Prove that a boundary point of a set $S \subset \mathbb{R}$ is either a limit point of $S$ or an isolated point of $S$.
(1st Draft) Proof: Let $x \in \partial S$. Then by definition we have $N(x, \varepsilon) \cap{S} \neq \emptyset$ and $N(x, \varepsilon) \cap{S}^c \neq \emptyset$, $\forall$ $\varepsilon > 0$. If $N(x, \varepsilon) \setminus \{x\} \cap{S} \neq \emptyset$ then by definition $x$ is a limit point of $S$. Otherwise we must have $N(x, \varepsilon) \setminus{ \{x\} } \cap{S} = \emptyset$ which gives us that $x$ is an isolated point of $S$ by definition. Since these are the only two possibilities, conclude that $x$ is either a limit point or a boundary point of $S$. $\Box$
(2nd Draft) Proof: Let $x \in \partial S$. Then by definition we have $N(x, \varepsilon) \cap{S} \neq \emptyset$ and $N(x, \varepsilon) \cap{S}^c \neq \emptyset$, $\forall$ $\varepsilon > 0$. If for any $\varepsilon > 0$ we have $N(x, \varepsilon) \setminus{\{x\}} \cap{S} \neq \emptyset$ then $x$ is by definition a limit point of $S$. However, if for some $\varepsilon > 0$ we have $N(x, \varepsilon) \setminus{\{x\}} \cap{S} = \emptyset$ then $x$ is by definition an isolated point of $S$. Hence for any $\varepsilon > 0$ we have that $x$ is either a limit point or an isolated point of $S$. $\Box$
Is this proof sufficient to show the proposition is true?
Best Answer
The argumet needs a modification. You are fixing an $\epsilon$ so $N(x,\epsilon)\setminus \{x\} \cap S \neq \emptyset$ does not tell you that $x$ is a limit point. To correct this problem consider the following possibilities:
a) $N(x,\epsilon)\setminus \{x\} \cap S =\emptyset$ for some $\epsilon >0$
b) $N(x,\epsilon)\setminus \{x\} \cap S \neq \emptyset$ for any $\epsilon >0$