4

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.

Gerry Myerson
  • 3,404

3 Answers3

2

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.]

Andreas Blass
  • 71,833
  • Thank you for your help. I am still not sure that this works with the definition provided in Bishop's book see my PPS in the original post. you must be able to take an arbitrary element $x$ of $\mathbb{R}$ and add it to $Y$ and the resulting set has to be bounded.

    In the case of the $Y$ you gave I think, if you choose x=0 than if you can find a bound $C>0$ you have solved the goldbach conjecture.

     –  Dec 13 '20 at 12:38
  • @Oliver3347 Thanks for adding the definitions to the question. Your comment shows why we can't say that $Y$ is bounded in $\mathbb R$. Yet it is bounded as a metric space in its own right. – Andreas Blass Dec 13 '20 at 16:26
  • The $Y$ in my answer is bounded as a metric space but we don't know (unless we can settle the Goldbach conjecture) that it's bounded as a subset of $\mathbb R$. To prove that it's bounded as a metric space, I must (according to the definition in your P.P.S.) exhibit a real number $C>0$ such that $(\forall x,y\in Y),|x-y|\leq C$. I exhibit $C=1$ and observe that, for all $x,y$ in $Y$, we have $x=y$ and so $|x-y|=0\leq 1$, as required. – Andreas Blass Dec 14 '20 at 03:29
0

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.

Hanul Jeon
  • 27,376
-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 $d_j = d(y_{k_j},y_{k_{j + 1}})$.

Jürgen Sukumaran
  • 7,645