Is a closed interval $[a,b]$, or $[0,1]$ in particular, a convex set ?
I mean, let $\lambda\in [0,1]$. Then,
\begin{align}
a\lambda+(1-\lambda)b&=a\lambda+b -b\lambda
&=(a-b)\lambda+b
\end{align}
Where does it lie?
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Octagonal Monk
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2Yes, trivially. Segments in the real numbers are just intervals. – Rushabh Mehta May 23 '19 at 03:04
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A set in the real line is convex iff it is a (possibly extended, or trivial) interval. $(-\infty, 0], (0,1]$, etc. – copper.hat May 23 '19 at 04:40
1 Answers
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if $a \le x \le b$ and $a \le y \le b$,
since $\lambda$ and $1-\lambda$ are nonnegative,
$\lambda a \le \lambda x \le \lambda b$ and $(1-\lambda )a \le (1-\lambda)y \le (1-\lambda)b$,
Summing the equations, we have
$$a \le \lambda x + (1-\lambda )y \le b$$
Siong Thye Goh
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