1

Let $M = [0,1]^{[0,1]}$ Prove that the set of increasing functions $$ J := \{f \in M : \forall \space a,b \in [0,1], a \leq b : f(b) − f(a) \geq 0 \} $$ is a $d$-closed subset of $M$ where $d = d_∞ \colon M \times M \to \mathbb{R}^+_0$ is given by $$ d(f,g) = \sup \{|f(x)-g(x)| : x \in [0,1]\}. $$

I have already shown that the map $\phi_{a,b} \colon M \to \mathbb{R}$ where $\phi_{a,b}(f) = f(b) − f(a)$ for $a,b \in [0,1]$ is continuous, as it apparently can help with the proof.

We know $J \subseteq M$.

So we need to show that $f \in M : \forall \epsilon > 0 \space \exists x \in B_{\epsilon}(f) : \space x \notin J \implies f \in J$.

Is that correct? The Ball is defined using the metric $d$.

I can't seem to go further than that. Any ideas?

Jendrik Stelzner
  • 16,401
Gregory Peck
  • 1,717

1 Answers1

1

You are nearly finshed: If you have a sequence $(f_n)_{n \in \mathbb{N}}$ on $J$ and $f \in M$ with $d(f,f_n) \to 0$, then for all $a,b \in [0,1]$ with $a \leq b$ you have $\phi_{a,b}(f_n) \geq 0$ for all $n \in \mathbb{N}$, and thus by the continuity of $\phi_{ab}$ also $\phi(f) \geq 0$. So $f$ is increasing as well.

Alternatively notice that $$ J = \bigcap_{0 \leq a \leq b \leq 1} \phi_{ab}^{-1}([0,\infty)) $$ is by the continuity of the $\phi_{ab}$ an intersection of closed sets and thus closed.

Jendrik Stelzner
  • 16,401