To show that $S$ is a convex set, we need to demonstrate that if $x = (x_1, x_2, \ldots, x_n)$ and $y = (y_1, y_2, \ldots, y_n)$ are two points in $S$, then for any $\lambda \in [0, 1]$, the point $z = \lambda x + (1-\lambda)y$ also belongs to $S$.

Proof:

1. Definition of $S$:
   $S$ is defined as the set of all points $x \in \mathbb{R}^n$ satisfying the $m$ inequalities: $a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n \leq b_1,$ $a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n \leq b_2,$ $\vdots$ $a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n \leq b_m.$

   Alternatively, we can express these inequalities as: $A x \leq b,$ where $A$ is an $m \times n$ matrix with coefficients $a_{ij}$, $x$ is a vector in $\mathbb{R}^n$, and $b$ is a vector in $\mathbb{R}^m$.

2. Take two points in $S$:
   Let $x, y \in S$. By definition of $S$, they satisfy: $A x \leq b \quad \text{and} \quad A y \leq b.$

3. Form a convex combination:
   Let $z = \lambda x + (1-\lambda)y$, where $\lambda \in [0, 1]$. We will show that $z \in S$, i.e., $A z \leq b$.

4. Apply linearity of $A$:
   Substituting $z = \lambda x + (1-\lambda)y$ into $A z$, we have: $A z = A (\lambda x + (1-\lambda)y) = \lambda A x + (1-\lambda) A y.$

5. Use the inequalities for $x$ and $y$:
   Since $x \in S$, $A x \leq b$, and since $y \in S$, $A y \leq b$. Therefore: $\lambda A x + (1-\lambda) A y \leq \lambda b + (1-\lambda) b = b,$ because $\lambda \in [0, 1]$ and the inequality holds component-wise.

6. Conclusion:
   We have shown that $A z \leq b$, which means $z \in S$. Therefore, $S$ is a convex set.

Summary:

The convex combination $z = \lambda x + (1-\lambda)y$ satisfies all the inequalities defining $S$, so $S$ is convex.

Would you like further clarification or a graphical illustration of this proof?
Here are some related questions to consider:

1. Can you describe the geometric interpretation of convexity in $\mathbb{R}^n$?
2. What conditions on the matrix $A$ and vector $b$ might affect the shape of $S$?
3. How would the proof change if the inequalities were strict ($<$ instead of $\leq$)?
4. What is an example of a convex set defined by such inequalities in $\mathbb{R}^2$?
5. How is this result used in optimization problems?

Tip: Convex sets are fundamental in optimization because local minima in convex sets are also global minima.