1
$\begingroup$

Using the fact that the only type of discontinuity compatible with monotonic function is the jump discontinuity, I showed that $A:=\{a\in[0,1]:f\textrm{ is discontinuous at }a\}$ is enumerable for every $f\in X$.

But that didn't help, once for a fixed $f\in X$, $\pi^{-1}[]f(x)-\epsilon,f(x)+\epsilon[]$ is an open nhood of $f$ for every $x\in[0,1]$. I cannot see how a enumerable family of open nhoods of $f$ can be a local basis in this case.

$\endgroup$
1
$\begingroup$

Let $f\in X$ be an arbitrary function, $S=\{a_n\}$ be an enumerated countable set containing both the set $A$ of the discontinuity points of the function $f$ and a dense set $D\supset\{0,1\}$ of the segment $[0,1]$ (for instance, its rational points). We claim that the family $$U_n=\{g:\in X: |g(a_i)-f(a_i)|<1/n\mbox{ for each } 1\le i\le n \}$$ is a countable base at $f$. It suffices to show that for any point $x\in [0,1]$ and any natural number $m$ the standard subbase neighborhood $V_x=\{g:\in X: |g(x)-f(x)|<1/m\}$ contains a set $U_n$ for some $n$. If $x$ is a discontinuity point of the function $f$ then the claim follows from the inclusion $S\ni x$. Conversely, there exists a number $\delta>0$ such that $|f(x’)-f(x)|<1/(5m)$ provided $|x’-x|<\delta$. Since the set $S$ is dense in $[0,1]$, there exist numbers $p$ and $q$ such that $$x-\delta<a_p\le x\le a_q<x+\delta.$$ Assume that $n>\max\{p,q, 5m\}$ and $g\in U_n$. Then

$$|g(x)-f(x)|\le |g(x)-g(a_q)|+|g(a_q)-f(a_q)|+ |f(a_q)-f(x)| \le$$ $$|g(a_p)-g(a_q)|+|g(a_q)-f(a_q)|+ |f(a_q)-f(x)| \le$$ $$|g(a_p)-f(a_p)|+ |f(a_p)-f(a_q)|+ |f(a_q)-g(a_q)|+|g(a_q)-f(a_q)|+ |f(a_q)-f(x)|<$$ $$\frac 1{5m}+\frac 1{5m}+\frac 1{5m}+\frac 1{5m}+\frac 1{5m}=\frac 1m.$$

Thus $g\in V_x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.