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I have some confusion on (c) of Remark 1.38 (page 29) of Rudin's Functional Analysis (second edition) . Let me recap the relevant content here.

Let $X$ be a vector space and $\{p_k|k\in\mathbb{N}\}$ a countable separating family of seminorms on $X$. By Theorem 1.37 this countable family induces a topology $\mathcal{T}$ with a countable local base (and makes $(X, \mathcal{T})$ a locally convex space satisfying certain properties). Then by Theorem 1.24 $(X, \mathcal{T})$ is metrizable.

Then Rudin gives an explicit formula of a translation-invariant metric $d$ on $X$, which is $$d(x, y)=\max_k\frac{c_kp_k(x-y)}{1+p_k(x-y)},$$ where $(c_k)$ is a sequence of positive real numbers satisfying $\lim_{k\to\infty}c_k=0$. More importantly, this metric $d$ is compatible with $\mathcal{T}$ (meaning the metric topology generated by $d$ equals to $\mathcal{T}$). I can show that $d$ is a metric on $X$ that is translation-invariant, but I have a problem on understanding the following.

To show the metric topology generated by $d$ is compatible with $\mathcal{T}$ (I think) it suffices to show that the collection of balls $$B_r=\{x\in X|d(x, 0)<r\},$$ where $0<r<\infty$ forms a convex balanced local base for $\mathcal{T}$.

Let $r>0$. Then Rudin says $B_r$ can be written as $$B_r=\bigcap_{\substack{k\in\mathbb{N}\\ c_k>r}}\Big\{x\in X\bigg|p_k(x)<\frac{r}{c_k-r}\Big\},$$ and the intersection is finite, since for otherwise it would contradict the assumption $\lim_{k\to\infty}c_k=0$. I understand this bit. However, I don't understand what would happen if $r>0$ is so badly chosen, for example, $r>\sup_k\{c_k\}$? Since $(c_k)$ is a convergent sequence, it is bounded, say by some $M>0$. Then once we choose $r$ to be bigger than $M$ the family of sets that are going to be intersected would be empty. Did I miss something here?

Thank you.

P.S. I google what is the intersection of a family of empty sets, some said it is $X$, which makes sense here, but some said it is $\emptyset$. It seems to me that it depends on which convention and perhaps also which set theory one is using.

Ho Man-Ho
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  • Then you get the an empty intersection; this yields the whole space, which is an open set. – Mittens Jul 27 '21 at 18:25
  • Thx for your comment. But do you mean it depends on how one defines the intersection of a family of empty sets and which set theory one is using? If yes, then it does bother me. I understand functional analysis will go to set theory at some points, but this is not the one I would expect. Moreover. it seems we are forced to use this convention instead of others, because it would be wrong. I feel this is quite awkward. And I am not sure if Rudin should say something on this issue in his book. – Ho Man-Ho Jul 27 '21 at 19:07
  • It is not a convention to have the empty intersection of sets as the whole space. It follows naturally from the definition of intersection. Rudin's functional analysis is for people who already have gone through the ropes of set teary, point-set topology, real and complex analysis and loads of linear algebra. That is why no much is spent in the standard results of set theory (a little about the axiom of choice is always a good thing to discuss no matter the level, but no needed here). – Mittens Jul 27 '21 at 20:22

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