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I want to prove the interval $(1,5)$ is a convex set.

A convex set is a set having all the convex linear combinations of its point in it, where a convex linear combination is a linear combination of the form $X=(1-\alpha)X_1+ \alpha X_2$ where $\alpha$ lies between $0$ and $1$ and the sum of coefficients of above linear combination is equal to 1.

How can we prove the interval $(1,5)$ a convex set?

rschwieb
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hafsah
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  • Take any two $x_1, x_2$ from the interval and any $\alpha\in[0,1]$. Prove that $\alpha x_1 + (1-\alpha) x_2$ is in the interval. – 5xum Feb 11 '14 at 13:44
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3 Answers3

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You just have to prove that for every $\alpha \in (0,1)$ and every pair $x_1$, $x_2$ $\in (1,5)$, $\alpha x_1 + (1-\alpha)x_2$ also lies in $(1,5)$.

Suppose without loss of generality that $x_1 \leq x_2$. Then for every $\alpha \in (0,1)$ (which implies that both $\alpha$ and $1-\alpha$ are positive)

$\alpha x_1 + (1-\alpha)x_2 \leq \alpha x_2 + (1-\alpha)x_2 = x_2$

and

$\alpha x_1 + (1-\alpha)x_2 \geq \alpha x_1 + (1-\alpha)x_1 = x_1$.

Hence we find that for every $\alpha \in (0,1)$, $\alpha x_1 + (1-\alpha)x_2 \in [x_1,x_2]$.

Hence for every pair $x_1$, $x_2$, any convex-combination of $x_1$, $x_2$ lies in $[x_1,x_2]$. And since $x_1$, $x_2$ $\in (1,5)$, every convex-combination of any two points in $(1,5)$ also lies in $(1,5)$.

Martin
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  • an exrcise is given in a book that prove X=(-1/2,1/2,7/4) is the convex linear combination of X1,X2 and X3??? we did'nt provided with that X1,X2 and X3...either this question is wrong or there any way to prove it?? – hafsah Feb 14 '14 at 14:40
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If $x_1\leq x_2$ then $$x=(1-\alpha)x_1+\alpha x_2=x_1+\alpha(x_2-x_1)$$ If $\alpha \in [0,1]$ then $$0\leq \alpha(x_2-x_1)\leq x_2-x_1$$ so: $$x_1\leq x\leq x_1+(x_2-x_1)=x_2$$ Then $x_1,x_2\in (1,5)$ gives: $$1< x_1\leq x\leq x_2<5$$

drhab
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  • why we are taking x1<x2 here??when x2>x1 then alpha will be less then zero...which is against covexity condition that is alpha should lie between 0 and 1....is this rzn for taking x1<x2??? – hafsah Feb 13 '14 at 10:05
  • we do not take $x_1<x_2$ but $x_1\leq x_2$. If both are elements of $\mathbb R$ then you will have $x_1\leq x_2$ or $x_2\leq x_1$. You could say that the answer deals with case $x_1\leq x_2$ and it is quite obvious that the same way of proving can be handled in the other case. Just switch the roles of $x_1$ and $x_2$. – drhab Feb 13 '14 at 10:13
  • an exrcise is given in a book that prove X=(-1/2,1/2,7/4) is the convex linear combination of X1,X2 and X3??? we did'nt provided with that X1,X2 and X3...either this question is wrong or there any way to prove it?? – hafsah Feb 14 '14 at 15:02
  • It is not be the right way to go deeper into this in a comment on this answer. I advice you to post your problem as a new question. – drhab Feb 14 '14 at 15:13
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Hint :

  • $1<X_1<5\Rightarrow 1-\alpha<(1-\alpha)X_1<5(1-\alpha)$
  • $1<X_1<5\Rightarrow \alpha<\alpha X_2<5\alpha$

Can you now bound $(1-\alpha)X_1+ \alpha X_2$ just by adding above two inequalities?