Disconnected Topoological Space with Intermediate Value Property Does There exist a disconnected topological space with intermediate value property? Intermediate Value Property states that 'a topological space X is said to have intermediate value property if for every continuous function f: X to Y (where Y is ordered set with order topology) the following is true: If a, b belongs to X and there exist r in Y s.t. r lies between f(a) and f(b) then there exist c in X s.t. f(c) = r.
 A: One characterisation of disconnected spaces is
A topological space $X$ is disconnected if and only if there is a surjective continuous function $f\colon X \to Y$, where $Y = \{0,1\}$ is endowed with the discrete topology.
Proof: Let first $X$ be disconnected. Then by definition there are two disjoint nonempty open sets $U,V\subset X$ with $X = U \cup V$. Then the function
$$f(x) = \begin{cases}1 &, x \in U\\ 0 &, x \in V \end{cases}$$
is surjective, because $U \neq \varnothing \neq V$, and it is continuous because $f^{-1}(B)$ is one of the four open sets $\varnothing, U, V, X$ for any $B\subset Y$.
Conversely, if there is a surjective continuous $f\colon X \to Y$, then $U = f^{-1}(\{1\})$ and $V = f^{-1}(\{0\})$ are two disjoint open sets with $X = f^{-1}(\{0,1\}) = f^{-1}(\{0\}\cup\{1\}) = f^{-1}(\{0\}) \cup f^{-1}(\{1\}) = V\cup U$, so $X$ is disconnected.
Now any map $g$ from $Y$ to an ordered set with the order topology is continuous, and there certainly are such maps where there is a value between $g(0)$ and $g(1)$ (we can for example take $g\colon Y\to \mathbb{R}; 0 \mapsto 0,\, 1 \mapsto 1$). Then $g\circ f$ is a continuous mapping into an ordered set with the order topology that does not have the intermediate value property.
