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.

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