# Metric space where each continuous function has IVP is connected

The question:

Let $X$ be a space such that every continuous function $f:X\rightarrow\mathbb{R}$ does have the following property:

if $a<c<b$, $f(x) =a$, and $f(y) =b$, then there exists $z\in X$ such that $f(z) = c$. Prove that X is connected.

All continuous functions have the IVP which means that they (the continuous functions) do not map to a two point discrete space which implies connectedness.

Is this correct?

• "they do not map" ? What are "they"? Jun 21, 2016 at 6:31
• the continuous functions Jun 21, 2016 at 6:31
• Gary, are you there? Sep 28, 2019 at 5:22

I think what's missing in your proof is showing that, if $X$ is not connected, there is some continuous function that maps $X$ to a two point discrete space (hint: what can you say of the connected components of $X$?).

• I'm using the definition "X is connected if any continuous function from X to {0,1} is constant". Since the IVP property doesn't it follow immediate that any such function would be constant?! Jun 21, 2016 at 7:06
• With that definition and appropriate wording, I think your proof is fine. Jun 21, 2016 at 7:11

You are on the right track, but perhaps you should explain how your answer relates to functions from $X$ to the reals.

If $X=A\cup B$ where $A, B$ are each open and non-empty with $A\cap B=\emptyset$ , let $f(x)=0$ for $x\in A$ and $f(x)=1$ for $x\in B.$

If $U$ is an open subset of the reals then $f^{-1}U$ is one of $\emptyset, A,B,X$, so $f^{-1}U$ is open in $X,$ so $f$ is continuous.

Since $A\ne \emptyset \ne B$, there exists $x\in A$ and $y\in B,$ that is, $f(x)=0$ and $f(y)=1.$ Obviously no $z\in X$ satisfies $f(z)=1/2.$

So your condition on the set of continuous $f : X\to R$ implies that $X$ must be connected.... Note that the converse also holds, because jf $X$ is connected, then any continuous image of $X$ is connected.

Your conjecture is FALSE even in a much stronger version. I'll let you think about it but it's not completely trivial.

It's possible that space $$\ X\$$ which has more than one point has no non-trivial connected components (i.e. only 1-point components), and still your Darboux kind of property may hold.