If $I$ an interval and $x_1,x_2,x_3in I$, why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$?

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If $I$ an interval and $x_1,x_2,x_3in I$. Why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$ ?



Attempts : $t_3=1-t_1-t_2$ and thus $$t_1x_2+t_2x_2+t_3x_3=underbracet_1x_1+(1-t_1)x_3_in I+t_2(x_2+x_3).$$
Now, $t_1x_1+(1-t_1)x_3in I$, but this don't prove that $t_1x_1+(1-t_1)x_3+t_2(x_2+x_3)in I$. Any idea ?







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  • Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
    – Kavi Rama Murthy
    Aug 21 at 10:09















up vote
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down vote

favorite












If $I$ an interval and $x_1,x_2,x_3in I$. Why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$ ?



Attempts : $t_3=1-t_1-t_2$ and thus $$t_1x_2+t_2x_2+t_3x_3=underbracet_1x_1+(1-t_1)x_3_in I+t_2(x_2+x_3).$$
Now, $t_1x_1+(1-t_1)x_3in I$, but this don't prove that $t_1x_1+(1-t_1)x_3+t_2(x_2+x_3)in I$. Any idea ?







share|cite|improve this question




















  • Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
    – Kavi Rama Murthy
    Aug 21 at 10:09













up vote
1
down vote

favorite









up vote
1
down vote

favorite











If $I$ an interval and $x_1,x_2,x_3in I$. Why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$ ?



Attempts : $t_3=1-t_1-t_2$ and thus $$t_1x_2+t_2x_2+t_3x_3=underbracet_1x_1+(1-t_1)x_3_in I+t_2(x_2+x_3).$$
Now, $t_1x_1+(1-t_1)x_3in I$, but this don't prove that $t_1x_1+(1-t_1)x_3+t_2(x_2+x_3)in I$. Any idea ?







share|cite|improve this question












If $I$ an interval and $x_1,x_2,x_3in I$. Why $t_1x_1+t_2x_2+t_3x_3in I$ when $t_1+t_2+t_3=1, t_iin [0,1]$ ?



Attempts : $t_3=1-t_1-t_2$ and thus $$t_1x_2+t_2x_2+t_3x_3=underbracet_1x_1+(1-t_1)x_3_in I+t_2(x_2+x_3).$$
Now, $t_1x_1+(1-t_1)x_3in I$, but this don't prove that $t_1x_1+(1-t_1)x_3+t_2(x_2+x_3)in I$. Any idea ?









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asked Aug 21 at 10:04









Henri

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  • Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
    – Kavi Rama Murthy
    Aug 21 at 10:09

















  • Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
    – Kavi Rama Murthy
    Aug 21 at 10:09
















Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
– Kavi Rama Murthy
Aug 21 at 10:09





Assume $a<x_i<b$ for each $i$ and verify that $a<sum x_it_i<b$ . Similar argument works if $I$ is not open but closed or half-closed.
– Kavi Rama Murthy
Aug 21 at 10:09











2 Answers
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WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
$$
x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
$$
Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.






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    $$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$






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    • But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
      – Henri
      Aug 21 at 13:09






    • 1




      The two fractions add to 1.
      – Empy2
      Aug 21 at 13:11










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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    up vote
    7
    down vote



    accepted










    WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
    $$
    x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
    $$
    Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.






    share|cite|improve this answer


























      up vote
      7
      down vote



      accepted










      WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
      $$
      x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
      $$
      Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.






      share|cite|improve this answer
























        up vote
        7
        down vote



        accepted







        up vote
        7
        down vote



        accepted






        WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
        $$
        x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
        $$
        Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.






        share|cite|improve this answer














        WLOG, suppose $x_1 leqslant x_2leqslant x_3$, then
        $$
        x_1 =(t_1+t_2+t_3)x_1leqslant t_1 x_1 + t_2 x_2 + t_3 x_3 leqslant (t_1 + t_2 + t_3)x_3 = x_3.
        $$
        Since $[x_1, x_3]subset I$, so is $t_1 x_1 + t_2 x_2 + t_3x_3$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 21 at 10:09









        Stefan4024

        29.2k53276




        29.2k53276










        answered Aug 21 at 10:09









        xbh

        2,617114




        2,617114




















            up vote
            1
            down vote













            $$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$






            share|cite|improve this answer




















            • But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
              – Henri
              Aug 21 at 13:09






            • 1




              The two fractions add to 1.
              – Empy2
              Aug 21 at 13:11














            up vote
            1
            down vote













            $$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$






            share|cite|improve this answer




















            • But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
              – Henri
              Aug 21 at 13:09






            • 1




              The two fractions add to 1.
              – Empy2
              Aug 21 at 13:11












            up vote
            1
            down vote










            up vote
            1
            down vote









            $$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$






            share|cite|improve this answer












            $$t_1x_1+t_2x_2+t_3x_3=(t_1+t_2)left(fract_1t_1+t_2x_1+fract_2t_1+t_2x_2right)+t_3x_3$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 21 at 10:52









            Empy2

            31.9k12059




            31.9k12059











            • But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
              – Henri
              Aug 21 at 13:09






            • 1




              The two fractions add to 1.
              – Empy2
              Aug 21 at 13:11
















            • But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
              – Henri
              Aug 21 at 13:09






            • 1




              The two fractions add to 1.
              – Empy2
              Aug 21 at 13:11















            But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
            – Henri
            Aug 21 at 13:09




            But the problem is still the same : why $fract_1t_1+t_2x_1+fract_2t_1+t_2x_2in I$ ?
            – Henri
            Aug 21 at 13:09




            1




            1




            The two fractions add to 1.
            – Empy2
            Aug 21 at 13:11




            The two fractions add to 1.
            – Empy2
            Aug 21 at 13:11












             

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