# Prove that if a set $A \subset \mathbb R^n$ is connected, then it has the Intermediate Value Property.

Prove that if a set $A \subset \mathbb R^n$ is connected, then it has the Intermediate Value Property.

Intermediate Value Property: let $f$ be a real-valued continuous function on a domain $A$ if $a,b \in A$ and $f(a)<c<f(b)$ , then $c \in f(A)$

I know that $A \subset \mathbb R^n$ is connected meaning there exist no disjoint, non empty, open sets $U$ and $V$ such that $A=U \cup V$. This means the function $f:A \to \mathbb R$ should be continuous on $A$. therefore, $A$ should have the Intermediate Value Property. but I don't know hwo to prove it formally.

• What is the "intermediate value property"? – Batman Mar 7 '14 at 15:35
• The idea is that the image of a connected set under a continuous map is connected. – Carsten S Mar 7 '14 at 15:37
• Your definition of the Intermediate Value Property is inconsistent with your question: a connected set is not necessarily polygonally connected. Do you mean path-connected? That would make more sense. – TonyK Mar 7 '14 at 16:19
• You say in the last paragraph that $A$ is connected, which means that $f$ is continuous, therefore $A$ has the intermediate value property. None of that makes sense as written. – Andrés E. Caicedo Mar 7 '14 at 16:27

## 2 Answers

Note $f^{-1} (U)$ is open if $U$ is open in $\mathbb{R}$ because $f$ is a continuous function. If your set $A$ doesn't have a point $M$ for which $f(M)=0$. Let $N$ be the set of negative reals and $P$ the set of positive reals. Then the union of the opens $f^{-1} (N)\cap A \neq \emptyset$ and $f^{-1} (P)\cap A\neq \emptyset$ is $A$. So that means $A$ is disconnected. Which is a contradiction.

• you are using a contrapositive proof, not a proof by contradiction, but I understand this proof very well, thanks. – Diane Vanderwaif Mar 7 '14 at 16:40
• Ah, yes of course. – Kaladin Mar 7 '14 at 18:19

Perhaps it can help

A set $E \in \mathbb{R}$ is connected if and only if it has the following property: If $x \in E$, $y \in E$, and $x<z<y$, then $z \in E$.

• yes, I know this theorem, I can prove the converse but not the forward direction. – Diane Vanderwaif Mar 7 '14 at 15:42
• I think that you can proof by absurd. Instead of using 0, try to use $x=f(a)$ and $y=f(b)$. Suppose that it does not hold the intermediate value theorem. If it is connected, there is $f(a)<f(c)<f(b)$ and $f(c) \in E$. If $f()$ is continuous you can find an intermediate value. – Erivelton Geraldo Nepomuceno Mar 7 '14 at 15:53
• Can you explain it a little bit further? You suggest me to prove this by proof of contradiction, so I assume that $A$ is connected, $f$ is continuous on $A$, $f(a)<C<f(b)$ and $C$ not in $f(A)$. I still can't see how this help me. – Diane Vanderwaif Mar 7 '14 at 16:20
• I think in your affirmation you should include "a continuous function has the IVP" or affirm that $f(A) \in B$ which is also connected. Can you add one of these two affirmation? I mean, I think that it is possible to create a function over a connected domain but which its a non connected image. – Erivelton Geraldo Nepomuceno Mar 7 '14 at 16:30