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Let $Z$ be a compact metric space and $\rho_n$ a sequence of metrics with diameter 1 that converge to a metric $\rho$. Show that $\text{diam}(Z,\rho) \leq 1$ (where $\text{diam}(Z,\rho) := \sup_{\xi, \eta \in Z} \rho(\xi, \eta)$).

First of all: Can you give me an example where $diam(Z,\rho) < 1$?

And more importantly, could you check my proof?

Let $\xi_0, \eta_0 \in Z$ be such that $\rho(\xi_0, \eta_0) = \text{diam} (X, \rho)$. Clearly, for every $n\in \mathbb{N}$, we have $\rho_n (\xi_0, \eta_0) \leq \sup_{\xi, \eta \in Z} \rho_n(\xi, \eta)$. Taking the limit on both sides, we get

$$\lim_{n \to \infty} \rho_n (\xi_0, \eta_0) \leq \lim_{n \to \infty} \sup_{\xi, \eta \in Z} \rho_n(\xi, \eta) $$

and therefore

$$\text{diam} (Z, \rho) \leq 1.$$

Georgios
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1 Answers1

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First, the question is unclear - are we supposed to assume that the metrics $\rho_n$ and $\rho$ all induce the same compact topology? You say $Z$ is a compact metric space but don't specify what if anything this has to do with $\rho_n$ and $\rho$.

Second, luckily the problem mentioned above is irrelevant. Compactness is irrelevant. If $(\rho_n)$ is any sequence of metrics on any space, compact or not, the $\rho_n$ all have diameter $\le1$ and $\rho_n\to\rho$ then $\rho$ hhas diameter $\le 1$.

This is trivial. For any $x,y\in Z$ we have $$\rho(x,y)=\lim\rho_n(x,y)\le1,$$since each $\rho_n$ has diameter no larger than $1$. Hence $$\sup_{x,y\in Z}\rho(x,y)\le 1.$$

David C. Ullrich
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  • This statement was part of another proof where $Z$ is assumed to be compact. Thanks for the clarification that this assumption is not necessary. As for your proof: The metrics $\rho_n$ have diameter 1, not $\leq 1$. – Georgios Mar 08 '16 at 15:29
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    And if one proves [something] assuming that $\rho_n$ has diameter $\le1$ it follows that [something] holds in the case where $\rho_n$ has diameter $1$, because $1\le1$. – David C. Ullrich Mar 08 '16 at 15:32
  • I see. So is my proof unnecessarily complicated or even wrong? Also: Is it as trivial to find an example? Any hint? – Georgios Mar 08 '16 at 15:37
  • The things I said were unclear about your question were irrelevant to showing that the diameter of $\rho$ is $\le1$, but it seems to me they do make a difference regarding that example. You need to clarify the question. – David C. Ullrich Mar 08 '16 at 15:51
  • The metrics $\rho_n$ are supposed to be bi-lipschitz equivalent to each other. As for $\rho$, I can't see any such assumption in the original proof. It simply states that its diameter is $\leq 1$ and then proceeds by showing that it has diameter 1 (using other properties of the $\rho_n$'s). (The original proof I'm talking about is the proof to lemma 2.9 in this paper: http://maths.rkmvu.ac.in/~kingshook/isomex.pdf (p.5).) – Georgios Mar 08 '16 at 16:10