

   The speaker says, 'We will show that this is convex.'

2. Formula
   Observation

   The proof ends with the chain showing $a^T z \ge β$ for $z = (1-λ)x + λy$ with x,$y \in H$ and $λ \in [0,1]$.

For H := { x ∈ ℝ^n : $a^T x \ge β$ } with $a \ne 0$, the set H is convex: whenever x,$y \in H$ and $λ \in [0,1]$, the point $z = (1-λ)x + λy$ also belongs to H.

1. H is defined by a linear inequality $a^T x \ge β$

2. a ∈ ℝ^n and $a \ne 0$

3. β ∈ ℝ

4. x,$y \in H$

5. $λ \in [0,1]$

For all x,$y \in H$ and all $λ \in [0,1]$, the convex combination $z = (1-λ)x + λy$ lies in H.

1. Formula
   Observation

   The derivation on the board ends with So $a^T z \ge β$ ⇒ $z \in H$, so H is convex.

2. Audio
   Observation

   The lecturer says implying that z is in H, so H is convex.

If H = { x ∈ ℝ^n : $a^T x \ge β$ } with $a \ne 0$, then H is convex.

1. a ∈ ℝ^n and $a \ne 0$

2. β ∈ ℝ

3. x,$y \in H$

4. $λ \in [0,1]$

For all x,$y \in H$ and all $λ \in [0,1]$, the point (1-λ)$x + λy$ also lies in H.

1. Formula
   Observation

   The board writes Fact: The intersection of two convex sets is convex.

2. Audio
   Observation

   The lecturer states the same sentence aloud.

The intersection of two convex sets is convex.

1. Both sets are convex.

For any two convex sets $C_1$ and $C_2$, $C_1$ ∩ $C_2$ is convex.

1. Audio
   Observation

   The lecturer says if you apply this fact repeatedly, you can conclude that the intersection of any number of convex sets is also convex.

1. The board only writes the two-set version; the extension to any number of sets is spoken rather than fully written.

The intersection of any number of convex sets is convex.

1. Each set in the family is convex.

For any family of convex sets, their intersection is convex.

1. Audio
   Observation

   The lecturer says a polyhedron is simply the set of feasible solutions of a linear programming problem and therefore the set of feasible solutions is going to be convex because of this fact up here.

2. Formula
   Observation

   Earlier on the board, a polyhedron is defined as the intersection of finitely many halfspaces, and halfspaces were proved convex.

Every polyhedron is convex.

1. A polyhedron is an intersection of finitely many halfspaces.

2. Each halfspace is convex.

3. Intersections of convex sets are convex.

For every set that can be written as a finite intersection of halfspaces, the set is convex.

1. Audio
   Observation

   The intersection of two convex sets is convex.

2. Formula
   Observation

   Fact: The intersection of two convex sets is convex.

The intersection of two convex sets is convex.

1. The sets involved are convex.

For any two convex sets.

1. Audio
   Observation

   A polyhedron is simply the intersection of finitely many halfspaces, and that halfspaces are all convex.

2. Formula
   Observation

   Example shows intersection of two halfspaces resulting in a set satisfying both inequalities.

A polyhedron is a convex set.

1. A polyhedron is the intersection of finitely many halfspaces.

2. Halfspaces are convex.

3. The intersection of convex sets is convex.

For any polyhedron.

1. Animation
   Observation

   First a disk is shown with two interior points and a segment staying inside; then a kidney-shaped set is shown with two points whose segment exits the set.

2. Audio
   Observation

   The narration explains that convexity means every point on the segment between any two chosen points remains in the set, and that failure of this property makes a set non-convex.

1. Expression

   $$$$
   Explanation

   Pick two points in the candidate set.

   Justification

   This matches the informal definition stated in the audio.

   Shown in the video
2. Expression

   $$$$
   Explanation

   Draw the straight line segment joining them.

   Justification

   The segment represents all intermediate points between the two chosen points.

   Shown in the video
3. Expression

   $$$$
   Explanation

   If the whole segment stays inside the set, the pictured set passes the convexity test; if part of the segment leaves the set, it fails.

   Justification

   This is exactly the containment condition described verbally and later formalized on the board.

   Shown in the video

The animation distinguishes convex from non-convex sets by testing whether the segment between two chosen points remains wholly inside the set.

