Let's say we are given a graph. How do we determine from the graph that the function has intermediate value property? And also, why do we need to impose the condition 'monotone' so that a monotone function with intermediate value property is continuous ? Intuitively, I don see why the statement 'general function has intermediate value property is continous' is false.

  • $\begingroup$ By graph, do you mean the graph of a function? $\endgroup$ – copper.hat Mar 27 '13 at 5:55
  • $\begingroup$ @copper.hat: yeah $\endgroup$ – Idonknow Mar 27 '13 at 5:56
  • $\begingroup$ See also here. $\endgroup$ – Andrés E. Caicedo Jan 25 '14 at 7:45

Try $f(x) = \begin{cases} \sin \frac{1}{x} & x \neq 0 \\ 0 & x = 0\end{cases}$. This has the intermediate value property, but is not continuous at zero. Also, it is not monotone, so it serves as an example of why monotonicity is required.

Note that the graph is connected.

Note that if the graph is connected, it has the intermediate value property.

To see why, suppose a function $g$ does not have the intermediate value property. Then there exists $a,b, \gamma$ such that $a<b$ and $\gamma$ is in the smallest interval containing $g(a), g(b)$, but $\gamma \notin g([a,b])$. Without loss of generality, we may assume $g(a) < \gamma < g(b)$. Then let $U = \{ (x,y) | x < a \} \cup \{ (x,y) | x < b, \ y < \gamma \} $, and $V=\{ (x,y) | x >b \} \cup \{ (x,y) | x > a, \ y > \gamma \} $. Then both $U $ & $V$ are open & non-empty, and the graph of $g$ is contained in $U \cup V$. Hence the graph is disconnected.

Furthermore, if a function $g$ has the intermediate value property, the graph is connected.

To see this, suppose $g$ has the intermediate value property, but the graph is not connected. Then there are two non-empty, disjoint open sets $U,V$ such that the graph of $g$ is contained in $U \cup V$. We can assume without loss of generality that we have some $a < b$ such that $(a,g(a)) \in U$, and $(b,g(b)) \in V$. Let $b' = \sup \{x \ge a | (t,g(t)) \in U \ \forall t \in [a,x] \}$. Since $U$ is open, we must have $(b',g(b')) \in V$. Since $V$ is open, we have $B((b',g(b')),\epsilon) \subset V$ for some $\epsilon>0$. For convenience, we use the $\max$-norm on $\mathbb{R}^2$, hence we have $(b'-\epsilon,b'+\epsilon) \times (g(b')-\epsilon,g(b')+\epsilon) \subset V$.

Choose $a' = b'-\frac{\epsilon}{2}$, $\gamma = g(b')-\frac{\epsilon}{2}$. Then if $t \in [a',b')$, we have $(t,g(t)) \in U$, hence $|g(t) -g(b')| \ge \epsilon$, hence $g(s) \neq \gamma$ for all $s \in [a',b']$, which contradicts $g$ having the intermediate value property. Hence the graph is connected.

  • $\begingroup$ can you explain what is the function ? i don understand what is $1_{\lbrace 0 \rbrace}$ $\endgroup$ – Idonknow Mar 27 '13 at 6:06
  • $\begingroup$ I expanded the definition a little. $\endgroup$ – copper.hat Mar 27 '13 at 6:08
  • $\begingroup$ How do we know the function has intermediate value property? $\endgroup$ – Idonknow Mar 27 '13 at 6:11
  • $\begingroup$ Since $f$ is continuous for $x \neq 0$, it is clear that for any interval not containing $0$, $f$ has the IVP. On any interval (of non-zero length) containing $0$, $f$ takes all values in $[-1,1]$. Hence it has the IVP everywhere. $\endgroup$ – copper.hat Mar 27 '13 at 6:14
  • $\begingroup$ $f$ has IVP iff the domain of $f$ is bounded? $\endgroup$ – Idonknow Mar 27 '13 at 6:22

The famous example involves the derivative of the function $f$ defined by $f(x)=x^2\sin(1/x)$ if $x\ne 0$ and $f(0)=0$.

It is not difficult to show that $f$ is differentiable everywhere, but $f'(x)$ is not continuous at $0$. And, like any derivative, $f'(x)$ has the Intermediate Value Property.

Remark: The question about intuition is difficult to address. One problem is that continuous functions can be a lot weirder than any "can be drawn without lifting the pencil" function of our imagination. The compromise is an intuitive notion of smoothness, tempered by knowledge of standard counterexamples.


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