# Equivalent definitions of limit and Discontinuities of monotonic functions

I have two questions.

1. PMA, Rudin p.97 proves the existence of a monotonic real function $f$ defined on $(a,b)$ such that $f$ is continuous on $E$- a countable subset of $(a,b)$ and discontinuous on $(a,b) \setminus E$.

Then, it states that "It should be noted that the discontinuities of a monotonic function need not be isolated".

I don't understand how come the existence of above function implies that. Isn't $f\upharpoonright ((a,b)\setminus E)$ continuous on its domain$? 1. PMA, Rudin p.98 states an equivalent definition of limit defined by$\epsilon-\delta$. That is, when$f:E\rightarrow X$is a function from a metric space to another,$\lim_{t\to x} f(t) = A$iff [For every neighborhood$U$of$A$, there exists a neighborhood$V$of$x$such that$V\bigcap E≠\emptyset$and$t\in (V\bigcap E)\setminus \{x\} \Rightarrow f(t)\in U$. This is exactly the same as the definition by$\epsilon-\delta$. However, this should be equivalent to$\lim_{t\to x} f(t) = A$only if$x$is a limit point of$E$. Does the definition above has an information that$x$is a limit point of$E$? ## 2 Answers 1. As you cited,$f$is discontinuous on$(a,b)\setminus E$, so in this case, there are only countable continuity points. So, the quoted corollary is justified. 2. No, that$x$must be a limit point of$E$is not included, but perhaps should be. For example consider a constant$A$function, and$x$far away from$E$(say,$x$has a neighborhood$G$disjoint to$E$). Then for every neighborhood$U$we can consider$V:=G\cup E$. However, when one actually uses the definition, usually all the points in question are limit points of the given domain.. • The poster of the question has misread Rudin. A monotonic function has at most countably many points of discotninuity but$( a,b ) \setminus E$must be uncountable. – A.S Jun 12 '13 at 21:28 I'll try and answer the second question first as there seems to be something wrong with the first which I will address later. The definition you mention clearly does incorporate the condition that$ x $is a limit point, When you say that for every neighbourhood$U$of$A$,there exists a neighbourhood of$x$, i.e.$V$, such that$V\cap E$is non empty, this is almost as good as saying that$x$is a limit point of$E$. Its not explicit as it doesn't consider a deleted neighbourhood, but then again continuity is automatic for isolated points too which is allowed for here.The definition merely says that including$ \infty \$ as a limit is valid and nothing else.

As for the 1st question , from what I see, Rudin proves that a monotonic function has at most countable number of discontinuities. Then he gives an example of a function to show that these discontinuities need not be isolated. They can even be a set which is dense in the domain. I hope that helps.