Algebra and Topology

Let $(X, \mathbb{T})$ be a topological space. Consider the commutative ring with unity $\left(C(X,\mathbb{R}),+,\cdot\right)$ of all continuous functions $f: X \rightarrow \mathbb{R}$ . Prove that the topological space $(X, \mathbb{T})$ is connected if-f the ring $\left(C(X,\mathbb{R}),+,\cdot\right)$ is connected.

Solution



 Suppose that the topological space $\left(X,\mathbb{T}\right)$ is connected. Obviously, the elememts $\mathbb{O}=0_{C(X,\mathbb{R})}\,\,\,,\mathbb{1}=1_{C(X,\mathbb{R})}$ are central and idempotent elements of the ring $\left(C(X,\mathbb{R}),+,\cdot\right)$.

Let $f\in C(X,\mathbb{R})$ be a central and idempotent function, that is:
$$f\cdot g=g\cdot f\,,\forall\,g\in C(X,\mathbb{R})\,\,\,,f^2=f\cdot f=f$$

Then, $f(X)\subseteq \left\{0,1\right\}$ . If  $f$ is not constant, then $f(x_0)=0$ and $f(x_1)=1$ for some $x_0, \,x_1\in X, \;x_0\neq x_1$ .

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

We have that $A, \;B\neq \varnothing$ since $x_0\in A\,,x_1\in B$, $X=A\cup B,\;A\cap B=\varnothing$ and since the function $f$ is continuous and the sets $\left\{0\right\}, \;\left\{1\right\}$ are closed subsets of $\mathbb{R}$ , we also have that $A, \;B$  are closed subsets of $\left(X,\mathbb{T}\right)$, which means that $\left\{A,B\right\}$ is a partition of closed subsets of $\left(X,\mathbb{T}\right)$, a contradiction, since this topological space is connected.

Conversely, suppose that the ring $\left(C(X,\mathbb{R}),+,\cdot\right)$ is connected. Let $\left\{A,B\right\}$ be a partition of open subsets of $X$ , that is

$$A, \,B\in\mathbb{T}-\left\{\varnothing\right\}, \;X=A\cup B\,,A\cap B=\varnothing$$

Consider the function $f:X\rightarrow \mathbb{R}$  defined as $f(x)= \left\{\begin{matrix}
0 &,  &x \in A \\
 1&,  &x \in B
\end{matrix}\right.$.  The ring $\left(C(X,\mathbb{R}),+,\cdot\right)$ is commutative,

so: $f\cdot g=g\cdot f\,,\forall\,g\in C(X,\mathbb{R})$ . Also, if  $x\in X$, then :
$\color{gray}{\blacksquare} x\in A \Rightarrow f^2(x)=f(x)\cdot f(x)=0\cdot 0=0=f(x)$
$\color{gray}{\blacksquare}   x\in B \Rightarrow f^2(x)=f(x)\cdot f(x)=1\cdot 1=1=f(x)$

Thus $f^2=f$ . Let $x_0\in X$ and $W$ be a neighbourhood of  $f(x_0)$. Then there exists $\left(a,b\right)\subseteq \mathbb{R}$ such that  $f(x_0)\in\left(a,b\right)\subseteq W$ . If $x_0\in A, \;A\in\mathbb{T}$ , then : $f(x_0)=0$ and $f(A)=\left\{0\right\}\subseteq W$. If  $x_0\in B, \;B\in\mathbb{T}$  then $f(x_0)=1$ and $f(B)=\left\{1\right\}\subseteq W$. In any case, the function $f$ is continuous at  $x_0\in X$. Therefore, the function $f$ is continuous. This way, the function
$$f\in C(X,\mathbb{R})-\left\{\mathbb{O},\mathbb{1}\right\}$$such that

$f\cdot g=g\cdot f\,,\forall\,g\in C(X,\mathbb{R})$ and $f^2=f$, a contradiction, since the ring $\left(C(X,\mathbb{R}),+,\cdot\right)$ is connected.

So the topological space $\left(X,\mathbb{T}\right)$ is connected and the exercise comes to an end.

The exercise can also be found in mathimatikoi.org