The standard metric on $\Bbb R$ is $d_1(x,y)=|x-y|$. Over $(0,1)$ is proven that the metric $d_1$ is equivalent to the metric $d_2(x,y)=|\frac1x-\frac1y|$.

Now we want to prove that doesnt exists a metric $d_2$ on $\Bbb R$ such that is equivalent to $d_1$ and when it is restricted to $(0,1)$ is $d_2(x,y)=|\frac1x-\frac1y|$.

The criteria to be equivalent is that for any $\epsilon>0$ and any $x\in X$ exists $r,s>0$ such that

$$B_1(x,r)\subseteq B_2(x,\epsilon)\land B_2(x,s)\subseteq B_1(x,\epsilon)\tag{1}$$

where $B_j(x,\ell)$ is an open ball of center $x$, radius $\ell$ using the metric $d_j$ defined over $X$.

My thoughts: it is obvious that a direct extension of $d_2(x,y)=|\frac1x-\frac1y|$ to $\Bbb R$ is impossible because $1/0$ is undefined. Then I tried to define a picewise metric of the kind

$$d_2(x,y):=|f(x)-f(y)|=\begin{cases}f(x)=\frac1x,&\text{if }x\in(0,1)\\f(x)=h(x),&\text{otherwise }\end{cases}$$

and tested using $h(x)=x$, but this not result in a metric because the triangle inequality fails. But I found an expansion of $d_2$ on $\Bbb R$ just defining that $1/0:=0$.

Then to construct a proof for this exercise I must prove that for any $d_2$ expansion in $\Bbb R$ then this metric would be non-equivalent to $d_1$


I think that I finally found a way to write a formal and correct proof. Check it please.

Part one: Observe that the function $d_2(x,y)$ is not bounded in $(0,1)$ i.e.

$$\forall \epsilon>0,\exists x,y\in(0,1): d_2(x,y)=\left|\frac1x-\frac1y\right|>\epsilon$$

By the triangle inequality for any extension of $d_2$ on $\Bbb R$ we can write

$$d_2(x,y)\le d_2(x,z)+d_2(y,z)$$

for $x,y\in(0,1)$ and $z\notin(0,1)$. Then

$$\begin{align}\lim_{y\to 0^+}(d_2(x,y)\le d_2(x,z)+d_2(z,y))&\implies \lim_{y\to 0^+}\left|\frac1x-\frac1y\right|\le d_2(x,z)+\lim_{y\to 0^+}d_2(z,y)\\&\implies\color{red}{\lim_{y\to 0^+}d_2(z,y)=\infty}\end{align}\tag{2}$$

because $d_2(x,z)\in\Bbb R_{>0}$.

Part two: the logical negation of $(1)$ is

$$\exists z\in\Bbb R,\exists \epsilon>0,\forall r,s>0:B_2(z,r)\not\subseteq B_1(z,\epsilon)\lor B_1(z,s)\not\subseteq B_2(z,\epsilon)$$

Then we can set $z=0$ and $\epsilon=2$ and rewrite the above as

$$\forall r,s>0:(\exists y\in B_2(0,r):y\notin(-2,2))\lor(\color{red}{\exists y\in (-s,s):y\notin B_2(0,2)})\tag{3}$$

then observe that the statement colored in red in $(2)$ implies the statement colored in red in $(3)$. In other words: for any $\epsilon>0$ that we fix exists $y\in(0,1)$ enough close to zero such that $d_2(0,y)>\epsilon$.$\Box$

P.S.: the unique problem with this proof, supposing is correct, is that I used the notion of functional limit prior to any mention in the book.


The difference is that the metric $d_2$ can't make $\mathbb{R}$ a complete space.

A metric space $(X,d)$ is indeed complete iff every Cauchy sequence is convergent in $X$.

A sequence$(a_n) _{n \in \mathbb{N}} $ is said to be Cauchy's iff $\forall \epsilon\gt\ 0\ \exists\ \bar{n} \colon \forall n\geq \bar{n},\ d(a_n,a_m)\leq \epsilon $

Hint: what happens when you consider the sequences $(\frac{1}{n})_{n \in \mathbb{N}}$ and $(\frac{1}{n} +1)_{n \in \mathbb{N}}\ $? Are they Cauchy sequences with the metric $d_2$?

  • $\begingroup$ I will think about that but this exercise in the book is prior to the concept of Cauchy sequences or complete space, so it must exists other way to prove this. $\endgroup$ – Masacroso Oct 12 '16 at 12:08

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