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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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tvk
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3 Answers3

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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.]

user160609
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We can choose $V_{n+1}$ such that it is balanced inside $V'$.

Berci
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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? – tvk May 13 '14 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 '14 at 19:18
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I am confused by the question too, and also for the answers above, I am also confused by that $V^{'} $ will be a Balanced set. So, here is my suggestion.

Fix $\{V_n\}$ as the original countable balanced local base. Fix $n=1$ and $U_1 = V_1$. By the lemma, there is n.b.$W$ s.t. $W+W+W+W\subset V_1$. Then $\exists V_m \ s.t. \ V_m \subset W$ by the definition of local base. Let $U_2 =V_m$. Then do it inductively. We will get the desired set $\{U_n\}$.

Daaaaa
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