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 $f(x)=0\ \forall \ x\in U$ and $f(x)=1\ \forall \ x\in V$. 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.