# An easy proof of the Stone-Weierstrass Theorem

I've tried to understand the Stone-Weierstrass proof in many many books, pdf files, etc. and I still can't understand it, I mean I get stuck easily in some parts of the proof.

So if someone knows about a really really explained proof, I mean a proof with lots of details, please share it with me. I need it!

Thank you in advance.

Note: I'm taking a course of analysis, haven't take any course about topology.

• So the theorem must be in compact spaces. :) – Aaron Martinez Apr 27 '17 at 2:31
• Have you looked at real analysis book by Pugh? – Ken Apr 27 '17 at 2:34
• The explanations from this book are really nice. – Ken Apr 27 '17 at 2:34
• @Rose I'll check it, thanks – Aaron Martinez Apr 27 '17 at 2:41
• this kind of stuff really takes some time - at leas a couple of days - to really understand if this is your first exposition to real analysis. No need to rush or try to shortcut. Also, you really need to do some pen and paper work to get the gist of it, if you were just reading it like reading an essay. Math does not work like that. – dezdichado Apr 27 '17 at 2:42

Despite many people not liking his book for beginners, I remember Rudin's proof in Principles of Mathematical Analysis to be much easier to understand than his proof of the Weierstrass Theorem (admittedly, not saying much).

The proof is simple in it's general approach, but the devil is in the detail. I can flesh out the intuition, but you'll need to work through the details yourself.

When you want to prove that an algebra of some functions contains a nice set X of functions or is dense, you'll generally approach it this way:

use really basic functions you've either assumed already exist in your algebra, or can easily build, and combine them into more complicated functions. For example, the Stone-Weierstrass Theorem assumes that we have a non-zero constant function, and that our algebra separates points. Okay, if we only had the constant function, we would just have the set of all functions which can be built from that alone-namely multiples of that constant. This isn't a huge class of functions, and so we need more. How are we going to get non-linearity? If all out functions were constant, then adding and multiplying would still give us constants.

Thus, we need to assume a priori that there is some non-linear function in our algebra for us to eventually be dense in the set of continuous functions. This is why the separating points assumption is needed.

I like the presentation in General Topology by R. Engelking. Note: He uses "compact space" to mean "compact Hausdorff space". And he uses $I$ for the real interval $[0,1].$ And he writes $\{w_i\}$ to denote a sequence $(w_1,w_2,...)$.

Chapter 3, Section 2: Lemma 3.2.18 (The Dini Theorem) is stated and proved in order to prove Lemma 3.2.19, which is used to prove Lemma 3.2.20. (Note that the sequence $\{w_i\}$ in 3.2.20 is defined in 3.2.19)....Lemma 3.2.20 is used in the proof of Theorem 3.2.21: The Stone-W. Theorem.

Note. In his statement of the S.-W. theorem you can safely ignore the sentence within the brackets. The first sentence of the proof is a statement of the objective.

Altogether, from 3.2.18 through 3.2.21, less than 2 pages.

I really like Ransford's proof:

T.J. Ransford, A short elementary proof of the Bishop-Stone-Weierstrass Theorem, Math. Proc. Comb. Phil. Soc., 96, 309–311.

• what is that? seems difficult :( – Aaron Martinez Apr 28 '17 at 23:36
• @AaronMartinez can you please define what do you mean by an easy proof? – Tomek Kania Jun 8 '17 at 8:10
• something similar to the proof by Pugh's analysis book – Aaron Martinez Jun 8 '17 at 17:59