(a) Proof that the intersection of convex sets is convex

We are given that $C_1, C_2, \dots, C_m \subseteq \mathbb{R}^n$ are convex sets, and we need to show that their intersection $C = \bigcap_{i=1}^m C_i$ is also convex.

Proof:

A set $C \subseteq \mathbb{R}^n$ is convex if for any two points $x, y \in C$, the line segment joining $x$ and $y$ lies entirely within $C$. Mathematically, this means: $\lambda x + (1-\lambda) y \in C \quad \text{for all} \quad \lambda \in [0, 1].$

Now, let $x, y \in C = \bigcap_{i=1}^m C_i$. By definition of intersection, since $x, y \in C$, we have: $x, y \in C_i \quad \text{for all} \quad i = 1, 2, \dots, m.$ Because each $C_i$ is convex, for any $\lambda \in [0, 1]$, the point $\lambda x + (1-\lambda) y$ lies in each $C_i$. Thus, for each $i$, we have: $\lambda x + (1-\lambda) y \in C_i.$ Since the intersection $C = \bigcap_{i=1}^m C_i$ consists of points that belong to all $C_i$, it follows that: $\lambda x + (1-\lambda) y \in \bigcap_{i=1}^m C_i = C.$ Therefore, the line segment joining $x$ and $y$ lies entirely within $C$, which implies that $C$ is convex.

(b) Example where the union of two convex sets is not convex

Let us consider two intervals on the real line $\mathbb{R}$, which are convex sets. Let these intervals be: [ A = [0, 1] \quad \text{and} \quad B = [2, 3]. ] Both $A$ and $B$ are convex sets, as any interval in $\mathbb{R}$ is convex.

Now, consider their union: [ A \cup B = [0, 1] \cup [2, 3]. ] We need to check whether this union is convex. A set is convex if for any two points $x, y \in A \cup B$, the entire line segment joining $x$ and $y$ lies within $A \cup B$.

Let's take $x = 0.5 \in A$ and $y = 2.5 \in B$. The line segment joining $x$ and $y$ consists of all points of the form $\lambda x + (1-\lambda)y$ for $\lambda \in [0, 1]$. Specifically, consider the point halfway between $x$ and $y$: $z = \frac{0.5 + 2.5}{2} = 1.5.$ This point $z = 1.5$ does not lie in $A \cup B$, since it is not in either interval $[0, 1]$ or $[2, 3]$. Thus, the entire line segment joining $0.5$ and $2.5$ does not lie within $A \cup B$, and therefore $A \cup B$ is not convex.

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Further Insights:

If you'd like more details on convexity or examples of convex sets in other spaces, feel free to ask.

Here are some follow-up questions:

1. Can you prove that the union of two convex sets is convex if they are disjoint?
2. How does the convexity of a function relate to convex sets?
3. What are some applications of convex sets in optimization problems?
4. How would the convexity proof change if the sets $C_i$ are closed?
5. What is the difference between a convex set and a concave set in terms of their definitions?

Tip: When proving convexity, always remember that the condition for a set to be convex is that any convex combination of two points in the set must also lie in the set. This is the key to many such proofs!