Ngô Quốc Anh

March 13, 2015

Comparing topologies of normed spaces: The equivalency of norms and the convergence of sequences

Filed under: Uncategorized — Ngô Quốc Anh @ 23:01

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 X. Its topology, say \mathcal T is the collection of subsets of X which satisfies certain conditions. In the literature, each member of the collection \mathcal T is called an open set.

Regarding to topologies we have the following basic facts:

  • Given two topologies \mathcal T_1 and \mathcal T_2 on X, we say that \mathcal T_1 is stronger (or finer or richer) than \mathcal T_2 if \mathcal T_2 \subset \mathcal T_1.
  • Given a sequence (x_n)_n in X, we say that x_n converges to x in topology \mathcal T of X if for any neighborhood V of x, there exists some large number N such that x_n \in V for all n \geqslant N. (Here by the neighborhood V of x we mean that there exists an open set O of X, i.e. O is a member of the topology \mathcal T, such that x \in O \subset V.)

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 \mathcal T_1 and \mathcal T_2 on a topological space X, we are interested in comparing \mathcal T_1 and \mathcal T_2 in terms of the identity map \rm id : (X, \mathcal T_1) \to (X, \mathcal T_2).

Lemma 1. The identity map \rm id : (X, \mathcal T_1) \to (X, \mathcal T_2) is continuous if and only if \mathcal T_1 is stronger than \mathcal T_2.

Proof.

The proof is relatively easy. Indeed, if the map \rm id is continuous, then the preimage of any O_2 \in \mathcal T_2 is also a member of \mathcal T_1 which immediately implies that \mathcal T_1 includes \mathcal T_2.

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 X, two different norms generate the same topology if and only if they are equivalent.

Proof.

Assume that \|\cdot\|_1 and \|\cdot\|_2 are two norms on the normed space X. We denote by \mathcal T_i the topology generated by the norm \|\cdot\|_i with i=1,2.

If the two topologies \mathcal T_1 and \mathcal T_2 are the same, the two identities maps \rm id_1 : (X, \mathcal T_1) \to (X, \mathcal T_2) and \rm id_2 : (X, \mathcal T_2) \to (X, \mathcal T_1) are all continuous. Consequently, there would exist two positive constants C_1 and C_2 such that

\displaystyle \|{\rm id}_1 (x)\|_2 \leqslant c_1 \|x\|_1

and that

\displaystyle \|{\rm id}_2 (x)\|_1 \leqslant c_2 \|x\|_2.

Thus, \|x\|_1 \leqslant c_2 \|x\|_2 \leqslant c_1c_2 \|x\|_1 as claimed.

In the reverse, if the two norms \|\cdot\|_1 and \|\cdot\|_2 are equivalent, it is clear to see that the two topologies \mathcal T_1 and \mathcal T_2 generated by \|\cdot\|_1 and \|\cdot\|_2 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 X and let us assume that we have two different topologies \mathcal T_i with i=1,2 on X. First, we shall prove the following:

Lemma 3. If \mathcal T_1 is stronger than \mathcal T_2, then any sequence (x_n) converging to x in \mathcal T_1 topology also converges to x in \mathcal T_2 topology.

Proof.

Suppose that there is some sequence (x_n) converging to x in \mathcal T_1 topology, we need to prove that x_n also converges to x in \mathcal T_2 topology.

Take any neighborhood V of x in \mathcal T_2 topology, then there exists some O_2 \in \mathcal T_2 such that x \in O_2 \subset V. Since \mathcal T_1 is stronger than \mathcal T_2, we know that O_2 \in \mathcal T_1. Consider O_2 as a neighborhood of x in \mathcal T_1 topology, by definition, there exists some large number N such that $x_n \in O_2$ for all n \geqslant N. Thanks to the fact O_2 \subset V, we also obtain x_n \in V for all n\geqslant N in \mathcal T_2 topology. The proof is complete.

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

Lemma 4. If any sequence (x_n) converging to x in \mathcal T_1 topology also converges to x in \mathcal T_2 topology, then \mathcal T_1 is stronger than \mathcal T_2.

Proof.

To realize that \mathcal T_1 is stronger than \mathcal T_2, it suffices to prove that the identity map \rm id : (X, \mathcal T_1) \to (X, \mathcal T_2) is continuous, thanks to Lemma 1.

Indeed, we pick a point x \in X arbitrarily, we then prove that \rm id is continuous at x. In view of the sequential characterization of the continuity, it suffices to check whenever x_n \to x in \mathcal T_1 topology we then have {\rm id}(x_n) \to {\rm id}(x) in \mathcal T_2 topology. However, this is trivial since {\rm id}(x_n)=x_n and {\rm id}(x)=x 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:

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

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