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I tried to solve this problem:

Let $(X,d)$ a metric space. Show that $d$ and $\bar{d} =\min({d(x,y),1})$ are topologically equivalent metrics.

I proved that $\bar{d}$ is a distance, then I tried to show that every open ball on $(X,d)$ is contained on an open ball on $(X, \bar{d})$ and vice versa.

if $r<1$, and $\forall x \in X$ $B_r (x)=\bar{B}_r (x)$.

But if $r \ge1$, $\bar{B}_r (x)=X$, then I can't find a ball on $(X,d)$ such that $\bar{B}_r (x) \subseteq B_{r'} (x)$

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It's not true in general that every open ball for $\bar d$ will be contained in an open ball for $d$. Why are you trying to show it? –  Chris Eagle Oct 7 '12 at 6:51
    
I tried to show it, because in this case the two topologies should have the same neighbourhoods and hence the same open sets. Otherwise, how can I prove it? –  Madara Oct 7 '12 at 6:55
    
However, suppose that every neighbourhood on the $d$ metric contains a neighbourhood on the $\bar{d}$ metric and vice versa, if $A \subset X$ is open on $(X,d)$, it will an open set also on $(X,\bar{d})$,in fact $\forall \ x \in \ A \ \exists \ B_r(x) \subset A$, but if $B_r(x)$ contains a $\bar{B}_{r '}(x)$ then $A$ is an open set also on $(X,\bar{d})$ , hence $(X,d)$ and $(X,\bar{d})$ have the same open sets. –  Madara Oct 7 '12 at 7:08
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4 Answers 4

up vote 2 down vote accepted

To show that the topologies of $(X,d)$ and $(X,\overline{d})$ coincide, it suffices to show that every open set in $(X,d)$ is a $\textit{union}$ of open balls in $(X,\overline{d})$ and vice versa.

As you already showed, the open balls in $(X,\overline{d})$ are open in $(X,d)$.

On the other hand every open set $U$ in $(X,d)$ can be written as $U = \bigcup_{x \in U} B_{r_x}(x)$ for suitable $r_x > 0$. We can assume $r_x < 1$ without loss of generality. But then we have $B_{r_x}(x) = \overline{B}_{r_x}(x)$, hence $U = \bigcup_{x \in U} \overline{B}_{r_x}(x)$ which shows that $U$ is open in $(X,\overline{d})$.

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If $r<1$ then $\bar B_r(x)=B(r)$ and we are done.

If $r\ge 1$ then $\bar B_r(x)=X$. This is not an open ball, but the full space is always open, and that is enough.

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Let $i$ be the identity function on $X$. That is, $i(x)=x$ for every $x\in X$.

If $x_n$ is a sequence converging in $(X,d)$, then it converges in $(X,\overline{d})$.

If $x_n$ is a sequence converging in $(X,\overline{d})$, then it converges in $(X,d)$.

Therefore, $i$ is a homeomorphism. This means that as topological spaces $(X,\overline{d})$ and $(X,d)$ are equivalent.

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It is enough to show that any $\bar d$-neighborhood $\bar U$ of any $x\in X$ contains a $d$-neighborhood $U$ of $x$, and vice versa.

One way is trivial: Given an $x$ and an $\epsilon>0$ the condition $d(x,y)<\epsilon$ implies $\bar d(x,y)\leq d(x,y)<\epsilon$. It follows that the $d$-neighborhood $U$ of radius $\epsilon$ is contained in the given $\bar d$-neighborhood of radius $\epsilon$.

Conversely: Given an $x$ and an $\epsilon>0$ the condition $\bar d(x,y)<\epsilon':=\min\bigl\{\epsilon,{1\over2}\bigr\}$ implies $d(x,y)\leq{1\over2}$ and therefore $d(x,y)=\bar d(x,y)<\epsilon$. It follows that the $\bar d$-neighborhood $\bar U$ of radius $\epsilon'$ is contained in the given $d$-neighborhood $U$ of radius $\epsilon$.

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Nice, and if $\bar{d}=\frac{d(x,y)}{1+d(x,y)}$? The first part is trivial and $B_r (x) \subseteq \bar{B}_r (x)$ While in the second part: given $x$ and $r>0$, $\bar{d}(x,y)<\min{(r,\frac{1}{r+1})}=r '$ implies that $\frac{d(x,y)}{1+d(x,y)}<\frac{1}{r+1} \Rightarrow d(x,y)\le r$ then $\bar{B}_{r'} (x) \subseteq B_r (x)$. It's right? –  Madara Oct 7 '12 at 14:48
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