Continuous Real-Valued Function on Irreducible Space is Constant

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space which is irreducible.

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

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

Proof
By definition of $T = \left({S, \tau}\right)$ being irreducible:


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

For every $x \in \R$, define:


 * $L_x := f^{-1}\left[{\left({-\infty \,.\,.\, x}\right)}\right]$
 * $U_x := f^{-1}\left[{\left({x \,.\,.\, \infty}\right)}\right]$

By continuity of $f$, these are open in $T$.

They are also disjoint, because for each $s \in S$, $f(s) < x$ and $f(s) > x$ cannot hold simultaneously.

Suppose now that $f(s) = a$ and $f(s') = b$, for some $s, s' \in S$ and $a, b \in \R$ with $a < b$.

Define $c = \dfrac{a + b} 2$.

Then $a < c < b$, whence:


 * $s \in L_c$ and $s' \in U_c$

However, because $T$ is irreducible, it follows that either $L_c$ or $U_c$ must be empty.

This contradicts the existence of such $a$ and $b$.

Hence $f$ must be constant.