1. Formula
   Observation

   The board writes the line-segment definition and then the convex-set definition using $C \subseteq \mathbb{R}^n$ and $\forall x$,$y \in C$.

2. Audio
   Observation

   Speaker says first we have to define what a line segment is in n dimensions, then states the convexity condition.

1. Expression

   $$x,y \in \mathbb{R}^n$$
   Explanation

   Introduce two arbitrary points in the ambient space.

   Justification

   Needed before defining the segment algebraically.

   Shown in the video
2. Expression

   $$\{(1-\lambda)x+\lambda y : \lambda \in [0,1]\}$$
   Explanation

   Define the line segment between x and y as the set of interpolated points.

   Justification

   This is the written definition on the board.

   Shown in the video
3. Expression

   $$C \subseteq \mathbb{R}^n$$
   Explanation

   Let C be a subset of the same ambient space.

   Justification

   This sets up the object whose convexity is being defined.

   Shown in the video
4. Expression

   $$\forall x,y \in C,\; C \text{ contains the line segment between } x \text{ and } y$$
   Explanation

   State the convexity condition for C.

   Justification

   This is the formal definition written and spoken in the clip.

   Shown in the video

The clip converts the earlier picture-based intuition into a precise definition valid in any dimension n.

1. Formula
   Observation

   The board writes Take x, $y \in H$; Consider $z = (1-λ)x + λy$ for some $λ \in [0,1]$; Then $a^T z = a^T((1-λ)x + λy) = (1-λ)a^T x + λ a^T y \ge (1-λ)β + λβ = β$.

2. Audio
   Observation

   The speaker explains each step verbally: choose x and y in H, form z on the segment, test membership by computing $a^T z$, use distributivity, then use $a^T x \ge β$ and $a^T y \ge β$ together with nonnegativity of the coefficients.

3. Animation
   Observation

   Red annotations mark $a^T x \ge β$ and $a^T y \ge β$ under the expanded expression, and a green curve links the conclusion back to the membership condition.

1. Expression

   $$\text{Take } x,y \in H.$$
   Explanation

   Begin with two arbitrary points in the set H.

   Justification

   This instantiates the convexity criterion stated in the video: to prove convexity, consider any two points of the set.

   Shown in the video
2. Expression

   $$z=(1-\lambda)x+\lambda y,\quad \lambda\in[0,1].$$
   Explanation

   Form the general point on the line segment between x and y.

   Justification

   The speaker explicitly says z is an arbitrary point on the line segment between x and y.

   Shown in the video
3. Expression

   $$a^T z=a^T((1-\lambda)x+\lambda y).$$
   Explanation

   Substitute the definition of z into the left-hand side of the membership inequality.

   Justification

   This is direct substitution of z into the linear functional $a^T$(·).

   Shown in the video
4. Expression

   $$a^T z=(1-\lambda)a^T x+\lambda a^T y.$$
   Explanation

   Expand the expression into a weighted sum of $a^T x$ and $a^T y$.

   Justification

   The speaker says this follows by applying the distributive law.

   Shown in the video
5. Expression

   $$a^T x\ge \beta,\quad a^T y\ge \beta.$$
   Explanation

   Use the fact that both chosen endpoints belong to H.

   Justification

   By the definition of H, any point in H satisfies the defining inequality $a^T(\cdot)\ge\beta$.

   Shown in the video
6. Expression

   $$1-\lambda\ge 0,\quad \lambda\ge 0.$$
   Explanation

   Note that the coefficients in the convex combination are nonnegative.

   Justification

   This follows from $λ \in [0,1]$, as stated on the board and in the audio.

   Shown in the video
7. Expression

   $$(1-\lambda)a^T x\ge (1-\lambda)\beta,\quad \lambda a^T y\ge \lambda\beta.$$
   Explanation

   Multiply the endpoint inequalities by the corresponding nonnegative scalars.

   Justification

   The speaker explicitly says that multiplying both sides of $a^T x \ge β$ by the nonnegative constant 1-λ preserves the inequality, and uses the same reasoning for the second term.

   Shown in the video
8. Expression

   $$a^T z\ge (1-\lambda)\beta+\lambda\beta=\beta.$$
   Explanation

   Add the two lower bounds and simplify.

