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Show that a subset $I$ of $\textbf{R}$ is an interval if and only if whenever $a,b\in I$ and $0\leq t\leq 1$ then $(1-t)a + tb\in I$.

MY ATTEMPT

We say that a subset $I$ of $\textbf{R}$ is an interval iff $a \leq c \leq b$ and $a,b\in I$ implies that $c\in I$. Based on such definition, we may proceed.

Let us prove the implication $(\Rightarrow)$ first.

Suppose that $I$ is an interval and $a,b\in I$ s.t. $a \leq x \leq b$.

We shall prove that $(1-t)a + tb\in I$ whenever $t\in[0,1]$.

In order to do so, let us consider that $t = (x-a)/(b-a)\in[0,1]$.

We have that $t\in[0,1]$ because $a\leq x\leq b$. Moreover, we do also have that

\begin{align*} x = a + x - a = a + (b-a)\times\frac{x-a}{b-a} = a + (b-a)t = (1-t)a+ bt\in I \end{align*}

Conversely, let us prove the implication $(\Leftarrow)$.

Let us suppose that $a,b\in I$ and $(1-t)a + tb\in I$ whenever $t\in[0,1]$.

Thus, if $a\leq x\leq b$, we have \begin{align*} t = \frac{x-a}{b-a}\in[0,1]\Rightarrow (1-t)a + bt = a + t(b-a) = a + \frac{x-a}{b-a}\times(b-a) = a + x - a = x \in I \end{align*} and we are done.

Could someone please check if the wording of my proof is correct?

Any contribution is appreciated.

user0102
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  • Your $(\implies)$ has to be done more carefully. You have to show that given any $t \in [0, 1]$, the element $tb + (1 - t)a$ must be in $I$. The way you have done currently doesn't make that clear. (Unless you show that as you vary $c \in [a, b]$, you get all possible $t$. Does this make sense to you?) – Aryaman Maithani Jul 08 '20 at 22:09
  • Thanks for the feedback! Yes, it does make sense to me. But, as you have mentioned, $c$ is not fixed. It may vary between $a$ and $b$. – user0102 Jul 08 '20 at 22:12
  • In the $(\impliedby)$ part, you'd have to take care of $a = b$ separately since you're dividing by $b - a$. (And just depending on the level at which you're attempting this, maybe you should justify why $\dfrac{c-a}{b-a} \in [0, 1]$). – Aryaman Maithani Jul 08 '20 at 22:13
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    Yes, $c$ varies but you must show that it then covers every $t \in [0, 1].$ The cleaner way to do it would be to first start with an arbitrary $t \in [0, 1]$ and then set $c = (1 - t)a + bt$. Then, you should argue that $a \le c \le b$. Using the fact that $I$ is an interval, then you conclude that $c \in I$ and thus, you are done. – Aryaman Maithani Jul 08 '20 at 22:14
  • I have substituted the $c$ by $x$. Hopefully it is clearer now. Since the function $t(x) = (x-a)/(b-a)$ is "linear" and $a < b$, we conclude that $t(x)\in[0,1]$ as it has been claimed. – user0102 Jul 08 '20 at 22:17
  • The issue is not that $t \in [0, 1]$ but the converse that every $t$ is indeed obtained. For example, suppose that in your definition, to begin with, you had $t \in [0, 2]$ instead. According to what you've done, your proof would still have gone through, right? However, that equivalence is false. – Aryaman Maithani Jul 08 '20 at 22:20
  • @AryamanMaithani I've tried to fix it. Could you please double-check my arguments? If you think it still deserves any corrections, please let me know. – user0102 Jul 08 '20 at 22:29
  • The error is still pretty much the same. You've written "We shall prove that $(1-t)a + tb\in I$ whenever $t\in[0,1]$. $$ $$

    In order to do so, let us consider that $t = (x-a)/(b-a)\in[0,1]$." However, how do you know that every $t \in [0, 1]$ is of the form $(x-a)/(b-a)$? (For some $a \le x \le b$, presumably.)

     – Aryaman Maithani Jul 08 '20 at 22:32
  • You should try to prove that $(1-t)a+tb\geq a$ ie $tb\geq ta$ ie $b\geq a$ which is already the case. Similarly prove that $(1-t)a+tb\leq b$ and here you will need the fact $t\in[0,1]$. – Paramanand Singh Jul 09 '20 at 00:48
  • @ParamanandSingh Thanks for the comments. Could you please provide a full answer? – user0102 Jul 09 '20 at 00:50

1 Answers1

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The issue with your approach has been described in comments to your question, but for completeness sake let us put it here also. You are supposed to show that for every $t\in[0,1]$ the number $(1-t)a+tb$ lies in interval $[a, b] $. But then you take up $t$ in some specific form $t=(x-a) /b-a$. And you write that $t\in[0,1]$ because $a\leq x\leq b$. Well, you had to prove the other way round that if $t\in[0,1]$ then $x\in[a, b] $.

Just do it directly. Let $t\in [0,1]$ and we need to show $$(1-t)a+tb\geq a$$ ie $$tb\geq a-(1-t)a=ta$$ which clearly holds if $t=0$ and if $t>0$ this is equivalent to $b\geq a$ which is already part of hypothesis. And thus we have shown $$(1-t)a+tb\geq a$$ Next we also need to show $$(1-t)a+tb\leq b$$ or $$(1-t)a\leq (1-t)b$$ Clearly this holds for $t=1$ and if $0\leq t <1$ then we can see (via dividing by positive number $1-t$) that it is equivalent to $a\leq b$ which is already given. This completes the proof of first part.

For the second part let us assume that for every $t\in[0,1]$ the number $(1-t)a+bt\in I$. If $a\leq c\leq b$ then we can write $$c=(1-t) a+tb$$ where $t=(c-a) /(b-a) $. Clearly $$0\leq c-a\leq b-a$$ and hence $0\leq t\leq 1$ and then by our assumption $c=(1-t)a+tb\in I$ and therefore $I$ is an interval. Also note that if $a=b$ then this has to be handled specially.

Paramanand Singh
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