Lemma on bounding one metric by a multiple of another - clarification

Question

My book (Principles of Topology by Fred Croom) states the following lemma: "Let $d_1$ and $d_2$ be two metrics for the set $X$ and suppose that there is a positive number $c$ such that $d_1(x,y) \le cd_2(x,y)$ for all $x,y\in X$. Then the identity function $i: (X,d_2) \to (X,d_1)$ is continuous."

My question is simple: must $c$ be a constant independent of $x$ and $y$, or is it a not-necessarily-constant positive number? Experience suggests the former, but the lack of the word 'constant' and two possible interpretations of the wording make me question it. I worked out a proof that every metric space is homeomorphic to a bounded metric space in which it works to set $c=d(x,y)$, and this is the reason that I want to get this straight.

score 1 · Accepted Answer · answered Jun 06 '12 at 03:21

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The constant $c$ is independent of the points $x$ and $y$. To see why this is important, let $d_1$ be the usual metric on $\Bbb R$, and let $d_2$ be the discrete metric given by $$d_2(x,y)=\begin{cases}0,&\text{if }x=y\\1,&\text{if }x\ne y\;.\end{cases}$$ Let $f:\Bbb R\to\Bbb R:x\mapsto x$ be the identity map. It’s easy to see that $f$ is not continuous as a function from $\langle\Bbb R,d_1\rangle$ to $\langle\Bbb R,d_2\rangle$. On the other hand, if we set $c(x,y)=\max\{1,|x-y|\}$ for each $x,y\in\Bbb R$, it’s easy to check that $$d_1(x,y)=|x-y|\le c(x,y)\cdot1=c(x,y)d_2(x,y)$$ for all $x,y\in\Bbb R$.

answered Jun 06 '12 at 03:21

Brian M. Scott

616,228

I understand the first part. But for the second ("On the other hand..."), what exactly does that tell me? – Alex Petzke Jun 06 '12 at 22:30
@Alex: It shows you that there’s a way to choose the constants $c$, depending on $x$ and $y$, so that $d_1(x,y)$ is always bounded by $cd_2(x,y)$ even though the identity map is not continuous. – Brian M. Scott Jun 06 '12 at 22:37
I understand, thanks. – Alex Petzke Jun 07 '12 at 02:46

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

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In this case, they are indicating that $c$ is a constant independent of $x,y$. You are correct, though, that it isn't a necessary condition, only sufficient.

For more, see this question.

edited Apr 13 '17 at 12:20

Community

1

answered Jun 06 '12 at 03:20

Cameron Buie

102,994

I'm not quite pulling together what you're saying...can you explain more? I can't quite figure out how your two sentences aren't contradictory. – Alex Petzke Jun 06 '12 at 22:29
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If you follow the link to the next question, it tells you some other conditions under which the above-defined identity function is continuous. What I mean by "[it is a] suffficient [condition]" is that, if we make the assumption that $c$ is an independent constant, then the identity is continuous. However, there are other conditions (some strictly weaker, some independent) that yield the same conclusion, which is what I meant by "it isn't a necessary condition." Does that help? If not, we may be having a vocabulary miscommunication. What is your math background? – Cameron Buie Jun 07 '12 at 01:44
If I implied that I thought it wasn't a necessary condition but only sufficient, I didn't realize that I did so, hence the confusion. I'm fine now. Thanks. – Alex Petzke Jun 07 '12 at 02:48

Lemma on bounding one metric by a multiple of another - clarification

2 Answers2