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I feel confused by the proof of the following theorem in Rudin 2/e:

Theorem 1.24 If $X$ is a topological vector space (t.v.s.) with a countable local base, then there is a metric $d$ on $X$ s.t.

  1. $d$ is compatible with the topology of $X$,
  2. the open balls centered at $0$ are balanced, and
  3. $d$ is invariant: $d(x+z,y+z)=d(x,y), \forall x,y,z \in X$.

(This is the same theorem as in this question here.)

I quote the beginning of the proof where I got confused:

Proof. By Theorem 1.14 (which states that in a t.v.s. every neighborhood of $0$ contains a balanced neighborhood of $0$), $X$ has a balanced local base $\{V_n\}$ s.t. $$V_{n+1}+V_{n+1}+V_{n+1}+V_{n+1}\subset V_n , \forall n=1,2,\ldots $$

My question: I understand that we can make the countable local base balanced by choosing a balanced neighborhood inside each base element, but how can I guarantee that $V_{n+1}+V_{n+1}+V_{n+1}+V_{n+1}\subset V_n$ always holds? I feel that this might be related to the following result (p.10):

If $W$ is a neighbhorhood of $0$ in $X$, then there is a neighborhood $U$ of $0$ which is symmetric and which satisfies $U+U\subset W$.

However this only helps me find (balanced) neighborhoods $V'$ s.t. $V'+V'+V'+V'\subset V_n$, but how can I ensure that this $V'$ is actually $V_{n+1}$?

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3 Answers 3

up vote 1 down vote accepted

As the OP observes, one starts with a balanced countable basis $\{V_n\}$.
We will then construct a new countable base $\{U_n\}$ such that $U_{n+1} + U_{n+1} + U_{n+1} + U_{n+1} \subset U_n.$ Relabelling the $U_n$ as $V_n$ answers the OP's question.


To begin with, define $U_1 := V_1$. Next, as described by the OP, one produces a balanced n.h. $V'$ such that $V' + V' + V' + V' \subset U_1.$ Now define $U_2 := V_2 \cap V'.$ Note that $U_2$ is balanced, is contained in $V_2$, and satisfies $U_2 + U_2 + U_2 + U_2 \subset U_1$.

We now continue inductive. At the $n$th step, we have produced $U_n$ which is balanced, is contained in $V_n$, and satisfies $U_n + U_n + U_n + U_n \subset U_{n-1}$. To define $U_{n+1}$, we then construct (as the OP describes) a balanced $V'$ such that $V' + V' + V' + V' \subset U_n,$ and define $U_{n+1} := V' \cap V_{n+1}$. This completes the induction, and hence the construction.

(Note that since the $V_n$ forms a n.h. basis, and each $U_n \subset V_n$, we have that the $U_n$ also form a n.h. basis.)

[Also: this is essentially hsc's answer. For some reason, perhaps because of it's phrasing, that answer was voted down, and attracted a comment which seemed to misinterpret it as a new question, rather than an answer to this question. Hopefully this more detailed explanation will make things clear.]

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We can choose $V_{n+1}$ such that it is balanced inside $V'$.

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Thanks! But how can I ensure that $\{V_k\}$ chosen as such is a local base of this topology? I think some base elements in the original countable local base might be "skipped" in such a choice, resulting in that $\{V_k\}$ is not a local base or that it is the local base of a different topology? –  Fang Jing May 13 at 17:07
    
@FangJing: If you look at the answer by hsc, and my answer (which gives more details), you will get an answer to the question raised in your comment. –  user160609 Jun 29 at 19:18

How about taking the intersection of the V' (induced by V_{n}) and V_{n+1} as the new V_{n+1}? By the way, thanks for your posting, I am reading Rudin too and have similar questions, haha.

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If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. –  Shaun Jun 28 at 13:13
    
Hi hsc, Your suggestion is correct. I formalized it a little in the answer I posted (while giving you credit); I hope that's okay. –  user160609 Jun 29 at 19:16
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Hi @Shaun, hsc is suggesting an answer to the question, not really asking a new question. –  user160609 Jun 29 at 19:17
1  
@user160609 Good job! –  hsc Jun 29 at 20:59
1  
@Shaun Never mind~ –  hsc Jun 29 at 20:59

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