# Standard bounded metric induces same topology

Theorem: If $(X,d)$ is a metric space and $d' = \min (d(x,y), 1)$ is the standard bounded metric then $d$ and $d'$ induce the same topology.

Equivalently, for all $x_0$ there are $a,b$ such that for all $y$: $a d'(x_0,y) \le d(x_0,y) \le b d'(x_0, y)$. Clearly, $a=1$. But: how to determine $b$? The statement appears to be false: $d'$ is bounded while $d$ is not. Yet, see here on page 3. Thank you.

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Well, you can simply use the fact that $d$ and $d'$ agree on small balls. By the definition, $A\subset X$ is open iff for any $x\in A$ there exists $r>0$ such that $B(x,r)\subset A$. Equivalence of metric is sufficient for the equivalence of the induced topologies, but is not necessary. –  Ilya Jan 11 '13 at 14:13
@Ilya Thank you, you have answered my question very well. –  user54938 Jan 11 '13 at 14:17
Dear @user54938, it is usually best not to delete questions once they havee been answered, even if only in comments. –  Mariano Suárez-Alvarez Jan 11 '13 at 14:57

Well, you can simply use the fact that $d$ and $d'$ agree on small balls. By the definition, $A\subset X$ is open iff for any $x\in A$ there exists $r>0$ such that $B(x,r)\subset A$.
Suppose that for some $x_0\in A$ and $r>0$ it holds that $B_{d'}(x_0,r)\subset A$. Then for any $q\in (0,r)$ it holds that $B_{d'}(x_0,q) \subset A$. In particular, it holds for $q = \min(\frac12,r)<1$ but then $$B_{d'}(x_0,q) = B_d(x_0,q).$$ In similar lines you can show the converse. Equivalence of metric is sufficient for the equivalence of the induced topologies, but is not necessary.
You wrote *if there exists a d-ball then there exists such a d'-ball because $d'\le d.$* But it is the other way round: Since $d(x,y)\le r$ implies $d'(x,y)\le r$, the set $B_d(x,r)$ is contained in $B_{d'}(x,r)$. –  Stefan Hamcke Jan 11 '13 at 15:37
I think it is a mistake that can likely happen, if you don't pay attention. But the smaller the metric, the larger the $\epsilon$-balls become. –  Stefan Hamcke Jan 11 '13 at 17:14