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If $c \in \mathbb{R}$ and $S$ is a set, define $c S=\{c \mathbf{x} : \mathbf{x} \in S\} .$ Let $S$be a convex set and suppose $c>0$ and $d>0 .$ Prove that $c S+d S=(c+d) S .$

S is a convex set and $c S=\{c \mathbf{x} : \mathbf{x} \in S\}$ for $c \in \mathbb{R}$ .So by this definition, write$c S+d S=\{c x+d y : x, y \in S\}$$(c+d) S=\{(c+d) z : z \in S\}$Let $x=y=z$ to infer the following:$(c+d) S \subset c S+d S$This implies that $c x+d y \in(c+d) S$ for all $x, y \in S .$So $c S+d S \subseteq(c+d) S \quad \ldots \ldots(1)$Define $z=\frac{c}{c+d} x+\frac{d}{c+d} y$Note that $\frac{c}{c+d}+\frac{d}{c+d}=1$ for $c, d>0$ holds true.So $z$ is a convex combination of $x$ and $y .$As $S$ is convex, $z \in S$ and henceforth $(c+d) z \in(c+d) S$Then $(c+d) z=c z+d z$ and therefore $(c+d) z \in c S+d S$So $(c+d) S \subseteq c S+d S \ldots \ldots(2)$From $(1)$ and $(2),$ we have $c S+d S=(c+d) S$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 5

Polytopes

Vectors

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so given a conduct set and to positive numbers, C and D were asked to show that CS plus D s going to be equal to C plus D s. So if we in other words, stretch the elements of a convex set by sea and stretch the elements of a complex set ID and then add thes two new sets together we get a set, which is three elements of s stretched by C plus D. So we'll let s be convex set. Let's see, be greater than zero and d be greater than zero. Now let X be an element of CS plus ts and we confined S one and s two in s such that X is equal to see X one are sorry CS one plus D s to then it follows. The X is also equal to C plus d times. See over C plus d. We can divide by C plus Decency plus D is greater than zero since both c and D or greater than zero s one plus de over C plus D s to now because S is a convex set. It follows that see, oversee plus D s one plus D over C plus D. As to linear combination of elements from s also lies an s. So it follows that X is an element of C plus D s. So we've proven one direction. That is, that C s plus D s is a subset of C plus D s. Now to prove the other direction Similar exercise. So we're going toe Let X be in the set C plus D s. This means we confined. A vector s in the set s such that X is equal to C plus D. Yes, and of course, this is equal to CS plus D s. After distributing the scaler, we could say, even distributing the vector across the scale Er's And this is clearly an element of C S plus D s. So X is an element of CS plus ts and therefore it follows that C plus D S is a subset of CS plus D s. Combining this statement and the previous statement we get that CS plus D s equals C plus D s, which is what we wanted to prove

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