Algebra and topology

[Papapetros Vaggelis]

Let \(\displaystyle{\left(X,\mathbb{T}\right)}\) be a topological space. Consider the commutative ring with unity

\(\displaystyle{\left(C(X,\mathbb{R}),+,\cdot\right)}\) of all continuous functions \(\displaystyle{f:X\longrightarrow \mathbb{R}}\) .



Prove that the topological space \(\displaystyle{\left(X,\mathbb{T}\right)}\) is connected if, and only if the ring \(\displaystyle{\left(C(X,\mathbb{R}),+,\cdot\right)}\)

is connected.



Note: An associatove ring with unity \(\displaystyle{\left(R,+,\cdot\right)}\) is connected, if the only central and idempotent elements are \(\displaystyle{0_{R},1_{R}}\) .

[Papapetros Vaggelis]

Here is a solution.

Suppose that the topological space \(\displaystyle{\left(X,\mathbb{T}\right)}\) is connected.

Obviously, the elememts \(\displaystyle{\mathbb{O}=0_{C(X,\mathbb{R})}\,\,\,,\mathbb{1}=1_{C(X,\mathbb{R})}}\)

are central and idempotent elements of the ring \(\displaystyle{\left(C(X,\mathbb{R}),+,\cdot\right)}\) .

Let \(\displaystyle{f\in C(X,\mathbb{R})}\) be a central and idempotent function, that is:

\(\displaystyle{f\cdot g=g\cdot f\,,\forall\,g\in C(X,\mathbb{R})\,\,\,,f^2=f\cdot f=f}\) .

Then, \(\displaystyle{f(X)\subseteq \left\{0,1\right\}}\) . If the function \(\displaystyle{f}\) is not constant, then

\(\displaystyle{f(x_0)=0}\) and \(\displaystyle{f(x_1)=1}\) for some \(\displaystyle{x_0\,,x_1\in X\,,x_0\neq x_1}\) .

Consider the sets \(\displaystyle{A=\left\{x\in X: f(x)=0\right\}=f^{-1}\left(\left\{0\right\}\right)\,\,,B=\left\{x\in X: f(x)=1\right\}=f^{-1}\left(\left\{1\right\}\right)}\) .

We have that \(\displaystyle{A\,,B\neq \varnothing}\) since \(\displaystyle{x_0\in A\,,x_1\in B}\)

\(\displaystyle{X=A\cup B\,,A\cap B=\varnothing}\) and since the function \(\displaystyle{f}\) is continuous and the sets

\(\displaystyle{\left\{0\right\}\,,\left\{1\right\}}\) are closed subsets of \(\displaystyle{\mathbb{R}}\) , we also get that (\displaystyle{A\,,B}\)

are closed subsets of \(\displaystyle{\left(X,\mathbb{T}\right)}\), which means that \(\displaystyle{\left\{A,B\right\}}\) is a partition of

closed subsets of \(\displaystyle{\left(X,\mathbb{T}\right)}\), a contradiction, since this topological space is connected.

Conversely, suppose that the ring \(\displaystyle{\left(C(X,\mathbb{R}),+,\cdot\right)}\) is connected.

Let \(\displaystyle{\left\{A,B\right\}}\) be a partition of open subsets of \(\displaystyle{X}\), that is

\(\displaystyle{A\,,B\in\mathbb{T}-\left\{\varnothing\right\}\,,X=A\cup B\,,A\cap B=\varnothing}\). Consider the function

\(\displaystyle{f:X\longrightarrow \mathbb{R}\,,x\mapsto f(x)=\begin{cases}
0\,\,,x\in A\\
1\,\,,x\in B
\end{cases}}\)

Obviously, \(\displaystyle{f\neq \mathbb{O}\,,f\neq \mathbb{1}\,\,\,,f(X)=\left\{0,1\right\}}\) .

The ring \(\displaystyle{\left(C(X,\mathbb{R}),+,\cdot\right)}\) is commutative,

so: \(\displaystyle{f\cdot g=g\cdot f\,,\forall\,g\in C(X,\mathbb{R})}\) . Also, if \(\displaystyle{x\in X}\), then :

\(\displaystyle{\bullet: x\in A\implies f^2(x)=f(x)\cdot f(x)=0\cdot 0=0=f(x)}\)

\(\displaystyle{\bullet: x\in B\implies f^2(x)=f(x)\cdot f(x)=1\cdot 1=1=f(x)}\)

and thus: \(\displaystyle{f^2=f}\) . Let \(\displaystyle{x_0\in X}\) and \(\displaystyle{W}\) be a region

of \(\displaystyle{f(x_0)}\). There exists \(\displaystyle{\left(a,b\right)\subseteq \mathbb{R}}\) such that

\(\displaystyle{f(x_0)\in\left(a,b\right)\subseteq W}\) . If \(\displaystyle{x_0\in A\,,A\in\mathbb{T}}\), then :

\(\displaystyle{f(x_0)=0}\) and \(\displaystyle{f(A)=\left\{0\right\}\subseteq W}\). If \(\displaystyle{x_0\in B\,,B\in\mathbb{T}}\)

then \(\displaystyle{f(x_0)=1}\) and \(\displaystyle{f(B)=\left\{1\right\}\subseteq W}\). In any case, the function \(\displaystyle{f}\)

is continuous at \(\displaystyle{x_0\in X}\). Therefore, the function \(\displaystyle{f}\) is continuous. This way, the function

\(\displaystyle{f\in C(X,\mathbb{R})-\left\{\mathbb{O},\mathbb{1}\right\}}\) such that

\(\displaystyle{f\cdot g=g\cdot f\,,\forall\,g\in C(X,\mathbb{R})}\) and \(\displaystyle{f^2=f}\), a contradiction, since the

ring \(\displaystyle{\left(C(X,\mathbb{R}),+,\cdot\right)}\) is connected.

So, the topological space \(\displaystyle{\left(X,\mathbb{T}\right)}\) is connected.