Prove topological equivalence between two distances

Question

Let $d(x,y)=|x-y|$ and $h(x,y)=|x-y|^{\frac{1}{3}}$ be two distances on $\mathbb{R}$.

I want to prove that they are topologically equivalent

I supposed that $f(x)=x$ and $g(x)=x^{\frac{1}{3}}$

so we can say that

$f(x)\leq g(x)$ $\forall x\in [0,1]$ and $f(x)\geq g(x)$ $\forall x\geq 1$

Will this information lead me somewhere ? If yes, how would I proceed ?

Answer 1

Here, I use the term "$d$-ball" to indicate a set of the form $\bigl\{y\in\Bbb R:d(x,y)<r\bigr\}$--specifically, I'd call that the $d$-ball centered at $x$ with radius $r$--and similarly use the term "$h$-ball" to indicate a set of the form $\bigl\{y\in\Bbb R:h(x,y)<r\bigr\}.$

To show topological equivalence, we want to show the following:

1. Given any $d$-ball $B$ and any $y\in B,$ there exists some $h$-ball $B'$ such that $y\in B'\subseteq B.$
2. Given any $h$-ball $B$ and any $y\in B,$ there exists some $d$-ball $B'$ such that $y\in B'\subseteq B.$

Your first observation makes one direction fairly easy. Pick an arbitrary $d$-ball $B,$ with center $x$ and radius $r,$ and take any $y\in B.$ Find a radius $r'$ such that the $d$-ball centered at $y$ with radius $r_0$ is a subset of $B.$ Call this $d$-ball $B_0.$ Now, using your first observation, show that by choosing a radius $r'$ such that $0<r'<1$ and $r'\le r_0,$ then the $h$-ball centered at $y$ of radius $r'$ is then a subset of $B_0$ containing $y,$ so you've proved 1.

To prove 2, we proceed similarly, but have more work to do. Start with an arbitrary $h$-ball $B,$ with center $x$ and radius $r,$ and take any $y\in B.$ Your goal is to find some $r'>0$ such that whenever $d(y,z)<r',$ we have $h(x,z)<r.$ In other words, we need $|x-z|^{\frac13}<r$ whenever $|y-z|<r'.$

By triangle inequality, we have $$|x-z|^{\frac13}\le |x-y|^{\frac13}+|y-z|^{\frac13}$$ right away, and recalling our choice of $y,$ we have that $|x-y|^{\frac13}<r,$ so that $0<r-|x-y|^{\frac13}.$ Consequently, if we can ensure that $|y-z|^{\frac13}<r-|x-y|^{\frac13},$ we'll have $|x-z|^{\frac13}<r$ as desired. Can you take it from there?

Answer 2

Observe that for any sequences $x_n$ and $y_n$, $$|x_n - y_n| \to 0 \iff |x_n - y_n|^{1/3} \to 0$$ as $n\to\infty$.