# Axiom of Choice, Continuity and Intermediate Value Theorem

I am trying to understand a proof I read in Herrlich's book Axiom of Choice. For those who know the book, it is theorem 4.54 on page 74. The part I am interested in reads:

(9) A function $f:X \rightarrow \mathbb{R}$, defined on some subspace $X$ of $\mathbb{R}$, is continuous iff it is sequentially continuous. $\Rightarrow$ (10) CC($\mathbb{R}$).

CC($\mathbb{R}$) being the axiom of countable choice for the reals.

The proof goes as follows (I quote):

By Theorem 3.8 it suffices to show that every unbounded subset $A$ of $\mathbb{R}$ contains an unbounded sequence. Let $h: \mathbb{R} \rightarrow (0,1)$ be a homeomorphism. Without loss of generality, 0 is an accumulation point of h[A]. Define $X = h[A] \cup \{0\}$ and $f: X \rightarrow \mathbb{R}$ by $f(x) = \begin{cases} 0, & \text{ if } x \in h[A] \\ 1, & \text{ if } x = 0\\ \end{cases}$. Then $f$ is not continuous, thus, by (9), not sequentially continuous. Thus there exists a sequence $(b_n)$ in $h[A]$ that converges to 0. Consequently $(h^{-1}(b_n))$ is an unbounded sequence in A. \

For the proof to work out $h[A]$ must have an accumulation point in 1 or in 0. Now I tried to figure out why this must be the case. The person I spoke to about it said, that, since $h$ is a homeomorphism, $h$ must be an order isomorphism. If that is the case everything is fine, because then, because of the monotony, any unbounded set must have 0 or 1 as an accumulation point.

Now the part I am worrying about is the proof that if $h$ is a homeomorphism, then it is an order isomorphism. In prooving that $h$ is an order isomorphism one point that needs to be proven, is that $h$ is monotone. This can be done easily by using the continuity and bijectivity of $h$, and the intermediate value theorem.

In standard mathematics this is fine of course, but in standard mathematics we have the axiom of countable choice, which we are not supposed to use here. The point I am worried about is the use of the intermediate value theorem. I am not sure if one needs the axiom of countable choice in order to proove it. In the proof I know, it is not used directly. What aroused my suspicion though, is that the intermediate value theorem cannot be proven in constructive mathematics. Of course there are many other principles apart from the axiom of choice that contructive mathematics reject, but it is still puzzling.

Now I have 3 questions that are kind of related:

1. Does someone know another, easier way of showing that $h[A]$ has either the accumulation point 0 or 1?

2. Does someone know if the intermediate value theorem is based in some way on the axiom of choice?

3. Is there any mistake in the reasoning above?

I am of course willing to specify anything I mentioned, it just seemed to me that the post was getting very long. Thanks for the reading!

• 2. I think that the intermediate value theorem holds without axiom of choice. It follows from the fact that continuous image of an (generalized) interval is an interval. That follows from the fact that continuous image of connected space is connected and that connected subsets of $\mathbb{R}$ are exactly generalized intervals. – user87690 Jul 29 '13 at 17:48

You don’t need any part of choice for the intermediate value theorem. Let $I\subseteq\Bbb R$, and let $f:I\to\Bbb R$ be continuous. Suppose that $f[I]$ is not order-convex. Then there is an $\alpha\in\Bbb R\setminus f[I]$ such that $(\leftarrow,\alpha)\cap f[I]\ne\varnothing\ne(\alpha,\to)\cap f[I]$. It follows that
$$g:f[I]\to\Bbb R:x\mapsto\begin{cases} 0,&\text{if }x<\alpha\\ 1,&\text{if }x>\alpha \end{cases}$$
is continuous, $g\circ f:I\to\{0,1\}$ is a continuous surjection, and $I$ is not connected and hence not order-convex. (Note that by Theorem $4.52(4)$ of Herrlich’s book, the connected subsets of $\Bbb R$ are precisely the order-convex subsets even in $\mathsf{ZF}$.) Thus, if $I$ is order-convex, then so is $f[I]$, and that’s essentially the intermediate value theorem.
I doubt that there’s a much simpler proof that $h[A]$ must accumulate at either $0$ or $1$.
• @user80703: I don't think that it actually simplifies the proof significantly: if $h$ is a homeomorphism, then either $h$ or $1-h$ is an order-isomorphism anyway. – Brian M. Scott Aug 12 '13 at 12:37