The aim of this note is to derive some connections between topologies of normed spaces in terms of the equivalency of norms and the convergence of sequences.

**Topological space and its topology**: First, we start with a topological space, call . Its topology, say is the collection of subsets of which satisfies certain conditions. In the literature, each member of the collection is called an open set.

Regarding to topologies we have the following basic facts:

- Given two topologies and on , we say that is
*stronger*(or finer or richer) than if . - Given a sequence in , we say that
*converges*to in topology of if for any neighborhood of , there exists some large number such that for all . (Here by the neighborhood of we mean that there exists an open set of , i.e. is a member of the topology , such that .)

The key ingredient to compare topologies is to make use of the identity map. In the following part, we state a result which shall be used frequently in this note.

**Topologies under the identity map**: Given two topologies and on a topological space , we are interested in comparing and in terms of the identity map .

Lemma 1. The identity map iscontinuous if and only ifis stronger than .

*Proof*.

The proof is relatively easy. Indeed, if the map is continuous, then the preimage of any is also a member of which immediately implies that includes .

Having Lemma 1 in hand, we now try to compare topologies using norms.

**Two norms generate the same topology if and only if they are equivalent**. We are now interested in comparing topologies generated by different norms. We shall prove the following result:

Lemma 2. On a normed space , two different norms generate the same topology if and only if they are equivalent.

*Proof*.

Assume that and are two norms on the normed space . We denote by the topology generated by the norm with .

If the two topologies and are the same, the two identities maps and are all continuous. Consequently, there would exist two positive constants and such that

and that

Thus, as claimed.

In the reverse, if the two norms and are equivalent, it is clear to see that the two topologies and generated by and respectively are identical.

In the final part, we characterize topologies using the notion of convergent sequences.

**Topologies in terms of convergent sequences**: We now compare topologies using convergent sequences. Consider the topological space and let us assume that we have two different topologies with on . First, we shall prove the following:

Lemma 3. If is stronger than , then any sequence converging to in topology also converges to in topology.

*Proof*.

Suppose that there is some sequence converging to in topology, we need to prove that also converges to in topology.

Take any neighborhood of in topology, then there exists some such that . Since is stronger than , we know that . Consider as a neighborhood of in topology, by definition, there exists some large number such that $x_n \in O_2$ for all . Thanks to the fact , we also obtain for all in topology. The proof is complete.

We are now interested in the reverse direction of Lemma 3. We shall prove:

Lemma 4. If any sequence converging to in topology also converges to in topology, then is stronger than .

*Proof*.

To realize that is stronger than , it suffices to prove that the identity map is continuous, thanks to Lemma 1.

Indeed, we pick a point arbitrarily, we then prove that is continuous at . In view of the sequential characterization of the continuity, it suffices to check whenever in topology we then have in topology. However, this is trivial since and and our hypothesis.

**Conclusion**: What we have done in this topic is to realize that *identical topologies*, *same convergent sequences* with the same limits, and the *equivalency of norms* are all equivalent. Therefore, one can use one of these three notions to define the identical topologies:

- They have the same collection of open sets, or
- They have equivalent norms, or
- They have convergent sequences with the same limits.

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