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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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  • $\begingroup$ 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. $\endgroup$ – Ilya Jan 11 '13 at 14:13
  • $\begingroup$ @Ilya Thank you, you have answered my question very well. $\endgroup$ – user54938 Jan 11 '13 at 14:17
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    $\begingroup$ Dear @user54938, it is usually best not to delete questions once they havee been answered, even if only in comments. $\endgroup$ – Mariano Suárez-Álvarez Jan 11 '13 at 14:57
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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 $B(x,r)\subset A$.

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.

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    $\begingroup$ 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 $B_{d'}(x,r)$. $\endgroup$ – Stefan Hamcke Jan 11 '13 at 15:37
  • $\begingroup$ @StefanH.: maybe there is some confusion. Hope now it's better $\endgroup$ – Ilya Jan 11 '13 at 15:45
  • $\begingroup$ 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. $\endgroup$ – Stefan Hamcke Jan 11 '13 at 17:14
  • $\begingroup$ What would be the elements of a basis element in standard bounded metric with radius greater than 1? Will they for the whole space ? $\endgroup$ – Cosmic Sep 30 '19 at 16:58
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In Topology by James Munkres, a different approach is taken:

Now we note that in any metric space, the collection of $\epsilon$-balls with $\epsilon<1$ forms a basis for the metric topology, for every basis element containing $x$ contains such an $\epsilon$-ball centered at $x$. It follows that $d$ and $\bar{d}$ induce the same topology on $X$, because the collection of $\epsilon$-balls with $\epsilon<1$ under these two metrics are the same collection. (page 122)

That last sentence was a bit confusing to me at first, so let me expand with what I think that means. Clearly, the topology induced by $\bar{d}$ is coarser than the topology induced by $d$. Now consider an open set $U$, an arbitrary element $y\in U$, and the ball $B_d(y,\epsilon)\subset U$ in the topology induced by $d$. We're given $\exists \epsilon$ such that $B_d(y,\epsilon)=\{b|d(b,y)<\epsilon\}\subset U$. If $\epsilon<1$ then $B_d(y,\epsilon)=B_\bar{d}(y,\epsilon)$, so clearly $B_\bar{d}(y,\epsilon)\subset U$. And if $\epsilon>1$, $B_\bar{d}(y,\epsilon)=\{b|d(b,y)<1\}\subset \{b|d(b,y)<\epsilon\}=B_d(y,\epsilon)\subset U$. Hence, the topology induced by $\bar{d}$ is also finer than the topology induced by $d$. So they induce the same topology.

Of course I'm studying this now, so take the last paragraph with a pound of salt.

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