Converse of intermediate value theorem

Statement

Suppose $X$ is a topological space that satisfies the conclusion of the intermediate value theorem: For any continuous function $f:X\to \mathbb{R}$, and two elements $x_1,x_2\in X$ such that $f(x_1)<f(x_2)$, $f(X)$ must contain $[f(x_1),f(x_2)]$.

Then, $X$ is a connected space.

Related facts

Converse

• Intermediate value theorem

Proof

Given: A topological space $X$ such that for any continuous function $f:X\to \mathbb{R}$, and two elements $x_1,x_2\in X$ such that $f(x_1)<f(x_2)$, $f(X)$ must contain $[f(x_1),f(x_2)]$.

To prove: $X$ is connected.

Proof: Suppose not, i.e., suppose $X$ is not connected. Then, $X$ is a union of two nonempty disjoint open subsets $U$ and $V$. Consider the function $f:X\to \mathbb{R}$ defined by $f(x)=0\mathrm{\forall }x\in U$ and $f(x)=1\mathrm{\forall }x\in V$. This is a continuous function (in fact, all its fibers are open).

Pick $x_1\in U$ and $x_2\in V$. By our construction, $f(x_1)<f(x_2)$, so by the given data, $f(X)$ should contain the interval $[f(x_1),f(x_2)]=[0,1]$. But this contradicts the fact that the image of $f$ is the two-element set $\{0,1\}$.

Thus, our original assumption that $X$ is not connected cannot hold. Hence, $X$ must be connected.