# 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.

• 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-Álvarez 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
• @StefanH.: maybe there is some confusion. Hope now it's better – Ilya Jan 11 '13 at 15:45
• 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
• What would be the elements of a basis element in standard bounded metric with radius greater than 1? Will they for the whole space ? – Cosmic Sep 30 '19 at 16:58

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.