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Let $\mathscr{A}$ be an algebra of bounded complex functions. (Or if necessary, continuous and domain of functions is compact)

Definition:

$\mathscr{B}$ is uniformly closed iff $f\in\mathscr{B}$ whenever $f_n\in \mathscr{B} (n=1,2,\cdot)$ and $f_n\rightarrow f$ uniformly.

$\mathscr{B}$ is the uniform closure of $\mathscr{A}$ iff $\mathscr{B}$ is the set of all functions which are limits of uniformly convergent sequences of members of $\mathscr{A}$.

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Let $\mathscr{B}$ be a uniform closure of $\mathscr{A}$.

How do i prove that $\mathscr{B}$ is uniformly closed in ZF?

Does Stone-Weierstrass theorem require choice since it is critical to prove Stone-Weierstrass Theorem?

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    How do you define the uniform closure?2012-12-17
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    $\mathscr{B}$ is the uniform closure of $\mathscr{A}$ if $\mathscr{B}$ is the set of all functions which are limits of uniformly convergent sequences of members of $\mathscr{A}$.2012-12-17
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    I just edited my question. Is there a term to distinguish these two definitions (one is on the post and the other is $\overline{\mathscr{A}}$)? The relation looks very similar to that between sequential continuity and $\epsilon-\delta$ continuity.2012-12-17

2 Answers 2

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I would answer for the first question.

Theorem: Let $\mathscr{B}$ be a uniform closure of $\mathscr{A}$. (Where $\mathscr{A}$ - algebra consisting of bounded functions). Then $\mathscr{B}$- uniformly closed algebra.

Proof: If $f\in\mathscr{B}$ and $g\in\mathscr{B}$, then there are uniformly convergent sequences $f_n\in\mathscr{A}$ and $g_n\in\mathscr{A}$ such that $f_n\to f$ and $g_n\to g$. Since the functions are bounded, we can write: $$f_n+g_n\to f+g$$ $$f_ng_n\to fg$$ $$cg_n\to cg$$ Where $c$ is constant from the field.

So $f+g\in\mathscr{B} $,$fg\in\mathscr{B} $,$cg\in\mathscr{B}$. So $\mathscr{B}$ is algebra.

Let $f_n$ is uniformly convergent sequence of elements from $\mathscr{B}$. There are functions $g_n$ such that $|f_n(x)-g_n(x)|<\frac{1}{n}$. If $f_n\to f$ then it is clear that $g_n\to f$, so (by definition of $\mathscr{B} $) $f\in\mathscr{B}$, so $\mathscr{B} $ is uniformly closed.

$\blacksquare$

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    Why choosing $g_n$ is not a choice?2012-12-17
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    @Katlus: The existence of such sequence $g_n$ is guaranteed from the definition, and the choice of sequence does not affect the outcome. Since you are interested in finite number of choices it's fine.2012-12-17
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    I don't really understand. Please let me know where i'm thinking wrong. Fix $n\in\mathbb{N}$. Then there exist (maybe) infinitely many sequences in $\mathscr{A}$ convergent to $f_n$. Isn't this choosing $g_n$ for each $n\in\mathbb{N}$?2012-12-17
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    I still don't understand why constructin $\{g_n\}$ is not a countable choice. Please explain me the algorithm to construct that sequence $\{g_n\}$?? Sure $f$ is a limit point of $\mathscr{A}$, but how does that imply there actually exists such sequence2012-12-17
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    @Asaf Please explain me why $\{g_n\}$ is guranteed from the definition. It seems countable choice is definitely in need here.. I'm stuck2012-12-17
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    @Katlus: Recall the definition in your question. $f\in\mathscr B$ if and only if there is a sequence $f_n$ in $\mathscr B$ which converging to $f$ pointwise. Then if we assume that $f,g$ were taken from $\mathscr B$ then there are such sequences. This is a similar question, on your side, to the one with the D-finite set of reals. Recall that without countable choice closure and sequential closure may differ (in metric spaces). The question you need to ask yourself is whether or not the closure is equal to the sequential closure. But coming from the given definitions, I don't see issues. [cont]2012-12-17
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    [...] as the definition is essentially the sequential closure. So if you were to ask whether the sequential closure of a set is sequentially closed, the answer is trivially yes. You could, and perhaps should, review your own definitions and questions and possibly revise your question to point out where you think that choice sneaks into the game. Perhaps you want to give two definitions of closure, or so, in which case you wish to show equivalence and there choice might lay. But as far as the question in its current form goes -- there is no issue.2012-12-17
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    @Asaf Thank you for your time to help me. I read over and over my definitions and your comments, but there is a still problem. In order to show that $\mathscr{B}$ is uniformly closed, we have to show that "For a given uniformly convergent sequence $\{f_n\}$ in $\mathscr{B}$ such that $f_n\rightarrow f$, there exists $\{g_n\}$ in $\mathscr{A}$ such that $g_n\rightarrow f$". (I believe you made a typo in your comment. It is $\mathscr{A}$ ,not $\mathscr{B}$).2012-12-17
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    So let $\{f_n\}$ be a uniformly convergent sequence in $\mathscr{B}$ such that $f_n\rightarrow f$. Now, we have to construct a sequence in $\mathscr{A}$ which is convergent to $f$. It can be shown that $f$ is a limit point of $\mathscr{A}$. Thus we need to choose elements for each $B(f,1/n)\cap \mathscr{A}$, which i think is definitely a countable choice. I cannot figure out an algorithm to construct this $\{g_n\}$, which is not trivial to me, but rather i would be very surprised if it is true.2012-12-17
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    @Asaf Plus, this is on wikipeida that "It's possible to be $[A]_{\text{ seq}}\subsetneq [[A]_{\text{ seq}}]_{\text{ seq}}$. Link; http://en.m.wikipedia.org/wiki/Sequential_space2012-12-17
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    @Katlus, under choice metric spaces are sequential spaces which is what I meant in my comment (hence the remark in parenthesis saying "in metric spaces"). I finally understand your question now, and yes it would seem that choice might be involved somehow. I'll give it some thought.2012-12-17
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    @Asaf Thank you and now i feel so safe :)2012-12-17
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I think that $\mathscr{B}$ is the uniform closure of $\mathscr{A}$, for any $x$, $\mathscr{B}$ is the uniform closure of $\mathscr{A}$.

Recall the closure $\overline{E}$ of the set $E$ is closed. Here, $f_n(x) \rightarrow f(x)$ for any $x$ is holded. So $\mathscr{B}$ is uniform closed.