2 definitions and a theorem relating them

[dextercioby]

Homework Statement

Let E be a vector space and p,q 2 norms on it. By definition p,q induce the same topology on E, iff they assign the same neighborhood basis to the 0 vector. *QUESTION: What does the bolded part mean ? Does this mean that, if whatever A included in the system (=basis?) induced by p, A is included in a certain B in the system induced by q and viceversa, i.e. any B in the system induced by q is included in a certain A induced by p ? If so, then see my arguments below.

The 2nd definition would be:

E a vector space and p, q norms on it. p, q are said to be equivalent iff there exist c, C>0 such as

[tex] c p(x) \leq q(x) \leq C p(x), ~\forall x \in E [/tex]

The theorem would be the equivalence of the 2 definitions, namely

Theorem: p,q are equivalent iff they induce the same topology on E.

The problem would be to prove the theorem.

The Attempt at a Solution

Assume p, q equivalent. Then if [itex]\epsilon >0 [/itex], [itex] \{ x\in E, p(x) \leq \epsilon/c \}[/itex] is a n-hood system of 0 wrt p. Let A be one of the n-hoods in this system. Then [itex] A\subseteq \{x\in E, q(x) \leq\epsilon \} [/itex] ?? That would mean that [itex] \{x\in E, q(x) \leq\epsilon \} [/itex] is also a n-hood system of 0 induced by q. Using now the 2nd inequality would mean

Let [itex] \epsilon>0 [/itex]. Then [itex] \{x\in E, q(x)\leq \epsilon/C \} is a n-hood system of 0 induced by q and let B be a n-hood from it. Does it mean (using the 2nd inequality in the definition of equivalence) that:

[tex] B\subseteq \{x\in E, p(x)\leq \epsilon\}[/tex] ??

If so, then the 2 systems of n-hoods would be the same (??) so that p,q induce the same topology.

This would be half the proof. How's the other half ?

 

[Evalesco]

To prove the other half of the theorem, assume that p, q induce the same topology on E. That would mean that the systems of n-hoods induced by p and q are the same. Let \epsilon>0. Then, for each n-hood A in the system induced by p, A\subseteq \{x\in E, q(x)\leq\epsilon \}. This means that c p(x) \leq q(x)\leq \epsilon, ~\forall x \in A. Taking the supremum over all x in A implies that c p(x) \leq q(x)\leq \epsilon, ~\forall x \in E. By taking supremum over all \epsilon>0, we get the desired result that there exist c, C>0 such that c p(x) \leq q(x) \leq C p(x), \forall x \in E.

Thus, the theorem is proved.