What does the sentence "The only sub-algebras of $\mathbb{R}^{2}$ are $0,\mathbb{R}^{2},\mathbb{R}(0,1),\mathbb{R}(1,0),\mathbb{R}(1,1)$" mean?

Question

I started studying functional analysis a couple of days ago, I have reached the Stone-Weierstrass theorem which is stated in my lecture notes as

Let $X$ be a compact metric space, $A\subseteq C(X)$ a sub-algebra that contains the constant functions and separates between the points, then $\overline{A}=C(X)$

And the proof begins with a lemma that claims

The only sub-algebras of $\mathbb{R}^{2}$ are $0,\mathbb{R}^{2},\mathbb{R}(0,1),\mathbb{R}(1,0),\mathbb{R}(1,1)$

I don't understand the theorem statement and the notation in the lemma - the notes never defined what an algebra or sub-algebra is (searching "algebra" in Google yields many results and it seems there is more than one meaning to this term) , and I am unfamiliar with the notation $\mathbb{R}(0,1)$ etc'.

Can someone please explain, in this context, what is the meaning of an algebra (or sub-algrbra) and of the notation ?

Answer 1

For any $\langle a,b\rangle\in\Bbb R^2$, $\Bbb R\langle a,b\rangle=\{\langle ra,rb\rangle:r\in\Bbb R\}$, so the last three sub-algebras in that list correspond to the $y$-axis, the $x$-axis, and the diagonal line $y=x$, respectively. The definition of algebra that you want appears to be this one: $\Bbb R^2$ under ordinary vector addition is an additive Abelian group, a ring (as a product of two rings), and an $\Bbb R$-module. You’ll find the appropriate notion of sub-algebra at the link.

Answer 2

‘Algebra’ and ‘subalgebra’ seem to be used in the sense of (associative commutative but not necessarily unital) algebras over $\mathbb{R}$. These are commutative rngs which are equipped with a compatible $\mathbb{R}$-vector space structure; more precisely, such a thing is a $\mathbb{R}$-vector space $A$ equipped with an extra binary operation $\times$ such that:

• $\times$ is associative and commutative.
• $(r x) \times y = x \times (r y)$ for all $x$ and $y$ in $A$ and $r$ in $\mathbb{R}$.
• $(x + y) \times (z + w) = x \times z + x \times w + y \times z + y \times w$ for all $x$, $y$, $z$, and $w$ in $A$.

The notation $\mathbb{R} \vec{v}$ refers to the vector subspace spanned by $\vec{v}$. Notice that when $\vec{v} \times \vec{v} = \vec{v}$, the subspace $\mathbb{R} \vec{v}$ is automatically a subalgebra. It is not hard to check that the only such $\vec{v}$ in $\mathbb{R}^2$ are $(0, 0), (0, 1), (1, 0), (1, 1)$.

Answer 3

As for the statement of the theorem,

Let $X$ be a compact metric space, [...]

so far so good hopefully

[...] $A\subseteq C(X)$ a sub-algebra [...]

$C(X)$ refers to the set of continuous, real-valued functions on $X$. $A$ is a subset of $C(X)$ so the elements of $A$ are continuous, real-valued functions on $X$. A family of real functions $\mathscr{A}$ is an algebra if $f,g\in\mathscr{A}$ implies that each of $f+g\in\mathscr{A}$, $fg\in\mathscr{A}$ and $cf\in\mathscr{A}$ hold for all $c\in\mathbb{R}$. You can verify that $C(X)$ is an algebra. Then if $A\subseteq C(X)$ is an algebra, then we can say $A$ is a subalgebra of $C(X)$.

[...] that contains the constant functions [...]

So every constant function from $X$ to $\mathbb{R}$ is in $A$

[...] and separates between the points [...]

An algebra $\mathscr{A}$ of functions on a set $Y$ separates points if for each $x,y\in Y$ there exists $f\in\mathscr{A}$ such that $f(x)\ne f(y)$.

[...] then $\overline{A}=C(X)$.

$\overline{A}$ is the uniform closure of $A$, which means that it is the set of all functions $f$ such that $f$ is the limit of a uniformly convergent sequence of functions in $A$. So what this is really saying is that every element of $C(X)$ is the limit of a uniformly convergent sequence in $A$.