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I'm currently studying in the book of Analysis of Terry Tao, amazing book by the way. In one exercise I'm not pretty sure about how can do it (I know that will be almost trivial but I'm stuck in it).

Definition: Let $\varepsilon >0$ and $x,y\in \mathbb{Q}$. We say that $y \text{ is } \,\,\varepsilon\text{ - close to } x \iff d(x,y)\le \varepsilon$; where $ d(x,y)$ is the distance of $y$ to $x$ define as $d(x,y)= \,\mid x-y\mid.$

Ex: Let $\varepsilon>0$. If $x$ and $y$ are $\varepsilon \text{ - close}$ to $x$, and $w$ is between $y$ and $z$, then $w$ is also $\varepsilon \text{ - close}$ to $x$.

So my question: is there a clever way to do it without divide the exercise by cases when $y\le w\le z$ and then when $z\le w\le y$? Sorry if I'm not put my attempt, everything I tried is completely wrong. Thanks in advance.

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    $\begingroup$ You can say "without loss of generality, we may assume that $x\leq y$". This basically says that "the same proof works if $y\leq x$" $\endgroup$
    – Prahlad Vaidyanathan
    Commented Sep 4, 2013 at 16:51
  • $\begingroup$ @Prahlad Vaidyanathan; Without loss of generality we may assume that $x \le y$. Either $y\le w\le z$ or $z\le w\le y$. If $y\le w\le z$, clearly $0 \le y-x \le w-x \le z-x $. Thus, $\vert w- x \vert = w-x$ and $\vert z-x \vert =z-x$. It follows that $\vert w- x \vert \le \vert z- x \vert \le \varepsilon $. On the hand, if $z\le w\le y$, either $w-x \ge 0$ or $\,w-x \le 0 $. If $w-x \ge 0$, then $\vert w-x\vert \le \vert y-x \vert \le \varepsilon$. If $w-x \le 0$, then $0\le\vert w-x \vert = x-w\le \vert x-z \vert \le \varepsilon$. Hence in either case $d(x,w) \le \varepsilon$. $\endgroup$
    – Jose Antonio
    Commented Sep 4, 2013 at 19:12

2 Answers 2

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One characterization/definition/property of "between" is that $d(x,w) + d(w,y) = d(x,y)$. The distances to $x$ and $y$ are bounded by the $xy$ distance.

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  • $\begingroup$ But why not $d(x,w) + d(w,y) \ge d(x,y)$ since, using the definition of distance as the book says, $d(x,y)=\vert x-y\vert =\vert (x-w)+(w-y)\vert\, \le \,\vert x-w \vert +\vert w-y \vert = d(x,w)+d(w,y)$ $\endgroup$
    – Jose Antonio
    Commented Sep 4, 2013 at 17:19
  • $\begingroup$ The triangle in-equality does hold if its stronger special case, with equality, applies. The equality has additional implications that the inequality does not, such as the conclusion of this exercise being true. $\endgroup$
    – zyx
    Commented Sep 4, 2013 at 17:48
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There is a typo in the original post, saying "if x and y are e-close to x", instead of "y and z" being e-close to x.

Claim: Let $\epsilon > 0$. If $y$ and $z$ are both $\epsilon$-close $x$, and $w$ is between $y$ and $z$ (i.e., $y \leq w\leq z$ or $z\leq w \leq y$), then $w$ is also $\epsilon$-close to $x$.

Pf. Suppose $y \leq w \leq z$. We need to show that $d(x, w) \leq \epsilon$. By definition it suffices to show that $|x - w| \leq \epsilon$, i.e. that, $-\epsilon \leq x - w \leq \epsilon$ or that $w \leq x + \epsilon$ and $x - \epsilon \leq w$. To prove the claim, we show that these last two inequalities hold.

Since $d(x, y) \leq \epsilon$, we have $-\epsilon \leq x - y \leq \epsilon$. From the last inequality we have that $x - \epsilon \leq y$, and since $y \leq w$, by transitivity, $x - \epsilon \leq w$, which shows that the second inequality holds.

On the other hand since $d(x, z) \leq \epsilon$, $-\epsilon \leq x - z \leq \epsilon$. It follows that $z \leq x + \epsilon$ and since $w \leq z$, by transitivity again, we conclude that $w \leq x + \epsilon$, which shows that the first inequality also holds and proves the claim for the case where $y \leq w \leq z$. The other case can be proved similarly.

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