3
$\begingroup$

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 $\bar{B}_r (x) \subseteq B_{r'} (x)$

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – Chris Eagle Oct 7 '12 at 6:51
  • $\begingroup$ 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? $\endgroup$ – Madara Oct 7 '12 at 6:55
  • $\begingroup$ 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. $\endgroup$ – Madara Oct 7 '12 at 7:08
4
$\begingroup$

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 $(X,\overline{d})$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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 $\epsilon$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Madara Oct 7 '12 at 14:48
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

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.