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Suppose that there are given two distance functions $d(x,y)$ and $d_1 (x,y)$ on the same space $S$. They are said to be equivalent if they determine the same open sets.

Show that $d$ and $d_1$ are equivalent if to every $\varepsilon$ there exists a $\delta>0$ such that $d(x,y)<\delta$ implies $d_1 (x,y)<\varepsilon$ and vice versa.

I have no idea how to handle this kind of problem.

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4 Answers 4

up vote 3 down vote accepted

Since open sets are the sums of open balls, it suffices to check if those metrices determine the same open balls. Let us denote by $B_d(x,r)$ an open (wrt $d$) ball with a centre in $x\in S$ and a radius $r\gt0$, i.e. $B_d(x,r)=\{y\in S\mid d(x,y)<r\}$.

We will prove an arbitrary open ball $B_d(x_0,r)$ is $d_1$-open. In order to do that, we will show that for every $x\in B_d(x_0,r)$ there exists $r_x$ such that $B_{d_1}(x,r_x)\subset B_d(x_0,r)$.
What we have is $$(*)\quad\forall_{a\in S} \forall_{\varepsilon>0}\exists_{\delta>0}\forall_{b\in S}\ \ d_1(a,b)\lt\delta\implies d(a,b)\lt\varepsilon.$$ (our $a$ is actually $x$ and $\delta$ will be desired $r_x$.)
As has been said, fix an arbitrary $x\in B_d(x_0,r), x\neq x_0$ and set $\varepsilon=\min{(d(x,x_0),r-d(x,x_0))}>0.$ From $(*)$ we know that there exists $\delta>0$ such that for all $y\in S$ $$(1)\quad d(x,y)<\min{(d(x,x_0),r-d(x,x_0))}\le r-d(x,x_0)$$ as long as $d_1(x,y)\lt\delta$.
Now we could reformulate (1) by adding $d(x,x_0)$ to both sides and applying triangle inequality: $$d(x_0,y)<d(x,y)+d(x,x_0)<r.$$ Thus, what we proved is: $$\forall_{x\in B_d(x_0,r)}\exists_{\delta>0} \forall_{y\in S}\ y\in B_{d_1}(x,\delta) \implies y\in B_d(x_0,r)$$ and, in other words, it means that every $x\neq x_0$ from $d$-ball is contained there with some small $d_1$-ball, that is $d$-ball is open in the sense of metric $d_1$. For $x_0$, the existence of such $d_1$-ball is ensured by $(*)$ directly (take $\varepsilon=r)$.
Entirely analogously, one can prove second part of the problem.

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I think this might help. I suppose you define an open set (in a metric space as)

DEFINITION 1 A set is open if it is a neighborhood of each of its points.

Now you can prove this simple lemma:

LEMMA 1 Every open ball is a neighborhood of each of its points.

PROOF Let $B=B(x;\epsilon)$ be an open ball around $x\in X$ with radius $\epsilon$. Let $y\in B$. We want to prove there is an open ball around $y$ contained in $B$. Let $\delta=\epsilon -d(x,y)$. We claim if we let $r<\delta$ then $B'=B'(y,r)\subset B(x,\epsilon)$. Indeed, let $x'\in B'$. Then $$\begin{align} d(x',x)&\leq d(x',y)+d(y,x) \\ &<\epsilon -d(x,y)+d(x,y)\\&=\epsilon. \end{align}$$

Maybe an image would have been nicer:

$\hspace{4.5 cm}$enter image description here

THEOREM 1 A set is open if and only if it is a union of open balls.

PROOF First suppose the set $O$ is open. Then it is a neighborhood of each of its points, so it contains, for each $x\in O$, an open ball $B(x,\epsilon_x)$ (the radius depends on this $x$, too), so we may write

$$O=\bigcup_{x\in O}B(x,\epsilon_x)$$

that is, it is a union of open balls.

Conversely, suppose the set $O$ is a union of open balls. Then each $x\in O$ is contained in some open ball of this set. But the previous LEMMA says this ball contains another open ball about this particular point. Since this point was arbitrarily chosen, we can conclude $O$ is a neighborhood of each of its points, so it is open.

The idea is that open balls "make up" open sets, they are the "basic" open sets (because themselves are open) that make up the "bigger ones".