   Justification

   Addition preserves inequalities, and (1-λ)$β+λβ$ simplifies to β because (1-λ)+$λ=1$.

   Shown in the video
9. Expression

   $$z\in H.$$
   Explanation

   Conclude that the arbitrary segment point z belongs to H.

   Justification

   The video's membership test says a point is in H exactly when $a^T$(point) ≥ β; the previous line proves this for z.

   Shown in the video

Since every convex combination $z=(1-λ)x+λy$ of points x,$y\in H$ also satisfies $a^T z\ge β$, the whole line segment between any two points of H lies in H, so H is convex.

1. Formula
   Observation

   The board shows $a^T z = a^T((1-λ)x + λy) = (1-λ)a^T x + λ a^T y \ge (1-λ)β + λβ = β$.

2. Audio
   Observation

   The lecturer says this implies z is in H, so H is convex.

1. Expression

   $$H := \{ x \in \mathbb{R}^n : a^T x \ge \beta \},\quad a\neq 0$$
   Explanation

   Start from the displayed definition of the halfspace H.

   Justification

   Given definition on the board.

   Shown in the video
2. Expression

   $$x,y\in H,\quad z=(1-\lambda)x+\lambda y,\quad \lambda\in[0,1]$$
   Explanation

   Take arbitrary points x and y in H and form their convex combination z.

   Justification

   Standard convexity test for sets.

   Shown in the video
3. Expression

   $$a^T z = a^T((1-\lambda)x+\lambda y)$$
   Explanation

   Apply the linear functional $a^T$(·) to z.

   Justification

   Substitution of the definition of z.

   Shown in the video
4. Expression

   $$= (1-\lambda)a^T x + \lambda a^T y$$
   Explanation

   Distribute $a^T$ over the sum and pull out scalar coefficients.

   Justification

   Linearity of the dot product / matrix-vector multiplication.

   Shown in the video
5. Expression

   $$\ge (1-\lambda)\beta + \lambda\beta$$
   Explanation

   Replace $a^T x$ and $a^T y$ by their lower bounds β using x,$y \in H$ and $λ \in [0,1]$.

   Justification

   Because x,$y \in H$, $a^T x \ge β$ and $a^T y \ge β$; multiplying by nonnegative scalars preserves the inequalities.

   Shown in the video
6. Expression

   $$= \beta$$
   Explanation

   Simplify the right-hand side.

   Justification

   (1-λ)$β + λβ = β$.

   Shown in the video
7. Expression

   $$a^T z \ge \beta \Rightarrow z\in H$$
   Explanation

   Since z satisfies the defining inequality of H, it belongs to H.

   Justification

   Definition of H.

   Shown in the video
8. Expression

   $$\text{Therefore } H \text{ is convex.}$$
   Explanation

   Every convex combination of two points of H remains in H.

   Justification

   Conclusion of the convexity criterion.

   Shown in the video

The halfspace H = { x ∈ ℝ^n : $a^T x \ge β$ } is convex.

1. Audio
   Observation

   The lecturer says applying the two-set intersection fact repeatedly gives convexity for intersections of any number of convex sets, then defines a polyhedron as an intersection of finitely many halfspaces.

2. Formula
   Observation

   The board writes Fact: The intersection of two convex sets is convex. and A polyhedron is the intersection of finitely many halfspaces.

1. Expression

   $$\text{Each halfspace is convex.}$$
   Explanation

   Use the earlier proposition proved on the board.

   Justification

   Claim claim-halfspace-convex.

   Shown in the video
2. Expression

   $$P = \bigcap_{i=1}^{m} H_i$$
   Explanation

   Write a polyhedron as a finite intersection of halfspaces $H_i$.

   Justification

   Definition of polyhedron given in the clip.

   Shown in the video
3. Expression

   $$H_1\cap H_2 \text{ is convex}$$
   Explanation

   Intersect the first two convex sets.

   Justification

   Fact that the intersection of two convex sets is convex.

   Shown in the video
4. Expression

   $$(H_1\cap H_2)\cap H_3,\ldots,\bigcap_{i=1}^{m}H_i \text{ are convex}$$
   Explanation

   Repeat the two-set intersection argument finitely many times.

   Justification

   Repeated application of the same intersection fact.

   Derived from the video

A polyhedron, being a finite intersection of convex halfspaces, is convex.

