# Is it possible to define a topology on a set such that continuity of a map out of this space implies that map is an isometry?

I understand that the use of defining a topology on a set is to define what it means for a map out of the space to be continuous. It is then natural to ask for which types of maps out of a set is it possible to define a topology on the set to allow only said maps to be continuous (e.g. any map continuity -> discrete topology, 'traditional' continuity -> euclidean topology).

This has not yet been addressed in the textbook I am currently reading so I decided to take it here.

My question is specifically about isometric maps, but any more general answer will suffice.

In order to talk about "isometry", you need a "metric".

In a general topological space, this is no metric and the topology is not necessarily metrizable.

If you do have a metric space $$(X,d)$$, then the natural topology is the metric topology induced by the metric $$d$$. One can then ask whether a continuous function with respect to this topology is an isometry.

But there is no reason for an arbitrary continuous map to be an isometry: you can always consider the constant map.

• Ok, but obviously the natural topology on such a metric space doesn't cut it. But is it possible to define some other topology on the metric space that does the job is my question. Commented Jan 14, 2021 at 13:21
• @KillianHeanue: whatever topology you define, you can always consider the constant map, which is continuous.
– user9464
Commented Jan 14, 2021 at 13:22
• that's not an issue as the constant map is an isometry. Commented Jan 14, 2021 at 13:23
• @KillianHeanue: no, the constant map is not an isometry. The constant map is not the same as the "identity" map.
– user9464
Commented Jan 14, 2021 at 13:24
• @KillianHeanue: sorry I have to go now. you can leave questions there and I may be back.
– user9464
Commented Jan 14, 2021 at 13:54