2
$\begingroup$

From the notes: Let $(X, d_1)$ and $(X, d_2)$ denote the two metric spaces, where X is compact and suppose both metrics induce the same topology. Then the identity maps $i : (X, d_1) \to (X, d_2)$ and $j : (X, d_2) \to (X, d_1)$ are continuous and since X compact, they are uniformly continuous. Fix $ε_1 > 0$. Then, by uniform continuity, we can subsequently find an $ε_2 > 0$ and $ε_3 > 0$ such that for all x, y ∈ X: $d_1(x, y) < ε_1 ⇒ d_2(x, y) < ε_2$ and $d_2(x, y) < ε_2 ⇒ d_1(x, y) < ε_3$

My problem is that my understanding of uniform continuity is the other way around, given $\epsilon_2$ we can find such an $\epsilon_1$ and so on. I tried to prove the above my contradiction,which would mean for every $\epsilon_2>0$ there are some x,y in X such that $d_1(x, y) < ε_1 ⇒ d_2(x, y) \geq ε_2$. I tried to show this contradicts uniform continuity of either i or j but I was not successful. Could someone help? Thanks.

$\endgroup$
  • $\begingroup$ Without knowing the full context in which these notes were presented, perhaps it was supposed to read: "Fix $\varepsilon>0$. Then by uniform continuity, we may subsequently find $\delta_1>0$ and $\delta_2>0$ such that for all $x, y \in X: d_1(x,y)<\delta_1 \implies d_2(x,y)<\varepsilon$ and $d_2(x,y)<\delta_2 \implies d_1(x,y)<\varepsilon$." Which would be fine because clearly there would just be a second Lebesgue's Number for the same number $\varepsilon>0$. I mean if we were to write out the proof showing that identity map is uniformly $(d_2, d_1)$-continuous. $\endgroup$ – Matt A Pelto Aug 10 '17 at 7:59
  • $\begingroup$ or another possibility: "Fix $\varepsilon>0$. Then by uniform continuity, we may subsequently find $\delta_1>0$ and $\delta_2>0$ such that for all $x,y \in X: d_2(x,y)<\delta_2 \implies d_1(x,y)<\delta_1$ and $d_1(x,y)<\delta_1 \implies d_2(x,y)<\varepsilon$." In this case we would write the second half of the proof by applying the definition of uniformly $(d_2,d_1)$-continuous to the positive number $\delta_1:=\lambda$ (instead of $\varepsilon>0$). $\endgroup$ – Matt A Pelto Aug 11 '17 at 6:50
  • 1
    $\begingroup$ Here are the notes, look at page 116, proof of number 1. staff.science.uu.nl/~kraai101/lecturenotes2009.pdf Thanks. $\endgroup$ – user172377 Aug 11 '17 at 7:10
  • $\begingroup$ Unless the author is sneaky and I am not looking carefully enough, it sounds like they have some sort of Lipschitz definition in mind. Perhaps: proofwiki.org/wiki/Definition:Lipschitz_Equivalence/Metrics -- IE; $\varepsilon_2:=k_1\varepsilon_1$ and $\varepsilon_3:=k_2\varepsilon_2$. Though I am now just speculating over what definition might have to be satisfied before we may so dubiously find $\varepsilon_2>0$ and $\varepsilon_3>0$. $\endgroup$ – Matt A Pelto Aug 11 '17 at 23:45
  • $\begingroup$ While the writing could be slightly more detailed or rather less allusive in a few spots, you might want to see the proof of Lemma 5.3 on page 9 here: math.uchicago.edu/~may/REU2014/REUPapers/Butt.pdf $\endgroup$ – Matt A Pelto Aug 11 '17 at 23:53
1
$\begingroup$

We first simplify notation by setting $K_1=(X,d_1)$ and $K_2=(X, d_2)$. We shall show that the identity map on $X$ is uniformly $(d_1,d_2)$-continuous.

Let $\varepsilon>0$ be given. For each $x \in K_2$ we define \begin{equation} U_x : = \displaystyle \left\{ y \in X : d_2(\,y,x)<\frac{\varepsilon}{2} \right\}. \end{equation} Since the identity map $f(x)=x$ is trivially $(d_1,d_1)$-continuous, we have that for every $x \in K_2$ the set $U_x$ is open in the compact metric space $K_1$ (Definition of Topologically Equivalent Metrics). So the family of open sets $\,\mathcal{U} := \{U_x : x \in K_2\}$ forms an open cover of $K_1$. Thus there is a number $\lambda>0$ such that for every $y \in K_1$ the open ball $B( y, \lambda):= \{z \in X: d_1(z,y)<\lambda\}$ is contained in one of the open sets of $\mathcal{U}$ (Lebesgue's Number Lemma). Therefore \begin{aligned} d_2(z,y) & \leq d_2(z,x) + d_2(x,y) \\ & < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon \; \; \:\; \text{ whenever }\, d_1(z,y)<\lambda \text{, and } z,y \in K_1. \end{aligned} Without loss of generality, the identity map is uniformly $(d_2, d_1)$-continuous as well.

$\endgroup$
  • $\begingroup$ proofwiki.org/wiki/Definition:Topologically_Equivalent_Metrics -- It sounds like y'all (:you all) are using Definition 2. Either definition of topologically equivalent metrics is fine, it is only needed to know that $U_x$ is open for every $x \in X$. In fact Definition 2 (the one you seem to cite) clearly says so much directly. I suspect Lebesgue's Number Lemma to be the difference in whatever you have seen. $\endgroup$ – Matt A Pelto Aug 9 '17 at 9:50
  • 1
    $\begingroup$ What does (d1,d2) uniformly continuous mean? Does it mean the metric for its domain is d1 and d2 for its range? $\endgroup$ – user172377 Aug 9 '17 at 11:15
  • $\begingroup$ We say that a function $f:(X,d_1) \to (Y, d_2)$ is uniformly $(d_1,d_2)$-continuous if and only if for all $\varepsilon>0$, there exists a $\delta>0$ such that $d_2\left(\,f(x),f(y)\right)<\varepsilon$ whenever $d_1(x,y)<\delta$, and $x,y \in X$. It is not exactly any different from the standard definition of uniformly continuous. I only say uniformly $(d_1,d_2)$-continuous to make the context more clear. $\endgroup$ – Matt A Pelto Aug 9 '17 at 18:13
  • $\begingroup$ So the answer to your second question is an affirmative yes. $\endgroup$ – Matt A Pelto Aug 9 '17 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy