This might be very easy, but I can not see how to prove that:
$[3,5]\times [1,2]\times \{1\} \subseteq \mathbb{R}^3 $ is a convex set.
Thank you for your help.
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It is a rectangle at $z=1$. – VIVID May 08 '21 at 21:03
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Perhaps it is easier to show that if $A_1,\ldots, A_n\subseteq \Bbb R$ are convex, then so is $A_1\times\cdots \times A_n\subseteq \Bbb R^n$? – Hagen von Eitzen May 08 '21 at 21:04
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@HagenvonEitzen Can we conclude from my previous comment; if we take any two points on that rectangle, the rectangle still contains the line connecting these two points? – VIVID May 08 '21 at 21:05
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I guess, it's the interval $[3,5]$? – user408858 May 08 '21 at 21:05
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@user408858 No, it is a 'lifting" of the rectangle $[3,5]\times [1,2]$ from the ground floor ($z=0$) to the first floor ($z=1$). – uniquesolution May 08 '21 at 21:09
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Well, in the text there was written the interval $[5,3]$... That's what I was addressing. – user408858 May 08 '21 at 21:11
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For any $p_1,q_1\in[3,5]$ and any $t\in[0,1]$, it holds $tp_1+(1-t)q_1\in[3,5]$
This holds, because $3\le p_1,q_1 \le 5$, therefore $3= t\cdot 3+(1-t)\cdot 3\le tp_1+(1-t)q_1\le t\cdot 5+(1-t)\cdot 5=5$, this means $tp_1+(1-t)q_1\in[3,5]$
The same arguments applies for the second and third component:
For any $p_2,q_2\in[1,2]$ and any $t\in[0,1]$, it holds $tp_2+(1-t)q_2\in[1,2]$
For any $p_3,q_3\in\{1\}$ and any $t\in[0,1]$, it holds $tp_3+(1-t)q_3\in\{1\}$
Therefore:
For any $p=(p_1,p_2,p_3),q=(q_1,q_2,q_3)\in[3,5]\times[1,2]\times{1}$ and any $t\in[0,1]$, it holds $tp+(1-t)q\in[3,5]\times[1,2]\times\{1\}$
user408858
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1Thank you, I had no idea how to interpreted the intervals, neither how to construct the affine combination. – ergch24 May 08 '21 at 21:21
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Sorry @user408858, hope this is clear. Is it possible to use the same values for the LHS and RHS of the inequality for instance $p_1 and q_1$ to probe that $tp_1+(1-t)q_1\in[3,5]$? Is clear to me that you choose the smallest value possible for $p_1 and q_1$ to take ( both $p_1 and q_1$ are equal to 3) in the LHS and the maximum possible ( both $p_1 and q_1$ are equal to 5) for the inequality to hold that the convex combination $\in [3, 5]$. Do I get it right? – ergch24 May 08 '21 at 23:23
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Yes, that's right. You could pick $p_1=q_1$, but then $tp_1+(1-t)q_1=p_1$, which is in $[3,5]$, of course. Essential is, that for any $t\in[0,1]$, both $t$ and $1-t$ are in $[0,1]$. Therefore the inequalities hold. – user408858 May 09 '21 at 06:45