Let $S$ be a subset of $\mathbb{R}$. If the diameter of $S$ is finite, that is, $\forall x, y \in S$, if $|x-y| \leq M$ for some non-negative real number $M$, why does it imply that $\forall x \in S, |x| \leq K$ for some $K \in \mathbb{R}$? Help me with the proof.
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4What is the definition of $|x|$ in a metric space? – Dave L. Renfro Oct 05 '16 at 16:21
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1What does |x| mean? Is there any relationship between |x| and |y| and |x-y|? – fleablood Oct 05 '16 at 16:27
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1And seriously, what does |x| mean? Not all metric spaces has |x| meaning anything. – fleablood Oct 05 '16 at 16:29
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I presume that since $S$ is a subset of $\mathbb{R}$, the OP is looking at the usual metric on $\mathbb{R}$, and $\vert x\vert$ means the usual absolute value (based on their comment to my answer below). – Noah Schweber Oct 05 '16 at 16:47
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EDIT:Since $S\subseteq\mathbb{R}$, I'm assuming that the metric on $S$ is (the restriction of) the usual one on $\mathbb{R}$, and $\vert x\vert$ is just the usual absolute value. In particular, we have $\vert x\vert=d(x, 0)$; note that although $0$ need not be in $S$, "$d(x, 0)$" still makes sense as a real number (just interpret $d$ in $\mathbb{R}$).
If this is correct, then the argument is just an application of the triangle inequality! We know $S$ has finite diameter $D$; so fix some element $s\in S$ (I'm assuming $S$ is nonempty, since otherwise it's trivial). Let $d=\vert s\vert$; then for $t\in S$ we have $$\vert t\vert=d(t, 0)\le d(t, s)+d(s, 0)\le D+d.$$
Noah Schweber
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@Vpd $d$ is a metric - by definition $d(a, b)$ is always finite. Also, you need to explain what you mean by "$\vert x\vert$". – Noah Schweber Oct 05 '16 at 16:41
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@Vpd Ah, so you are definitely using the usual metric on $\mathbb{R}$ to view $S$ as a metric space. In general, note that most metric spaces do not come equipped with a notion of absolute value! – Noah Schweber Oct 05 '16 at 16:48
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@Vpd Do you understand my answer re: $d(s, 0)$? (Note the line "let $d=\vert s\vert$", in conjunction with the fact that $\vert x\vert=d(x, 0)$.) – Noah Schweber Oct 05 '16 at 16:49
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$\forall x, y \in S, d(x,y)$ is finite ($d$ being the usual metric). But $0$ need not be an element of $S$. Then how can we say $d(s,0)$ is finite? – Oct 05 '16 at 16:51
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@Vpd If $d$ is the restriction to $S$ of the usual metric on $\mathbb{R}$, then $d(s, 0)$ still makes sense even if $0\not\in S$; just compute the distance in $\mathbb{R}$! (That is, I'm saying "$\forall x, y\in\mathbb{R}, d(x, y)$ is finite.") If $d$ is not the restriction to $S$ of the usual metric on $\mathbb{R}$, then we're back to square one about how $d$ and $\vert\cdot\vert$ interact. So: what is your definition of $d(x, y)$ for $x, y\in S$? – Noah Schweber Oct 05 '16 at 16:53
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