So, back to your question: suppose that we have two metrics, $d_1$ and $d_0$. We know that in each of the metric spaces associated to them, the open sets are determined precisely by the open balls of $d_0$, call them $B_0$, and the open balls of $d_1$ call them $B_1$. Now suppose we have an open ball $B_0$ in $(X,d_0)$:

$$B_0=B_0(x,r).$$

The hypothesis is that

For every $\epsilon$ there exists a $\delta >0$ such that $d_0(x,y)<\delta$ implies $d_1(x,y)<\epsilon$ and vice versa.

Putting this in terms of open balls, it says

Whenever we have a ball $B_1(x,\epsilon)$; there is a $\delta$ such that $B_0(x,\delta)\subset B_1(x,\epsilon)$ and vice versa.

This is saying that every open ball of $d_0$ about a point $x$ contains another open ball of $d_1$ about the same point, and vice versa. So let's use this to prove that the open sets of one metric space are the same as those in the other:

THEOREM 2 Let $d_0$ and $d_1$ be such that for every $\epsilon>0$ and $x\in X$, there is a $\delta$ such that $B_0(x,\delta)\subset B_1(x,\epsilon)$ and vice versa. Then a set $O$ is open $(X,d_0)$ if and only if it is open in $(X,d_1)$.

PROOF Let $O$ be open in $(X,d_1)$. Then by THEOREM 1 it is a union of open $d_1$-balls, that is $$O=\bigcup_{x\in O}B_1(x,\epsilon_x).$$ Let $x\in O$. Then $x\in B_1(x,\epsilon_x)$, so by hypothesis, there is a $\delta_x$ such that $B_0(x,\delta_x)\subset B_1(x,\epsilon_x)$. This means that each point of $O$ contains a $d_0$-ball around it, so by definition, $O$ is also open in $d_0$, since it is a "$d_0$" nbhd of each of its points. Conversely, let $O'$ be open in $d_0$. Then it is a union of open $d_0$-balls, that is

$$O'=\bigcup_{x\in O'}B_0(x,\epsilon'_x).$$

But again, for each point of $O'$, there is a $\delta'_x$ for which $B_1(x,\delta'_x)\subset B_0(x,\epsilon'_x)$. This means that each set $O'$ is a "$d_1$" nbhd of each of each of its points, so that is open in $(X,d_1)$, too. This concludes the proof.

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For the proof of Lemma 1, wouldn't the open ball itself have sufficed? –  WacDonald's Aug 17 '12 at 16:41
    
@WacDonald's It really depends on the approach. Sometimes people learn first about topological spaces and then about metric spaces, but it is also possible that metric spaces are first presented to grasp the ideas of "open", "closed", "nbhd", "closure", "limit point" and then present the topological spaces as a generalization of metric spaces. In the first case, if we define the basis of a topological/metric space to be the open balls around the points, then yes. However, in the latter case we must have an auxiliary definition of what we mean by open. Note that here the definition (...) –  Pedro Tamaroff Aug 17 '12 at 16:45
    
(...) of open set is a theorem in the general setting, and the theorem is a definition (the open sets are unions of basis elements). I wasn't sure the OP was familiar with the general setting, but if he were, this would serve as well. –  Pedro Tamaroff Aug 17 '12 at 16:47

If you're familiar with the concepts of topologies and topological spaces, I think you will find these other ideas useful. Recall that

DEFINITION. Given a topological space $(X,\mathfrak I)$, a collection $\mathscr B$ of subsets of $\mathfrak I$ is said to be a basis for the toplogy on $X$, or a basis for the open sets, or a basis for $\mathfrak I$, iff every element of $\mathfrak I$ is a union of elements of $\mathscr B$.

This will be useful later

LEMMA Let $\mathscr B$ be a collection of elements of $\mathfrak I$. Then $\mathscr B$ is a basis for $\mathfrak I$ if and only if for each $x\in X$ and $x\in O$ an open set, there is an element $B$ of $\mathscr B$ for which $x\subset B\subset O$.

PROOF First suppose $\mathscr B$ is a basis. Then each open set is a union of basic elements, say $O=\bigcup_{\alpha \in A} B_\alpha$, so given an open set $O$, and $x\in O$, it follows $x\in B_\alpha\subset O$ for some index. Conversely, suppose that for each $x\in X$ and $x\in O$ an open set, there is an element $B$ of $\mathscr B$ for which $x\subset B\subset O$. Then for each $x\in O$ we may set $$O=\bigcup_{x\in O}B_x$$

so each set is a union of elements of $\mathscr B$, and $\mathscr B$ is a basis.

