Let's analyze each statement:(1) The closed half spaces $H_1 = \{x | cx \ge z\}$ and $H_2 = \{x | cx \le z\}$ are convex sets. A set $S$ is convex if for any two points $x_1, x_2 \in S$ and any scalar $ \alpha \in [0, 1]$, the point $\alpha x_1 + (1-\alpha) x_2$ is also in $S$. Consider $H_1$. Let $x_1, x_2 \in H_1$ and $\alpha \in [0, 1]$. Then $cx_1 \ge z$ and $cx_2 \ge z$. We need to check if $c(\alpha x_1 + (1-\alpha) x_2) \ge z$. $c(\alpha x_1 + (1-\alpha) x_2) = \alpha c x_1 + (1-\alpha) c x_2$. Since $cx_1 \ge z$ and $cx_2 \ge z$, and $\alpha \ge 0$, $(1-\alpha) \ge 0$: $\alpha c x_1 \ge \alpha z$ and $(1-\alpha) c x_2 \ge (1-\alpha) z$. Adding these inequalities: $\alpha c x_1 + (1-\alpha) c x_2 \ge \alpha z + (1-\alpha) z = z$. Thus, $H_1$ is convex. Similarly, $H_2$ is convex. Statement (1) is correct.(2) Intersection of two convex sets is also a convex set. Let $S_1$ and $S_2$ be two convex sets. Let $S = S_1 \cap S_2$. Let $x_1, x_2 \in S$. Then $x_1, x_2 \in S_1$ and $x_1, x_2 \in S_2$. Since $S_1$ is convex, for any $\alpha \in [0, 1]$, $ \alpha x_1 + (1-\alpha) x_2 \in S_1$. Since $S_2$ is convex, for any $\alpha \in [0, 1]$, $\alpha x_1 + (1-\alpha) x_2 \in S_2$. Therefore, $\alpha x_1 + (1-\alpha) x_2$ is in both $S_1$ and $S_2$, so it is in $S_1 \cap S_2 = S$. Thus, the intersection of two convex sets is convex. Statement (2) is correct.(3) Let $A = \{x_1 - x_2 | x_1 \in S, x_2 \in T\}$, where S and T are two non-empty disjoint convex sets, then A is a convex set. Let $a_1, a_2 \in A$. Then $a_1 = s_1 - t_1$ for some $s_1 \in S, t_1 \in T$, and $a_2 = s_2 - t_2$ for some $s_2 \in S, t_2 \in T$. Let $\alpha \in [0, 1]$. Consider $\alpha a_1 + (1-\alpha) a_2$: $\alpha a_1 + (1-\alpha) a_2 = \alpha (s_1 - t_1) + (1-\alpha) (s_2 - t_2)$ $= (\alpha s_1 + (1-\alpha) s_2) - (\alpha t_1 + (1- \alpha) t_2)$ Since S is convex, $\alpha s_1 + (1- \alpha) s_2 \in S$. Let this be $s'$. Since T is convex, $\alpha t_1 + (1-\alpha) t_2 \in T$. Let this be $t'$. So, $\alpha a_1 + (1-\alpha) a_2 = s' - t'$, where $s' \in S$ and $t' \in T$. This means $\alpha a_1 + (1-\alpha) a_2 \in A$. Thus, A is a convex set. Statement (3) is correct. The condition that S and T are disjoint is irrelevant for the convexity of A itself, only for the properties of A in relation to S and T.(4) If S and T are any two convex sets in $ \mathbb{R}^n$, then for all scalars $\alpha$ and $ \beta$, the set $\alpha S + \beta T$ is not convex. The set $\alpha S + \beta T$ is defined as $\{\alpha s + \beta t | s \in S, t \in T\}$. Let $x_1, x_2 \in \alpha S + \beta T$. Then $x_1 = \alpha s_1 + \beta t_1$ for some $s_1 \in S, t_1 \in T$. And $x_2 = \alpha s_2 + \beta t_2$ for some $s_2 \in S, t_2 \in T$. Let $\gamma \in [0, 1]$. Consider $\gamma x_1 + (1-\gamma) x_2$: $\gamma x_1 + (1- \gamma) x_2 = \gamma (\alpha s_1 + \beta t_1) + (1-\gamma) (\alpha s_2 + \beta t_2)$ $= \gamma \alpha s_1 + \gamma \beta t_1 + (1-\gamma) \alpha s_2 + (1-\gamma) \beta t_2$ $= \alpha (\gamma s_1 + (1-\gamma) s_2) + \beta (\gamma t_1 + (1- \gamma) t_2)$Since S is convex, $\gamma s_1 + (1-\gamma) s_2 \in S$. Let $s' = \gamma s_1 + (1- \gamma) s_2$. Since T is convex, $\gamma t_1 + (1-\gamma) t_2 \in T$. Let $t' = \gamma t_1 + (1- \gamma) t_2$. So, $\gamma x_1 + (1-\gamma) x_2 = \alpha s' + \beta t'$, where $s' \in S$ and $t' \in T$. This means $\gamma x_1 + (1-\gamma) x_2 \in \alpha S + \beta T$. Therefore, $\alpha S + \beta T$ is always convex for any convex sets S and T and any scalars $\alpha, \beta$.The statement (4) says that $\alpha S + \beta T$ is not convex. This is false. Hence, statement (4) is incorrect.Special case: If $\alpha=1, \beta=0$, then we have $S$, which is convex. If $\alpha=0, \beta=1$, we have $T$, which is convex. If $\alpha$ and $\beta$ are positive, it is a sum of scaled convex sets. If $\alpha$ and $\beta$ are negative, it is still a combination that results in a convex set. For example, if $S$ is a square and $T$ is a disk, $\alpha S + \beta T$ will be a larger shape that is also convex.
Which of the following statement is not correct? (1) The closed half spaces $H_1 = \{x | cx \ge z\}$ and $H_2 = \{x | cx \le z\}$ are convex sets. (2) Intersection of two convex sets is also a convex set. (3) Let $A = \{x_1 - x_2 | x_1 \in S, x_2 \in T\}$, where S and T are two non-empty disjoint convex sets, then A is a convex set. (4) If S and T are any two convex sets in $\mathbb{R} ^n$, then for all scalars $\alpha$ and $\beta$, the set $\alpha S + \beta T$ is not convex. (5) Question not attempted
Asked in Asst. Statistical Officer (Agri. Deptt.) 2024 · Topic: Computer and Information Technology (Computational Physics: Numerical methods and algorithms for physical problems)
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