We often deal with sets in $\mathbb R$ such as solution sets of equations which are not simply open or closed intervals. To analyze the properties of the points in a subset of $\mathbb R$, we introduce some geometric concepts developed by Georg Cantor in late nineteenth century. We view real numbers as points on the line represented by $\mathbb R$.
An open neighborhood of a point $x$ in $\mathbb R$ is an open interval $(x-\varepsilon,x+\varepsilon)$ for some $\varepsilon>0$. Note $(x-\varepsilon,x+\varepsilon)$ is called an $\varepsilon$-neighborhood of $x$, denoted by $N_{\varepsilon}(x)$ or $B_{\varepsilon}(x)$.
In all the definitions in this section, $S\subseteq \mathbb R$. Some elements of $S$ are "deeply inside" $S$ called the interior points and defined as follows:
Definition (Interior point).
A point $x$ of $S$ is an interior point of $S$ if there is an open neighborhood of $x$ contained in $S$, i.e.,
$$\;\exists\; \varepsilon>0 \;s.t.\; (x-\varepsilon,x+\varepsilon)\subseteq S.$$
The set of all interior points of $S$ is the interior of $S$, denoted by $\operatorname{int}(S)$.
Note by the definition that $\operatorname{int}(S)\subseteq S$.
Example.
- Each point of $(a,b]$ except $b$ is an interior point. So $\text{int} (a,b]=(a,b)$.
First we show that $(a,b) \subseteq \text{int} (a,b]$. Let $x\in (a,b)$. Choose $\varepsilon=\min\{|x-a|,|b-x|\}$. Then $(x-\varepsilon,x+\varepsilon)\subseteq (a,b]$. So $x\in \text{int} (a,b]$. Thus $(a,b) \subseteq \text{int} (a,b]$.
To show $(a,b) = \text{int} (a,b]$, it remains to show that $b\notin \text{int} (a,b]$. Suppose to the contrary that $b\in \text{int} (a,b]$. Then
\[\;\exists\; \varepsilon>0 \;s.t.\; (b-\varepsilon,b+\varepsilon)\subseteq (a,b] \implies b+\varepsilon \leq b,\]
a contradiction. Thus $\text{int} (a,b]=(a,b)$.
- Each point of $\mathbb N$ is not an interior point. So $\operatorname{int}(\mathbb N)=\varnothing$.
Definition (Boundary point).
A point $x$ (not necessarily in $S$) is a boundary point of $S$ if every open neighborhood of $x$ contains one point of $S$ and one point not in $S$, i.e.,
$$\;\forall\; \varepsilon > 0 \;s.t.\; \mathrm{N}_{\varepsilon}(x) \cap S \neq \varnothing \text{ and } \mathrm{N}_{\varepsilon}(x)\cap (\mathbb R\setminus S) \neq \varnothing.$$
The set of all boundary points of $S$ is the boundary of $S$, denoted by $\partial(S)$.
Example.
- $a$ and $b$ are boundary points of $(a,b]$. So $\partial (a,b]=\{a,b\}$.
- Each point of $\mathbb N$ is a boundary point. So $\partial(\mathbb N)=\mathbb N$.
The following are a special kind of boundary points of a set.
Definition (Isolated point).
A point $x$ of $S$ is an isolated point of $S$ if there is an open neighborhood of $x$ which contains only one point of $S$, i.e.,
$$\;\exists\; \varepsilon>0 \;s.t.\; (x-\varepsilon,x+\varepsilon)\cap S =\{x\}.$$
Example.
- No point of $(a,b]$ is an isolated point.
- Each point of $\mathbb N$ is an isolated point. For each $n\in \mathbb N$, $(n-1,n+1)\cap \mathbb N =\{n\}$.
- An isolated point of a subset $S$ of $\mathbb R$ is a boundary point of $S$ (exercise).
The points of $S\subseteq \mathbb R$ that are not isolated points are called the accumulation points or limit points of $S$. But a limit point of $S$ does not necessarily belong to $S$.
Definition (Accumulation or limit point).
A point $x$ (not necessarily in $S$) is a limit point of $S$ if every open neighborhood of $x$ contains at least one point of $S$ other than $x$, i.e.,
$$\;\forall\; \varepsilon>0,\; (x-\varepsilon,x+\varepsilon)\cap
(S\setminus \{x\})\neq \varnothing.$$
The set of all limit points of $S$ is the derived set of $S$, denoted by $S'$.
Example.
- All points of $(a,b]$ and $a$ are limit points of $(a,b]$. So $(a,b]'=[a,b]$.
- No point of $\mathbb N$ is a limit point. So $\mathbb N'=\varnothing$.
- The only limit point of $S=\{\frac{1}{n} \;|\; n\in \mathbb N\}$ is $0$. So $S'=\{0\}$.
Note that $x$ is a limit point of a subset $S$ of $\mathbb R$ if and only if there is a sequence $\{x_n\}$ in $S\setminus \{x\}$ converging to the limit $x$ (Hint. choose $x_n\in N_{1/n}(x)\cap (S\setminus \{x\})$).
Definition (Closure).
The union of $S$ and its derived set $S'$ is the closure of $S$, denoted by $\overline{S}$ or $\operatorname{cl}(S)$. So $\overline{S}=S\cup S'$.
Example.
- Since $(a,b]'=[a,b]$, $\overline{(a,b]}=(a,b]\cup (a,b]'=[a,b]$.
- Since $\mathbb N'=\varnothing$, $\overline{\mathbb N}=\mathbb N\cup \mathbb N'=\mathbb N$.
- For $S=\{\frac{1}{n} \;|\; n\in \mathbb N\}$, $\overline{S}=S\cup S'=\{\frac{1}{n} \;|\; n\in \mathbb N\}\cup\{0\}$.
Last edited by Dr. Mallik
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