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Let be M a metric space with the property of Intermediate Value. Show that M is connected.

property of Intermediate Value: "all continuous function $f: M \longrightarrow \mathbb{R}$ that admit a positive value and a negative value, also admit a $c \in M$ such that $f(c) = 0$".

I think that I need to use the fact that open balls are connected (take an open ball with center $f(c)$ in $\mathbb{R}$ in this case) and suppose by absurd that $M$ is disconnected, but I dont know how use this exactly and if this right way to prove that $M$ is disconnected. Can anyone give me a hint on how to solve this problem? Thanks in advance!

George
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2 Answers2

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Hint: Consider any continuous function $f:M\to\{\pm 1\}$. Is $f$ constant? If not then what would happen? Can you use Intermediate Value Property of $M$ here?

I assume that you know the following theorem,

A metric space is $X$ connected iff for any continuous function $f:X\to \{\pm 1\}$, $f$ is constant.

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First, I believe the property you mentioned should hold for all continuous functions, not all functions in general. To answer your question, assume $M$ is not connected, and $M=U \cup V$ is a decomposition such that $V,U$ are open and disjoint. Define $f: M \rightarrow \mathbb{R}$ by $f(x)=-1$ for $x \in V$ and $f(y)=1$ for $y \in U$. This function is continuous (prove it!), but the property does not hold.

Orpheus
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