Continuous Real-Valued Function on Irreducible Space is Constant

Theorem
Let $T = \left({X, \vartheta}\right)$ be a topological space which is hyperconnected.

Let $f: X \to \R$ be a continuous real-valued function.

Then $f$ is constant, that is:
 * $\exists a \in \R: \forall x \in X: f \left({x}\right) = a$

Proof
Let $T = \left({X, \vartheta}\right)$ be hyperconnected.

Then by definition:


 * $\forall U_1, U_2 \in \vartheta: U_1, U_2 \ne \varnothing \implies U_1 \cap U_2 \ne \varnothing$

Let $a, b \in \R$.

Let $U_1 = f^{-1} \left({a}\right), U_2 = f^{-1} \left({b}\right)$.

As $f$ is continuous it follows that $U_1$ and $U_2$ are open in $T$.

As $\vartheta$ is a topology it follows by definition that $U_1 \cap U_2$ is also open in $T$.

But then from Image of Subset is Subset of Image/Corollary 2:
 * $f \left({U_1 \cap U_2}\right) \subseteq f \left({U_1}\right) = a$
 * $f \left({U_1 \cap U_2}\right) \subseteq f \left({U_2}\right) = b$

So by definition:
 * $\forall x \in U_1 \cap U_2: f \left({x}\right) = a$


 * $\forall x \in U_1 \cap U_2: f \left({x}\right) = b$

From definition of mapping it follows that $a = b$.

Hence the result.