Let $\bar{\mathcal{A}}$ denote the closure of $\mathcal{A}$ in $C^0 (X, \mathbb{R})$ according to the uniform convergence topology. We want to show that, if conditions 1 and 2 are satisfied, then $\bar{\mathcal{A}} = C^0 (X, \mathbb{R})$ .

First, we shall show that, if $f \in \bar{\mathcal{A}}$ , then $|f| \in \bar{\mathcal{A}}$ . Since $f$ is a continuous function on a compact space $f$ must be bounded - there exists constants $a$ and $b$ such that $a \le f \le b$ . By the Weierstrass approximation theorem, for every $\epsilon > 0$ , there exists a polynomial such that $|P(x) - |x|| < \epsilon$ when $x \in [a,b]$ . (By the way, one does not need the full-blown Weierstrass approximation theorem to show that $P$ exists - see the entry ``proof of Weierstrass approximation theorem'' for an elementary construction of $P$ ) Define $g: X \to \mathbb{R}$ by $g(x) = P(f(x))$ . Since $\bar{\mathcal{A}}$ is an algebra, $g \in \bar{\mathcal{A}}$ . For all $x \in X$ , $|g(x) - |f(x)|| < \epsilon$ . Since $\bar{\mathcal{A}}$ is closed under the uniform convergence topology, this implies that $|f| \in \bar{\mathcal{A}}$ .

A corollary of the fact just proven is that if $f,g \in \bar{\mathcal{A}}$ , then $\max (f,g) \in \bar{\mathcal{A}}$ and $\min (f,g) \in \bar{\mathcal{A}}$ . The reason for this is that one can write $$\max (a,b) = {1 \over 2} \left( a+b + |a-b| \right)$$ $$\min (a,b) = {1 \over 2} \left( a+b - |a-b| \right)$$

Second, we shall show that, for every $f \in C^0 (X, \mathbb{R})$ , every $x \in X$ , and every $\epsilon > 0$ , there exists $g_x \in \bar{\mathcal{A}}$ such that $g_x \le f + \epsilon$ and $g_x > f$ . By condition 1, if $y \neq x$ , there exists a function ${\tilde h}_{xy} \in \mathcal{A}$ such that ${\tilde h}_{xy} (x) \ne {\tilde h}_{xy} (y)$ . Define $h_{xy}$ by $h_{xy} (z) = p {\tilde h}_{xy} (z) + q$ , where the constants $p$ and $q$ have been chosen so that $$h_{xy} (x) = f(x) + \frac{\epsilon}{2}$$ $$h_{xy} (y) = f(y) - \frac{\epsilon}{2}$$ By condition 2, $h_{xy} \in \mathcal{A}$ . (Note: This is the only place in the proof where condition 2 is used, but it is crucial since, otherwise, it might not be possible to construct a function which takes arbitrary preassigned values at two distinct points of $X$ . The necessity of condition 2 can be shown by a simple example: Suppose that $\mathcal{A}$ is the algebra of all continuous functions on $f$ which vanish at a point $O \in X$ . It is easy to see that this algebra satisfies all the hypotheses of the theorem except condition 2 and the conclusion of the theorem does not hold in this case.)

For every $y \ne x$ , define the set $U_{xy}$ as $$U_{xy} = \{ z \in X \mid h_{xy} (z) < f(z) + \epsilon \}$$ Since $f$ and $h_{xy}$ are continuous, $U_{xy}$ is an open set. Because $x \in U_{xy}$ and $y \in U_{xy}$ , $\left\{ U_{xy} \mid y \in X \setminus \{x\} \right\}$ is an open cover of $X$ . By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset $\{y_1, y_2, \ldots, y_n \} \subset X$ such that $X = \bigcup_{m=0}^n U_{xy_m}$ . Define $$g_x = \min (h_{xy_1}, h_{xy_2}, \ldots, h_{xy_n}).$$ By the corollary of the first part of the proof, $g_x \in \bar{\mathcal{A}}$ . By construction, $g_x(x) = f(x) + \epsilon / 2$ and $g_x < f + \epsilon$ .

Third, we shall show that, for every $f \in C^0 (X, \mathbb{R})$ and every $\epsilon > 0$ , there exists a function $g \in \bar{\mathcal{A}}$ such that $f \le g < f + \epsilon$ . This will complete the proof becauase it implies that $\bar{\mathcal{A}} = C^0 (X, \mathbb{R})$ . For every $x \in X$ , define the set $V_x$ as $$V_x = \{ z \in X \mid g_{x} (z) > f(x) \}$$ where $g_x$ is defined as before. Since $f$ and $g_x$ are continuous, $V_x$ is an open set. Because $g_x(x)=f(x)+ \epsilon /2 > f(x)$ , $x \in V_x$ . Hence $\left\{ V_x \mid x \right\}$ is an open cover of $X$ . By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset $\{x_1, x_2, \ldots x_n \} \subset X$ such that $X = \bigcup_{m=0}^n V_{x_n}$ . Define $g$ as $$g (z) = \max \{ g_{x_1} (z), g_{x_2} (z), \ldots, g_{x_n} (z) \}$$ By the corollary of the first part of the proof, $g \in \bar{\mathcal{A}}$ . By construction, $g > f$ . Since $g_x < f + \epsilon$ for every $x \in X$ , $g < f + \epsilon$ .