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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 $E$?

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2 Answers 2

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.

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  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..

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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. –  Andrew Salmon Jun 12 '13 at 21:28

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