Given a set , there are always at least two topologies that can be placed on : the discrete topology and the trivial topology. The discrete topology is the topology where every set is open; the collection of open sets in the topology is simply the collection of all sets, . If is a map between topological spaces where has the discrete topology, then will be continuous, regardless of how it maps the elements of to . The trivial topology is the bare minimum for a topology: the empty set and the entire set . For any map , then, will be continuous as and . Regardless of the topology on , those two sets will be open, so is continuous.

Suppose now that is a topological space where is any topology. If we can place a topology on by using the topology of . Let be a collection of subsets of where ( is an open subset of ) if for some . We call the *subspace topology* of . For instance if we take with the standard topology, we can place a topology on using this subspace topology. In such a case, sets like are considered open in this topology since , even though this is *not* considered an open subset of .

If and are topological spaces and is continuous, we’d like to know when is continuous when restricted to a subspace of . We can show that if define as the image of under , and assign the subspace topology in , then is continuous. Let be an open subset of . Then for some open subset of . We know . By definition and is an open subset of by the continuity of . That is, is an open subset of with the subspace topology, so the restriction of to is continuous.

In the case of , the subspace topology inherited from is the same as the discrete topology on . Let . Then is an open set with the subspace topology as . This gives that every singleton set is an open set and since any set will be a union of singletons, any set will be open.

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