# 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. Jan 14, 2021 at 13:21
• @KillianHeanue: whatever topology you define, you can always consider the constant map, which is continuous.
– user9464
Jan 14, 2021 at 13:22
• that's not an issue as the constant map is an isometry. 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
Jan 14, 2021 at 13:24
• @KillianHeanue: sorry I have to go now. you can leave questions there and I may be back.
– user9464
Jan 14, 2021 at 13:54