A perfect set results by removing the intervals in such a way as to create no isolated points. Here is a statement in Zygmund's Measure and Integral that confuses me on Page 8:

Any closed set in $\mathbb{R}^1$ can be constructed by deleting a countable number of open disjoint intervals from $\mathbb{R}^1$. A perfect set results by removing the intervals in such a way as to create no isolated points; thus, we would not remove any two open intervals with a common endpoint.
Definition A perfect set $C$ is a closed set each of whose points is a limit point of $C$.

I don't understand - why removing the intervals results a prefect set?
 A: I'll try to show how, by starting with $\mathbb{R}$ and removing only non-overlapping open intervals, no two of which share an end point, the resulting set is a perfect set.
First of all, an easier definition of a perfect set to use (in this case) is a subset $A$ of a metric space $X$ which has no isolated points. An isolated point of $A$ is a point $x\in A\subset X$ such that there exists an open set $U\subset X$ such that $U\cap A=\{x\}$. You may like to try and prove that this is an equivalent definition yourself.
Now, suppose that the set $A$ constructed as described above had an isolated point $x$ (and so was not a perfect subset of $\mathbb{R}$). Let $U$ be such that $U\cap A=\{x\}$ and let $B_{\epsilon}(x)$ be an open interval centered at $x$ with radius $\epsilon$ where $\epsilon$ is small enough so that $B_{\epsilon}(x)\subset U$.
Now, as $B_{\epsilon}(x)\setminus\{x\}$ has empty intersection with $A$ and is an open set (equal to the union of two disjoint open intervals $(x-\epsilon,x)\cup (x,x+\epsilon)$), and we have not removed any overlapping open intervals from $\mathbb{R}$, at some point an open interval $(x-\alpha,x)$ was removed with $\alpha\geq\epsilon$ and also $(x,x+\beta)$ with $\beta\geq\epsilon$. However, this contradicts the condition that we removed open interval no two of which shared an end point. We conclude that no such isolated point can exist.
A: I'll just give a few of the definitions of related terms. Let $S$ be a subset of a topological space $X$.
Let $x\in X$.
Then:


*

*$x$ is in the closure of $S$ iff every open set containing $x$ contains an element of $S$.

*$x$ is a limit point of $S$ iff every open set containing $x$ contains an element of $S\setminus \{x\}$. That is, iff $x$ is in the closure of $S\setminus \{x\}$

*$x$ is an interior point of $S$ iff there is an open set $U$ such that $x\in U\subseteq S$. That is, iff $x$ is not in the closure of $X\setminus S$.

*$x$ is an isolated point of $S$ iff for some open set $U$, $U\cap S = \{x\}$.
