Show that the set S = { x/|x| <=2} is a convex set 

To show that the set \( S = \{ x \mid |x| \leq 2 \} \) is a convex set, we need to verify that for any two points \( x_1 \) and \( x_2 \) in \( S \), the line segment connecting these two points also lies entirely within \( S \).

### Definition of Convexity
A set \( S \) is convex if for any two points \( x_1, x_2 \in S \), and for any \( \lambda \in [0,1] \), the point 
\[
x_\lambda = \lambda x_1 + (1 - \lambda) x_2
\]
is also in \( S \). In other words, the line segment between \( x_1 \) and \( x_2 \) lies within the set \( S \).

### The Set \( S = \{ x \mid |x| \leq 2 \} \)
The set \( S \) consists of all points \( x \) such that the Euclidean norm (or absolute value in the one-dimensional case) of \( x \) is less than or equal to 2. Geometrically, this represents a disk (or ball) of radius 2 centered at the origin in any Euclidean space.

### Proof of Convexity

Let \( x_1, x_2 \in S \). Then we know that
\[
|x_1| \leq 2 \quad \text{and} \quad |x_2| \leq 2.
\]
We need to check whether the point
\[
x_\lambda = \lambda x_1 + (1 - \lambda) x_2
\]
also satisfies \( |x_\lambda| \leq 2 \) for \( \lambda \in [0,1] \).

By the triangle inequality, we have:
\[
|x_\lambda| = |\lambda x_1 + (1 - \lambda) x_2| \leq |\lambda x_1| + |(1 - \lambda) x_2|.
\]
Using the fact that norms are positively homogeneous (i.e., \( |c x| = |c| |x| \) for a scalar \( c \)), we get:
\[
|x_\lambda| \leq \lambda |x_1| + (1 - \lambda) |x_2|.
\]
Since \( |x_1| \leq 2 \) and \( |x_2| \leq 2 \), we substitute these into the inequality:
\[
|x_\lambda| \leq \lambda \cdot 2 + (1 - \lambda) \cdot 2 = 2(\lambda + 1 - \lambda) = 2.
\]

Thus, \( |x_\lambda| \leq 2 \), which implies that \( x_\lambda \in S \).

### Conclusion
Since for any \( x_1, x_2 \in S \) and any \( \lambda \in [0,1] \), the point \( x_\lambda \) also belongs to \( S \), we conclude that the set \( S = \{ x \mid |x| \leq 2 \} \) is convex.

Would you like to dive deeper into this proof or have any specific questions about convex sets?

### 5 Related Questions:
1. What is the geometrical interpretation of convexity in higher dimensions?
2. Can the union of two convex sets be convex? Why or why not?
3. How would you check if a set defined by inequalities is convex?
4. How does the convexity of a function differ from the convexity of a set?
5. Can a set be convex if it is unbounded? Provide examples.

### Tip:
Remember, convex sets are closely related to optimization problems. Many optimization problems have easier solutions when the feasible region (the set of possible solutions) is convex.

Learn how to prove the convexity of the set S = { x | |x| <= 2 } using the definition of convexity and the triangle inequality. This mathematical problem involves verifying that for any two points x1, x2 in S, the line segment connecting them lies entirely within S. Suitable for advanced undergraduate students.

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