Question about Bishop's "constructive analysis" for people familiar with bishop style constructive analysis only

Question

It has been a while since i read Bishop's book "constructive analysis", recently I dug it out of my book shelve and started to read. I came around this observation on the top of page 85.

"A subset Y of $\mathbb{R}$ can be bounded as a metric space but not bounded as a subset of $\mathbb{R}$."

I have tried to figure out how this can be and came up with nothing so far. Can anybody give me a hint?

PS: As far as I understand Bishop is referring to the metric induced by the absolute value.

PPS: The definition provided: A metric space $(X, p)$ is called bounded if there exists a real number $C > 0$, called a bound for $(X, p)$, such that $p(x, y) \leq C \forall x,y \in X$. A subset $Y$ of a non void metric space $X$ is bounded if, for all (equivalently, some) $x$ in $X$, the set $Y\cup\{x\}$ with the induced metric p* is a bounded metric space.

Answer 1

Let $Y=[0,\infty[$ and define the distance $d(x,y) = |\tan^{-1}(x) - \tan^{-1}(y)|$ for any $x,y\in Y$. Then $d(x,y)\leq |\tan^{-1}(x)| + |\tan^{-1}(y)| \leq \frac{\pi}{2} + \frac{\pi}{2} = \pi$. So the metric space $Y$ is bounded.

Assume $Y$ is unbounded. Then there is a sequence $\{y_k\}_{k\in\mathbb{N}}$ such that $|y_k|\to\infty$. Assume the metric space $(Y,d)$ is bounded with $d(x,y) = |x-y|$ for any $x,y\in Y$ and with some center $a\in \mathbb{R}$. Take $d_k = d(a,y_k)$. We have $d_k \geq ||a|-|y_k|| \to \infty$. So $d_k\to\infty$.

Some definitions of bounded metric space don't take a center, that's fine too. Since $y_k\to \infty$, we can pick a subsequence $\{y_{k_j}\}_{j\in\mathbb{N}}$ such that $||y_{k_j}| - |y_{k_{j+1}}||\to\infty$ and redefine $d_j = d(y_{k_j},y_{k_{j + 1}})$.

Answer 2

It is not an actual answer, but a bit long to be a comment.

Bishop's claim is refutable in the classical context; that is, every subset of $\mathbb{R}$ which is bounded as the metric space with the inherited metric is also a bounded subset of $\mathbb{R}$: let $X\subseteq\mathbb{R}$ is an inhabited subset which forms a bounded metric space with a bound $r$. Take any $a\in X$, then $X\subseteq \overline{B}(a,r)$.

Let $x\in \mathbb{R}$ be any real, and we will see that $X\cup\{x\}$ is a bounded metric space under the usual metric. Since $X$ is a bounded (in the classical sense), the set $$\{|x-y|: y\in X\}$$ is also bounded, so it has the supremum $M$. Then $\max(M,r)$ bounds $X\cup\{x\}$ as a metric space.

Thus we have to understand Bishop's claim as claiming the existence of a weak counterexample.

Answer 3

Let $Y=\{n\in\mathbb N:n\text{ is the first counterexample to the Goldbach conjecture}\}$. We don't know whether $Y$ has an element, but if it has one then it has only that one. That is, for any two elements of $Y$, the distance between them is $0$. So (assuming that Bishop's definition of "bounded metric space" is reasonable, i.e., that it's what I would use) I conclude that $Y$ is bounded as a metric space (under any metric). But to be able to say that $Y$ bounded as a subset of $\mathbb R$, we would need (again assuming a reasonable definition) to exhibit real numbers $c$ and $r$ (the center and radius of a metric ball) and prove that all elements of $Y$ are are within $r$ of $c$. And we don't know how to do that. [You might think we could be sneaky and take $c$ to be the unique element of $Y$ if $Y$ has an element and $0$ otherwise, because then any positive $r$ would work. But that "definition" of $c$ isn't constructive; we don't know which of the two cases holds. Or, as a constructivist would say, we don't know that $Y$ is inhabited or empty.]