1

Let $S$ be a half-open interval $(0,1]$. If we define $d$ on $S$ by

$$d(x,y) := \left|{1\over x} - {1\over y}\right|\;,$$

then show that $d$ is a metric on $S$. Also, prove that $(S,d)$ is a complete metric space.

I think that this problem would begin by showing the $3$ properties of a metric space. And then show that it is Cauchy?

Brian M. Scott
  • 616,228
cele
  • 2,247
  • 1
    No, once you’ve shown that $d$ is a metric, you need to show that every Cauchy sequence in $\langle S,d\rangle$ converges. Being Cauchy is a property of sequences, not of metrics or spaces. You can try to do that directly, but it might be easier to find an isometry between this space and the space $[1,\to)$ with the usual metric. – Brian M. Scott Oct 20 '13 at 23:48
  • @BrianM.Scott I know, I just meant to give an idea of what the space looked like. Take the wheel. =) – Pedro Oct 20 '13 at 23:51
  • @Kathyrn : I would go with Brian's hint. I think he means the space $[1,\infty)$. – Stefan Smith Oct 20 '13 at 23:53
  • @Pedro: I’ll admit that I started to make a somewhat similar suggestion, but then I decided that it had too much potential to confuse. – Brian M. Scott Oct 20 '13 at 23:53
  • @Stefan: Yes, $[1,\to)$ is synonymous with $[1,\infty)$. – Brian M. Scott Oct 20 '13 at 23:57
  • @BrianM.Scott how to find the isometry ? – Myshkin Oct 21 '13 at 03:10
  • 1
    @Sade: I’d prefer not to say until Kathryn’s had a chance to think about it, but the function that you need for the isometry can be found somewhere in the definition of the metric $d$. – Brian M. Scott Oct 21 '13 at 03:13
  • $f:(0,1]\to [1,\rightarrow),x\mapsto {1\over x}$ will do the job. Thank you. – Myshkin Oct 21 '13 at 03:20
  • @Kathryn: I’ll be happy to offer more help if you can be a bit more specific. Can you show that $d$ has any of the defining properties of a metric? – Brian M. Scott Oct 21 '13 at 03:49
  • well, i know how to show the properties of a metric, as far as d(x,y)=0 iff x=y and d(x,y)=d(y,x) for every x,y in X and the third being the triangle inequality. I do not understand what the definition of isometric is asking me to prove, therefore the concept of a completion of a metric space is unclear as well – cele Oct 21 '13 at 03:54
  • according to my text, an isometry is when a function f:(X,d) onto (Y,p) between 2 metrics and that if p(f(x),f(y))= d(x,y) holds for all x,y in X. This definition is confusing to me. – cele Oct 21 '13 at 03:57
  • Okay; give me a few minutes, and I’ll write up an answer to get you at least started in the right direction. – Brian M. Scott Oct 21 '13 at 04:01

1 Answers1

2

Just as two topological spaces are identical in terms of topological structure precisely when there is a homeomorphism between them, two metric spaces are identical in terms of metric structure precisely when there is an isometry between them. In both cases you could say that they’re the same space under a different name. My suggestion in the comments was that $S$ with the metric $d$ is structurally identical to the ray $[0,\to)$ with the usual metric. The idea is that a point $x\in S$ corresponds to $\frac1x\in[1,\to)$. This map $x\mapsto\frac1x$ really is an isometry: if $x,y\in S$, the $d$-distance between $x$ and $y$ is exactly the same as the ordinary distance between $\frac1x$ and $\frac1y$ in $[1,\to)$. And the space $[1,\to)$ with the usual metric is complete, so we can use this correspondence to show that $\langle S,d\rangle$ is complete as well.

Suppose that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence in $\langle S,d\rangle$. This means that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d(x_k,x_n)<\epsilon$ whenever $k,n\ge m$, i.e., that

$$\left|\frac1{x_k}-\frac1{x_n}\right|<\epsilon\qquad\text{whenever}\qquad k,n\ge m_\epsilon\;.$$

We want to show that the sequence converges to some $x\in S$. Look at the corresponding sequence $$\left\langle\frac1x_n:n\in\Bbb N\right\rangle\tag{1}$$ in $[1,\to)$. For convenience of notation let $y_n=\frac1{x_n}$, so that $(1)$ is just $\langle y_n:n\in\Bbb N\rangle$. Is this a Cauchy sequence in $[1,\to)$ with the usual metric? Absolutely: if $k,n\ge m_\epsilon$ (that’s the same $m_\epsilon$ as before), then

$$|y_k-y_n|=\left|\frac1{x_k}-\frac1{x_n}\right|=d(x_k,x_n)<\epsilon\;.$$

And $[1,\to)$ is complete, so $\langle y_n:n\in\Bbb N\rangle$ converges to some limit $y$.

At this point I’ve done most of it; can you use $y$ to find the limit of the original sequence $\langle x_n:n\in\Bbb N\rangle$ in $S$ and show that what you find really is the limit?

Brian M. Scott
  • 616,228
  • oh, ok, i think i now get what it means when they say that two metrics have the same isometry, that makes perfect sense now. give me a minute to wrap my mind around the rest. – cele Oct 21 '13 at 04:26
  • @Kathryn: It can’t: $0$ isn’t in $S$. You know that $\langle y_n:n\in\Bbb N\rangle\to y$, so for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $|y_n-y|<\epsilon$ whenever $n\ge m_\epsilon$. That means that $$\left|\frac1{x_n}-y\right|<\epsilon$$ whenever $n\ge m_\epsilon$. Can you use this to find an $x\in S$ such that $$d(x_n,x)=\left|\frac1{x_n}-\frac1x\right|<\epsilon$$ whenever $n\ge m_\epsilon$? – Brian M. Scott Oct 21 '13 at 04:40