# Density In The Theorem/Proof of The Stone-Weierstrass Theorem

Yet again, I have a question that I could use some help with. Note that almost everything can be found in C. Pugh's, Real Mathematical Analysis (soft-cover, 2nd Edition, ISBN: 978-1-4419-2941-9); furthermore, although it is a book-preview, what I'm about to refer to as well as question can (hopefully, as it did for me) be found here.

My question concerns the very beginning of the proof of the Stone-Weierstrass Theorem in Pugh's book (proof of Theorem 19, PG. 224 in the actual book, or PG. 235 in the book-preview link above). The theorem from the text is restated below.

Stone-Weierstrass Theorem: If $M$ is a compact metric space, and $\mathcal{A}$ is a function algebra in $C^{0}M$ that vanishes nowhere and separates points, then $\mathcal{A}$ is dense in $C^{0}M$.

As far as the definitions of a function algebra, the function algebra vanishing at a point $p\in M$, as well as the function algebra separating points in $M$, please see the link above (they are available in the preview for me, but I omitted them for the sake of saving space - I will gladly state them if need be). Moreover, for some compact metric space $M$, $C^{0}M$ is shorthand notation for the set $C^{0}\big(M,\mathbb{R}\big)=\big\{ \phi~|~\phi:M\rightarrow\mathbb{R},~\phi\text{ is continuous on }M\big\}$, i.e., $C^{0}M$ is the set of continuous, real-valued functions on the compact metric space $M$.

Now for my question. In the very beginning of Pugh's proof of the theorem above, we assume $\mathcal{A}\subset C^{0}M$ is a function algebra that vanishes nowhere and separates points (clearly), Then, we show $\mathcal{A}$ is dense in $C^{0}M$, to which the proof asserts that this can be shown by proving for any $F\in C^{0}M$ and any $\varepsilon>0$, there exists a $G\in\mathcal{A}$ such that for all $x\in M$ we have $F(x)-\varepsilon<G(x)<F(x)+\varepsilon$ (equivalently, $\big|G(x)-F(x)\big|=\big|F(x)-G(x)\big|<\varepsilon$). Now, earlier in the book -- which isn't part of the book-preview for me, unfortunately (if it helps, the definition is on PG. 98 in Pugh's text) -- Pugh defines density of a set which is restated below (I added the ambient topological space part for reference purposes).

Definition: Let $M$ be a metric (or topological) space. If $S\subset M$ and $\overline{S}=M$, then $S$ is dense in $M$.

I've been trying to connect the statement at the beginning of the Stone-Weierstrass proof with this definition above. In other words, in the proof of the Stone-Weierstrass Theorem in Pugh's text, we need to show that $\overline{\mathcal{A}}=C^{0}M$, correct? Furthermore, how does this translate to, or how do we connect this definition with, for any $F\in C^{0}M$ and any $\varepsilon>0$, there exists a $G\in\mathcal{A}$ such that for all $x\in M$ we have $F(x)-\varepsilon<G(x)<F(x)+\varepsilon$?

First I started working with the fact that (where this can be shown) $\lim\mathcal{A}=\overline{\mathcal{A}}$ where $\lim\mathcal{A}$ is the limit set of the function algebra $\mathcal{A}$ (i.e., the set of limit points of $\mathcal{A}$). Then, I recalled the example that $\mathbb{Q}$ is dense in $\mathbb{R}$ being equivalent to saying that for any pair of distinct real numbers $p<r\in\mathbb{R}$, there exists a rational number $q\in\mathbb{Q}$ such that $p<q<r$.

I'm thinking that I'm on the right track by going from the example that $\mathbb{Q}$ is dense in $\mathbb{R}$; and the connection I'm trying to make is possibly similar to showing that $\mathbb{Q}$ is dense in $\mathbb{R}$ is equivalent to saying that for some $q\in\mathbb{Q}$, $p<q<r$ whenever $p<r\in\mathbb{R}$...just in a more abstract scene?

In sum, any help with making this connection will be greatly appreciated. I apologize that this is a bit lengthy, but I wanted to be precise, and thank you in advanced for any recommendations, suggestions, answers, etc., as well as for spending time on this. As a personal remark, check out the proof, as it is pretty neat - especially how he finds a functions that supersolves the expression $F(x)-\varepsilon<G(x)<F(x)+\varepsilon$.

• The space $C^0 M$, equipped with the supremum norm $\| f\| = \sup_x |f(x)|$, is a normed space. Then the density of $\mathcal{A}$ is with respect to the resulting topology. – Sangchul Lee Dec 28 '15 at 0:06
• @SangchulLee : Thank you for your comment. See me comment below in terms of my mix-up; essentially I made things to hard on myself when trying to wrap myself around Pugh's assertion in his proof. In any case, I knew that the connection highly depends on the fact that we use the subspace metric coupled with $\mathcal{A}$, and that there had to be an easier way than using a sequence of functions in $\mathcal{A}$ – Procore Dec 28 '15 at 3:02

If for all $x$ you have $F(x)-\varepsilon<G_{\varepsilon}(x)<F(x)+\varepsilon$(as you noted this is the same as $\lVert F-G_{\varepsilon} \rVert<\varepsilon$) for any epsilon then $\lim_{\varepsilon \rightarrow 0}G_{\varepsilon}=F$, so $F$ is indeed a limit point. Since $F$ was arbitrary we get $\lim\mathcal{A}=F$.
• Thank you for clearing this up! What started everything is that there was a proof I did that uses the Stone-Weierstrass Theorem. I, coincidentally, used almost the same argument you posted above -- overall, I wasn't sure if I was correct. I kept thinking since WNTS $\lim\mathcal{A}=\overline{\mathcal{A}}=C^{0}(M,R)$, then I would have to choose a sequence $(F_{n})_{_{n\geq 1}}$ so that the equivalent statement stems from $F_{n}\rightrightarrows F$, as $n\rightarrow+\infty$. Then I realized that I didn't quite clearly/rigorously understand Pugh's proof in the first place. Thanks again! – Procore Dec 28 '15 at 2:49
Another way to look at the result is to start with $C^0M$ is a metric space with the metric $d(f,g)=\sup_{x\in M}|f(x)-g(x)|$. Then, we want to say that $\mathscr A$ is dense in $C^0M$, so we want $C^0M=\overline{\mathscr A}$.
In other words, we want to show that every $F\in C^0M$ is a point of $\mathscr A$ or a limit point of $\mathscr A$. To say that $F\in C^0M$ is a limit point of $\mathscr A$, with the language of the metric, is to say that for each $\varepsilon>0$, there is some $G\in \mathscr A$ such that $d(F,G)=|F(x)-G(x)|<\varepsilon$ $(x\in M)$, and that's just what Pugh does.