1. Formula
   Observation

   e.g. { [$x_1$; $x_2$] : $x_1 + 2x_2$ >= 5 } ∩ { [$x_1$; $x_2$] : -2x_1 + $x_2$ >= 0 } = { [$x_1$; $x_2$] : $x_1 + 2x_2$ >= 5, -2x_1 + $x_2$ >= 0 }

1. Expression

   $$\left\{ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} : x_1 + 2x_2 \ge 5 \right\} \cap \left\{ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} : -2x_1 + x_2 \ge 0 \right\}$$
   Explanation

   Start with the intersection of two specific halfspaces defined by linear inequalities.

   Justification

   Given in the video example.

   Shown in the video
2. Expression

   $$= \left\{ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} : \begin{aligned} x_1 + 2x_2 &\ge 5 \\ -2x_1 + x_2 &\ge 0 \end{aligned} \right\}$$
   Explanation

   Combine the conditions into a single set definition requiring both inequalities to hold simultaneously.

   Justification

   Definition of set intersection.

   Shown in the video

The intersection of these two halfspaces forms a polyhedron defined by the system of linear inequalities.

1. Diagram
   Observation

   A circular disk is drawn, shaded, and labeled 'Convex'.

2. Animation
   Observation

   Two red points inside the disk are connected by a red segment fully contained in the disk.

3. Audio
   Observation

   Speaker says in two dimensions, disks are convex and explains that no matter which two points you pick on the disk, the line segment between them lies within the disk.

Illustrate why a two-dimensional disk is convex.

1. A filled disk in the plane.

2. Two arbitrary points chosen inside the disk.

Show that the segment between the chosen points remains in the set.

1. Expression

   $$$$
   Explanation

   Choose two points inside the disk.

   Justification

   This corresponds to the universal quantifier in the definition of convexity.

   Shown in the video
2. Expression

   $$$$
   Explanation

   Draw the straight segment joining them.

   Justification

   The segment represents all intermediate points required by the definition.

   Shown in the video
3. Expression

   $$$$
   Explanation

   Observe that the segment lies entirely within the disk.

   Justification

   This is directly shown by the drawing and stated in the narration.

   Shown in the video

The disk is convex.

The visual test shows the connecting segment stays inside the shaded region for the displayed choice of points, matching the spoken general claim.

1. Diagram
   Observation

   A kidney-shaped region is drawn, shaded, and labeled 'Not convex'.

2. Animation
   Observation

   Two red points in different lobes are joined by a red segment that passes outside the region; a green arrow marks the outside portion.

3. Audio
   Observation

   Speaker says if you pick these two points, the line segment between them has points that are not in the set, so this is not convex.

Illustrate why the drawn non-disk region is not convex.

1. A kidney-shaped planar region.

2. Two points chosen in different lobes of the region.

Find a pair of points whose connecting segment is not contained in the set.

1. Expression

   $$$$
   Explanation

   Pick two points in separate lobes of the region.

   Justification

   This provides a counterexample pair for the universal convexity condition.

   Shown in the video
2. Expression

   $$$$
   Explanation

   Draw the straight segment between them.

   Justification

   The segment is the object required by the definition of convexity.

   Shown in the video
3. Expression

   $$$$
   Explanation

   Notice that part of the segment lies outside the shaded region.

   Justification

   This is explicitly indicated by the drawing and the green arrow.

   Shown in the video

The region is not convex.

The existence of two points in the set whose segment leaves the set falsifies the convexity condition.

1. Formula
   Observation

   Written: Consider the set H := { $x \in \mathbb{R}^n$ : $a^T x \geq \beta$ }.

2. Audio
   Observation

   Speaker says let's look at one special case and reads the definition of H.

1. The clip stops before any verification or proof about H.

2. No numerical values for a or $\beta$ are given.

Introduce a concrete subset of $\mathbb{R}^n$ defined by a linear inequality.

1. A vector a and scalar $\beta$ are implicitly present.

2. The ambient space is $\mathbb{R}^n$.

Define the set H for later convexity discussion.

1. Expression

   $$H := \{x \in \mathbb{R}^n : a^T x \geq \beta\}$$
   Explanation

   Write H as the collection of all x satisfying the inequality.

   Justification

   This is exactly the formula shown on the board.

   Shown in the video