Prove: A topological space $X$ has the intermediate value property if and only if it is connected.
$\impliedby$: We will prove the negation (no intermediate value property $\implies$ not connected):
$X$ does not have the intermediate value property so there exists $f:X\to \mathbb{R}$ continuous and $a,b\in X$, and $t\in \mathbb{R}$ such that $f(a)<t<f(b)$ for every $x\in X$ such that $f(x)\neq t$.
We define $U=f^{-1}((t,\infty)),V=f^{-1}((-\infty,t))$ so $V,U$ are open as $f$ is continuous (which means that the preimage of an open image is an open image), non empty (as $a\in V$, $b\in U$), disjoint ($(t,\infty),(-\infty,t)$) and their union is all $X$ (we assumed that for all $x\in X$, $f(x)\neq t$).
So $X$ is not connected .
$\implies:$ We will prove the negation (not connected $\implies$ no intermediate value property):
$X$ is not connected so there are $U,V$ open, non empty, disjoint, such that $U \cup V = X$, and we will define $f:X\to \mathbb{R}$ as
$$
f(x)=
\begin{cases}
0 & \text{if $x \in U$}, \\
1 & \text{if $x \in V$}.
\end{cases}
$$
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Why is this function continuous?
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It does not have the intermediate value property as there is no $x\in X$ such that $0 <f(x)< 1$?
Best Answer
The function is continuous because the preimage of every open set is open. In particular can be either empty, either $X$, either $U$ or $V$.
Moreover, $f$ does not have the intermediate value property, as you said, thus you proved the result.