I am using Bishop’s Theorem in the version given by Wikipedia¹: Let $\mathfrak{A}$ be a closed subalgebra of the Banach space $C(X,ℂ)$ of continuous complex-valued functions on a compact Hausdorff space $X$. Suppose that $f∈C(X,ℂ)$ has the following property: $f|_S∈\mathfrak{A}_S$ for every maximal set $S⊂X$ such that all real functions of $\mathfrak{A}_S$ are constant. Then $f∈\mathfrak{A}$. I apply this to $X$ being some neighbourhood of the origin and: $$ \mathfrak{A} = \big\langle ~ (x,y)↦xy; ~ (x,y)↦x^2y ~ \big\rangle,$$ where $⟨⟩$ denotes the generated set. Then $S^*=\left\{ (x,y) \,\middle |\, x=0 \lor y=0 \right\}$ is the only maximal subset on which all real functions of $\mathfrak{A}_{S^*}$ are constant and that has more than one point². My problem is that, according to Bishop’s theorem, the following functions must be in $\mathfrak{A}$, as they are in $\mathfrak{A}_{S^*}$ (being constantly $0$ on $S^*$): $(x,y) ↦ x y^2$: My problem with this is that I only get a higher power of $y$ by multiplying two generating functions with each other, which in turn yield a higher power of $x$ (than $1$). $(x,y) ↦ x^3 y$: My problem with this is that I only get an $x^3$ in the term by multiplying two generating functions, which in turn yield a higher power of $y$. My questions are: Did I make a mistake somewhere and the above functions are not in $\mathfrak{A}$? If not, how can I explicitly see that they are in $\mathfrak{A}$? What sequence in $\mathfrak{A}$ is converging to the functions in question? ¹ According to some sources, there is an additional requirement on $\mathfrak{A}$ to contain constant functions. But this does not solve my problem. ² If we had another such subset, there would have to be $(x_1,y_1), (x_2,y_2) \notin S^*$, such that $(x_1,y_1) ≠ (x_2,y_2)$ and $x_1 y_1 = x_2 y_2$ and $x_1^2 y_1 = x_2^2 y_2$. However, this implies: $$x_1 = \frac{x_1^2 y_1}{x_1 y_1} = \frac{x_2^2 y_2}{x_2 y_2} = x_2,$$ and from that $y_1 = \frac{y_1 x_1}{x_1} = \frac{y_2 x_2}{x_2} = y_2$, which contradicts with the requirements. Login To add answer/comment