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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.

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    $\begingroup$ No. Let $X = U \cup V$, where $U,V$ are disjoint nonempty open sets. Let $f\colon X \to \mathbb{R}$ be given by $f(x) = 1$ for $x\in U$ and $f(x) = 0$ for $x\in V$. $\endgroup$ – Daniel Fischer Aug 7 '14 at 16:53
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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.

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  • $\begingroup$ How to prove this characterisation of disconnected space. Rest everything is fairly good. $\endgroup$ – Sushil Aug 7 '14 at 17:23
  • $\begingroup$ Pretty much the same argument as in the comment, I'll add the proof. $\endgroup$ – Daniel Fischer Aug 7 '14 at 17:27
  • $\begingroup$ Hence in short: A connected topological space is equivalent to topological space with Intermediate Value Property. Am I right? $\endgroup$ – Sushil Aug 7 '14 at 17:41
  • $\begingroup$ In short: Yes. Exactly. $\endgroup$ – Daniel Fischer Aug 7 '14 at 17:47
  • $\begingroup$ Ok thanks that solves the problem completely $\endgroup$ – Sushil Aug 7 '14 at 17:50

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