Now, we have this useful criterion for comparing topologies in terms of their bases:

THEOREM Let $\mathscr B$ and $\mathscr B'$ be bases for the topologies $\mathfrak I$ and $\mathfrak I$ respectively, on $X$. Then the following statements are equivalent:

$(1)$ $\mathfrak I'$ is finer than $\mathfrak I$, that is $\mathfrak I\subset \mathfrak I'$

$(2)$ For each $x\in X$, and each $B\subset \mathscr B$ for which $x\in B$, there is a $B'\subset \mathscr B'$ for which $x\in B'\subset B$.

PROOF

$(2)\Rightarrow (1)$ We want to show that whenever $O\in \mathfrak I$ then $O\in \mathfrak I'$. Since $\mathscr B$ is a basis for $\mathfrak I$, given any point $x\in O$ there is a $B$ for which $x\subset B\in O$. Condition $(2)$ says that there is a $B'\in \mathscr B'$ for which $x\in B'\subset B \subset O$, that is $x\in B'\subset O$. Thus $O\subset \mathfrak I'$ by the previous LEMMA and the defintion of basis.

$(1)\Rightarrow (2)$ Let $x\in X$, $B\subset \mathscr B$, and $x\in B$. Now, $B\subset \mathfrak I$ ($B$ is itself an open set), and $\mathfrak I\subset \mathfrak I'$ by hypothesis, so $B\subset \mathfrak I$. Since $\mathfrak I'$ is generated by $\mathscr B'$, there is an element of $\mathscr B'$, call it $B'$; for which $x\in B'\subset B$.

So, the definition is

DEFINTION Two bases on a set $X$ are equivalent if they generate the same topology$\mathfrak I$, or collection of open sets. (Note a metric defines a base, namely, the collection of all open balls)

An immediate corollary of the last theorem is:

COROLLARY Two bases $\mathscr B'$ and $\mathscr B$ are equivalent if and only if for each $B\in \mathscr B$ and $x\in B$ there is a $B'\in \mathscr B'$ for which $x\in B'\subset B$, and viceversa.

...which is what you're asked to prove.

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You are missing a quantifier. The two distance functions determine the same open sets if for all $x\in S$ and all $\varepsilon>0$ there is $\delta>0$ such that for all $y\in S$, $d(x,y)<\delta$ implies $d_1(x,y)<\varepsilon$, and vice versa.

To show this, recall the definition of open sets: A set $U\subseteq S$ is open (with respect to a distance function $d$) if for all $x\in U$ there is $\varepsilon>0$ such that for all $y\in S$, $d(x,y)<\varepsilon$ implies $y\in U$.

Now take an open set wrt $d$, use your condition, and show that it is open wrt to $d_1$. This should be straight forward.

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Thanks for your reply! But I can't prove that whether $d_1 (x,y)$ tends to zero implies $d(x,y)$ tends to zero. –  Golbez Aug 17 '12 at 12:57
    
I think you don't have to explicitly prove that. Let $U$ be open wrt $d$ and let $x\in U$. For some $\varepsilon>0$, if $d(x,y)<\varepsilon$, then $y\in U$. By your condition, for some $\delta>0$, $d_1(x,y)<\delta$ implies $d(x,y)<\varepsilon$. So in particular, $d_1(x,y)<\delta$ implies $y\in U$. This shows that $U$ is open wrt. $d_1$. –  Stefan Geschke Aug 17 '12 at 13:13
2  
Moreover, it is actually not true that if $d_1(x,y)$ tends to zero then so does $d(x,y)$. Let $f:\mathbb R\to(-\pi/2,\pi/2);x\mapsto\arctan(x)$ and $S=\mathbb R$. Let $d_1(x,y)=|f(x)-f(y)|$ and $d(x,y)=|x-y|$. Let $x_n=n$ and $y_n=n+1$. Then $\lim_{n\to\infty}d_1(x_n,y_n)=0$ but $\lim_{n\to\infty}d(x_n,y_n)=1$. $d$ and $d_1$ are both metrics and they give the same open sets (and they satisfy your $\varepsilon$-$\delta$-condition). –  Stefan Geschke Aug 17 '12 at 13:27
    
Thanks for your Counterexample! –  Golbez Aug 17 '12 at 14:03

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