$\varepsilon$ - closeness property

Question

I'm studying Analysis from Terence Tao's book 'Analysis 1' and in an exercise he asks to prove seven properties regarding the notion of '$\varepsilon$ - closeness', which is defined as follows:

Def: Let $\varepsilon>0$, $x$, $y$ be rational numbers. We say that $y$ is $\varepsilon$-close to $x$ iff we have $d(y,x)\leq \varepsilon$.

I've managed to prove six out of the seven properties; the only one which I haven't been able to prove is the following one:

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

So, I would appreciate any hints about proving this property.

Best regards,

lorenzo.

Answer 1

Drawing a picture helps with this one. For one case, let's assume $y \leq w \leq z \leq x$.
$$x - w \leq x - y$$ So then $|x - w| \leq |x - y| \leq \epsilon$. You can then run through the other possibilities in a similar way.

Answer 2

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.