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


$\mathscr{B}$ 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 $\mathscr{A}$.


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?

  • $\begingroup$ How do you define the uniform closure? $\endgroup$ – Asaf Karagila Dec 17 '12 at 15:04
  • $\begingroup$ $\mathscr{B}$ 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 $\mathscr{A}$. $\endgroup$ – Katlus Dec 17 '12 at 15:18
  • $\begingroup$ 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. $\endgroup$ – Katlus Dec 17 '12 at 15:39

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.


  • 1
    $\begingroup$ Why choosing $g_n$ is not a choice? $\endgroup$ – Katlus Dec 17 '12 at 18:19
  • 1
    $\begingroup$ @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. $\endgroup$ – Asaf Karagila Dec 17 '12 at 18:57
  • $\begingroup$ 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 $n\in\mathbb{N}$? $\endgroup$ – Katlus Dec 17 '12 at 19:13
  • $\begingroup$ 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 sequence $\endgroup$ – Katlus Dec 17 '12 at 19:38
  • $\begingroup$ @Asaf Please explain me why $\{g_n\}$ is guranteed from the definition. It seems countable choice is definitely in need here.. I'm stuck $\endgroup$ – Katlus Dec 17 '12 at 21:16

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.


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