# 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.