We have
\begin{align} &A = \big\{ x \in \mathbb{R}^m: x_i \geq 0, \sum_{i=1}^{m}x_i = 1 \big\}\\ &B = \big\{ y \in \mathbb{R}^n: y_i \geq 0, \sum_{i=1}^{n}y_i = 1 \big\}. \end{align}
How can we show that they are convex?
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1Just apply the definition of convex set... – Guillermo García Sáez Jun 25 '22 at 09:53
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@Guillerminho77 care to elaborate? – algevristis Jun 25 '22 at 09:54
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The common topos of hemi hyper planes? – dmtri Jun 25 '22 at 09:56
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@dmtri do you mean semi hyper planes? – algevristis Jun 25 '22 at 09:57
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Why do you ask the same thing for A, B? Are not they the same thing? – dmtri Jun 25 '22 at 09:58
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Yes, bad language sorry. – dmtri Jun 25 '22 at 09:59
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@dmtri sorry mate that's all I have been given I have been asking this for 3 months now on stackexchange if you have a complete answer please write it so I can put this to rest. – algevristis Jun 25 '22 at 10:00
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Let us continue this discussion in chat. – algevristis Jun 25 '22 at 18:01
1 Answers
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We say that a set $C$ is convex if $$\forall x,y\in C,\;,\forall \lambda\in[0,1]\;, \lambda x+(1-\lambda)y\in C$$ Now take $x,y\in A$, and lets take $\lambda\in (0,1)$, the extreme cases $\lambda=0$ and $\lambda=1$ are trivial. Lets call $z:=\lambda x+(1-\lambda)y$. Is $z$ in the set $A$?. By definiton of set $A$, $z\in A\iff \sum_{i=1}^{m}z_i=1$. But for every $i=1,\cdots,m$, $z_i=\lambda x_i+(1-\lambda)y_i$. So: $$\sum_{i=1}^{m}z_i=\sum_{i=1}^{m}(\lambda x_i+(1-\lambda)y_i)=\sum_{i=1}^{m}\lambda x_i+\sum_{i=1}^{m}(1-\lambda)y_i=\lambda\sum_{i=1}^{m}x_i+(1-\lambda)\sum_{i=1}^{m}y_i=\lambda+1-\lambda=1$$ So $z\in A\implies A$ is convex.
The reasoning for $B$ is exactly the same, because are the same set for different dimensions
