5
$\begingroup$

I got stuck on this problem. Hope someone can give some hint to move on. Thanks.

Suppose $d_1(x,y) = |x-y|$, $d_2(x,y)=|\phi(x) - \phi(y)|$ where $\phi(x) = {x \over {1 + |x|}}$. Prove that $d_1$ and $d_2$ are metrics on $\mathbb{R}$ which induce the same topology, although $d_1$ is complete and $d_2$ is not.

For this problem, I can prove the part $d_1$ and $d_2$ are both metrics, $d_1$ is complete and $d_2$ is not. But I got stuck on the part to prove that two metrics induce the same topology (meaning that if $U$ is open in $(X,d_1)$, then it must be open in $(X, d_2)$).

More detail, for example, given $\epsilon > 0$, I need to find $\delta > 0$ such that if $d_2(x,y) < \delta$, then $d_1(x,y) < \epsilon$ and vice versa. Can anyone give me some hint to find those $\delta$. Thanks

$\endgroup$
4
$\begingroup$

The last statement is false: if you could always find such $\delta$ for each $\varepsilon$ such that it held for any pair $x,y$, and vice versa, the metrics would be uniformly equivalent. This implies that as uniform spaces they have the same uniformity and so their notion of completeness would also agree.

What you need to prove is that each ball with respect to $d_1$ contains one (with the same centre) with respect to $d_2$ and vice versa.

So you need

$$\forall x \in X \forall \varepsilon > 0 \exists \delta > 0 : B_{d_2}(x,\delta) \subseteq B_{d_1}(x,\varepsilon)\text{,}$$ and vice versa with $d_1$ and $d_2$ interchanged.

The inclusion can also be written as a quantifier statement and then gives:

$$\forall x \in X \forall \varepsilon > 0 \exists \delta > 0 : \forall y \in X: ( d_2(x,y) < \delta \rightarrow d_1(x,y) < \varepsilon )$$

(And the reverse too.) Note that the $\delta$ can depend on $x$ and in your statement it does not: you choose the same $\delta$ for all $x,y$.

As $d_2(x,y) = d(\phi(x),\phi(y))$ to prove this is closely related to showing $\phi$ continuous...

$\endgroup$
  • $\begingroup$ I see. So for the direction: $$\forall x \in X \forall \varepsilon > 0 \exists \delta > 0 : B_{d_1}(x,\delta) \subseteq B_{d_2}(x,\varepsilon)\text{,}$$, it is exactly the definition continuity of $\phi$, so we done. But the the converse direction, which is $$\forall x \in X \forall \varepsilon > 0 \exists \delta > 0 : B_{d_2}(x,\delta) \subseteq B_{d_1}(x,\varepsilon)\text{,}$$, can you show me how you choose $\delta$. I still got stuck on that. Sorry $\endgroup$ – le duc quang May 1 '16 at 16:52
  • $\begingroup$ @leducquang $\phi$ has an inverse $\psi$ (what is it?) and so $d_1(x,y) = d_1(\phi(\psi(x)), \phi(\psi(x))) = d_2(\psi(x), \psi(y))$... $\endgroup$ – Henno Brandsma May 1 '16 at 17:24
  • $\begingroup$ I see your point. I actually thought about that, but hesitated because there're 2 inverse functions based on condition $x >=0$ or not. But I reconsider, as long as those 2 functions continous in their domain and continuous in the intersection, it's ok. Thanks a lot $\endgroup$ – le duc quang May 1 '16 at 17:29
  • $\begingroup$ @leducquang Glad I could help! $\endgroup$ – Henno Brandsma May 1 '16 at 17:31

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.