Question

Do you know if the folowing statement is an equivalent form of AC or not ??

• *If $X$ is a compact metric space then every continuous function $f: X \longrightarrow \mathbb{R} $ is uniformly continuous *

If you know any references please let me know.

Answer 1

It seems to me that this is provable without using the axiom of choice.

Suppose that $X$ is a compact metric space and $f:X\to\mathbb{R}$ is continuous. Let's show it is uniformly continuous. Fix any $\epsilon\gt 0$. For each point $x\in X$, there is a small ball $B$ centered at $x$ such that $f(y)$ is within $\epsilon/2$ of $f(x)$ for all $y\in B$, and we may choose $B$ to have radius $1/{n_x}$, where we choose $n_x$ to the smallest positive natural number for which this radius has the desired property. Thus, we have a canonical choice of radius here, and so we do not need the axiom of choice to define the map $x\mapsto n_x$.

Consider now the family $\{ B_{1/{2n_x}}(x) \mid x\in X \}$, consisting of the inner core of each of those balls, with the radius of each of them shrunk to half. This is an open cover of $X$, and so by compactness there is a finite subcover, which consists of finitely many balls, having some minimal radius $1/{2n}$. Now, if $y$ and $z$ are within $1/{2n}$ of each other, then they are both within $1/{n_x}$ of the center $x$ of the ball in which $y$ sits in the subcover, and so $f(y)$ and $f(z)$ are within $\epsilon/2$ of $f(x)$, and hence within $\epsilon$ of each other, showing that $f$ is uniformly continuous.

Answer 2

Let us improve slightly on Joel's answer by avoiding not only choice but also excluded middle (which is used in assuming that the minima $n_x$ exist). In passing we also generalize to an arbitrary metric codomain. Since the various notions of compactness are not equivalent intuitionistically, we have to specify which one we mean. We mean by "compact" the Heine-Borel finite subcover property.

Theorem: If a map $f : X \to Y$ from a compact metric space to a metric space is continuous then it is uniformly continuous.

Proof. (No excluded middle, no choice.) Let $\epsilon > 0$ be given. Consider the family of open balls $$\mathcal{F} = \lbrace B(x,r) \mid x \in X, r > 0, \forall x', x'' \in B(x,2r) . d(f(x'), f(x'')) < \epsilon \rbrace.$$ Beware, we put $B(x,r)$ in $\mathcal{F}$ if the larger ball $B(x,2 r)$ is mapped by $f$ to a sufficiently small set.

Because $f$ is continuous, $\mathcal{F}$ covers $X$. By the Heine-Borel property it has a finite subcover $$B(x_1, r_1), \ldots, B(x_n, r_n).$$ Let $\delta = \min (r_1, \ldots, r_n)$. Suppose $d(y,z) < \delta$ for some $y, z \in X$. There is $i$ such that $d(x_i, y) < r_i$, hence $d(x_i, z) \leq d(x_i, y) + d(y, z) < r_i + \delta \leq 2 r_i$. Thus, since both $y$ and $z$ are contained in $B(x_i, 2 r_i)$ we conclude $d(f(y), f(z)) < \epsilon$. QED.

As usual, the constructive proof is also the most elegant one. The above proof is an easy adaptation that avoids unecessary use of choice of 4.3.31 and 4.3.32 of Engelking's famous General Topology. Further reading: Hajime Ishihara and Peter Schuster, Compactness under constructive scrutiny. Math. Log. Quart. 50, No. 6, 540 – 550 (2004).

Answer 3

Look Ma, no axiom of choice!

THEOREM 0   Let   $X$   be a compact space. Let   $\Phi$   be a non-empty family of closed subsets of   $X$,   $F := \bigcap \Phi$,   and   $G\supseteq F$   an open subset of   $X$.   Then there exists a finite   $\Phi_0\subseteq \Phi$   such that   $\bigcap\Phi_0\subseteq G$.

PROOF   Family   $\Gamma\ :=\ G\cup\{X\setminus A : A \in \Phi\}$   is an open covering of   $X$. Etc.   END of PROOF

As an instant corollary we get:

THEOREM 1   Let   $X$   be a compact space. Let   $\Phi$   be a non-empty family of closed subsets of   $X$,   $F := \bigcap \Phi$,   and   $G\supseteq F$   an open subset of   $X$.   Assume also that   $\Phi$   is linearly ordered by   $\subseteq$.   Then there exists   $A\in \Phi$   such that   $A \subseteq G$.

THEOREM 2   Let   $(X\ d)$   be a compact metric space. Let   $W$   be an open subset of   $X^2$,   such that   $\Delta_X\subset W$,   where   $\Delta_X := \{(x\ x):x\in X\}$.   Then there exists   $\delta > 0$   such that   $V_X(\delta)\subseteq W$,   where   $V_X(\delta) := \{(x\ y)\in X^2 : d(x\ y)\le\delta\}$.

PROOF   Apply Theorem 1 to   $X^2$ as a replacement of   $X$   of Theorem 1; etc.   END of PROOF

THEOREM 3   Let   $f : X\rightarrow Y$   be an arbitrary continuous function of a metric compact space   $(X\ d_X)$   into an arbitrary metric space   $(Y\ d_Y)$.   Then function   $f$   is uniformly continuous.

PROOF   Let   $\epsilon > 0$.   Let   $W\subseteq X^2$   be the inverse image of   $V_Y(\epsilon)$   under function   $f\times f$.   There exists, by Theorem 2,   $\delta > 0$   such that   $V_X(\delta)\subseteq W$.   Then   $d_Y(f(x')\ f(x''))\le\epsilon$   for every   $x'\ x''\in X$   such that   $d_X(x'\ x'')\le\delta$.   END of PROOF