# Question about equivalent metric spaces

I have studied that topologically equivalent metrics produce the same open and closed sets. They also produce same compact and connected subsets. Does it mean that topologically equivalent metrics have same open, closed or compact subsets? In what context topologically equivalent metrics differ from each other?

Intutively, how can we explain topologically equivalent metrics? Can we say that two topologically equivalent metrics have same topological properties?

Thanks and regards

## 4 Answers

Suppose that $d_0$ and $d_1$ are metrics on a set $X$. Each generates a topology: $d_0$ generates $$\tau_0=\left\{V\subseteq X:\forall x\in V\exists \epsilon_x>0\big(B_{d_0}(x,\epsilon_x)\subseteq V\big)\right\}\;,$$ and $d_1$ generates $$\tau_1=\left\{V\subseteq X:\forall x\in V\exists \epsilon_x>0\big(B_{d_1}(x,\epsilon_x)\subseteq V\big)\right\}\;,$$ where $B_{d_0}(x,\epsilon)=\{y\in X:d_0(x,y)<\epsilon\}$ and $B_{d_1}(x,\epsilon)=\{y\in X:d_1(x,y)<\epsilon\}$. We say that $d_0$ and $d_1$ are topologically equivalent iff $\tau_0=\tau_1$: the two metrics generate the same topology. Thus, the metric spaces $\langle X,d_0\rangle$ and $\langle X,d_1\rangle$ share all purely topological properties, because they are identical as topological spaces: they are both $\langle X,\tau\rangle$, where $\tau=\tau_0=\tau_1$. Thus, they have the same open sets, closed sets, compact sets, connected sets, discrete sets, etc.

They do not necessarily share specifically metric properties. For example, one of $\langle X,d_0\rangle$ and $\langle X,d_1\rangle$ may be complete and the other not. One may be bounded and the other not. One may be totally bounded and the other not. They may differ in anything that depends specifically on the metric, rather than on the topology.

• Sir, then can we say that equivalent metric spaces are homeiomorphic to each other? If yes, which map will determine homeomorphism between equivallent them? I guess, identity map. Am i right? May 29, 2012 at 10:35
• @srijan: Yes, they’re homeomorphic via the identity map. As I said, from a purely topological point of view they are not just homeomorphic: they are identical, with the same underlying set and the same topology. May 29, 2012 at 10:38
• Many many thanks to you. Now, things are cleared. May 29, 2012 at 10:40

In its simplest form, the fact that two metrics $d_1$ and $d_2$ are topologically equivalent means that given any point $x$ and a ball around $x$ in one of the metrics, it contains a ball around $x$ in the other.

That means that topological equivalence of metrics is a local condition; it only cares about small distances. Topological equivalence also means that any property that can be defined with open and / or closed sets is invariand under change of metric. This includes compactness, clopenness, convergence of sequenses, and continuity of functions in and out of the space.

Properties more closely related to the global properties of the metric, like uniform continuity are not , in general, kept intact.

• i think in place of clopeness , closedness should be there. Sorry i cant edit. May 29, 2012 at 10:42
• Nope, clopenness was the intended phrase. Has to do with connectedness. A set which is both open and closed is a union of connected components, and is called clopen. May 29, 2012 at 19:45

The difference isn't topological because the two metric space are equal up homeomorfism. But it can exist many difference of orther type. For example the euclidean plane and the hyperbolic plane are equivalents topologically, as metric spaces, but geometrically they are not essentially equivalent.

• I am sorry i didn't get you? How can you say that two metrics are equal up to homeomorphism? May 29, 2012 at 10:38

A theorem states: two metric spaces are equivalent iff they have the same topology. This means that they are equivalent up to topology (or equal up to homeomorphism). So every topological property is the same for both spaces, but other (non-topological) properties may differ.