# Neighborhood / Open Set / Continuity / Limit Points / Closure / Interior / Exterior / Boundary

Yao Yao on June 28, 2018

Quoted from Analysis II by Victor Guillemin, MIT and Topology for dummies by Damian Giaouris, Newcastle University with modification.

## Neighborhood

Definition: Given a point $x_i \in X$, and a real number $\epsilon > 0$, we define

where $d(\cdot, \cdot)$ is a mtric on $X$ (and thus $(X, d)$ is metric space).

We call $\Phi(x_i, \epsilon)$ the $\epsilon$-neighborhood of $x_i$ in space $X$. Obviously, $x_i \in \Phi(x_i, \epsilon)$.

Given a subspace $Y \subseteq X$ (已知 $\mathbb{R}^n$ 与 $\mathbb{R}^m$ 不存在 subspace 的关系 when $n \neq m$，所以 subspace 不应该有 dimension 上的变化), the $\epsilon$-neighborhood of $x_i$ in space $Y$ is just $\Phi(x_i, \epsilon) \cap Y$.

## Open and Closed Sets

Definition: A set $U$ in space $X$ is open if $\forall x_i \in U, \exists \epsilon_i > 0$ such that $\Phi(x_i, \epsilon_i) \subseteq U$.

• 比如开区间 (open interval) $(a, b)$ 就可以看做一个 open set；闭区间 $[a, b]$ 在 $a, b$ 两点上无法满足 open set 的要求。

Proposition: A neighborhood is not necessarily an open set.

Proof: Given a neighborhood $\Phi(x_i, \epsilon) = \lbrace x_j \in X \vert d(x_i, x_j) < \epsilon \rbrace$ and a point $x_j \in \Phi(x_i, \epsilon)$, $\forall \epsilon’$ we can define $\Phi(x_j, \epsilon’) = \lbrace x_k \in X \vert d(x_j, x_k) < \epsilon’ \rbrace$. However, according to triangle inequality, we only know that $d(x_i, x_k) \leq \epsilon + \epsilon’$. It’s not guaranteed that $d(x_i, x_k) < \epsilon$, i.e. not guaranteed that $x_k \in \Phi(x_i, \epsilon)$, i.e. not guaranteed that $\Phi(x_j, \epsilon’) \subseteq \Phi(x_i, \epsilon)$. $\blacksquare$

Proposition: Every open set is a union of neighborhoods.

Proof: Let $U$ be an open set. For each $x \in U$, let $N_x$ be the open neighborhood in $U$ containing $x$, which is guaranteed to exist by the definition. Consider $\mathcal{N} = \underset{x \in U}{\cup} N_x$. Since for every $x \in U$, $x \in N_x \subset \mathcal{N}$, we find $U \subset \mathcal{N}$. Since every $N_x \subset U$, we find $\mathcal{N} \subset U$. Therefore, $U = \mathcal{N}$ is a union of neighborhoods of points of $U$. $\blacksquare$

Proposition: Let $\lbrace U_i \vert i \in I \rbrace$ be a collection of open sets in space $X$, where $I$ is just a labeling set that can be finite or infinite. Then the set $\underset{i \in I}{\cup} U_i$ is open.

• 这里 labeling set 应该是 index set 的意思，即 a set whose members label or index members of another set. 比如 $\lbrace 1,2,3 \rbrace$ 就是 $\lbrace x_1, x_2, x_3 \rbrace$ 的 index set。
• 可以是 infinite union

Proposition: Let $\lbrace U_i \vert i = 1,\dots, N \rbrace$ be a finite collection of open sets in space $X$. Then the set $\overset{N}{\underset{i=1}{\cap}} U_i$ is open.

• 为什么不能是 infinite intersection? 反例：$\overset{\infty}{\underset{n=1}{\cap}} (-\frac{1}{n}, \frac{1}{n}) = \lbrace 0 \rbrace$
• 注意这里并不是取极限，而是因为 $\forall n>0$, $0 \in (-\frac{1}{n}, \frac{1}{n})$, hence $0 \in \underset{n}{\cap} (-\frac{1}{n}, \frac{1}{n})$

Definition: Define the complement of $A$ in $X$ to be $A^c = X - A$. The set $A$ is closed in $X$ if $A^c$ is open in $X$

## Clopen Sets

A clopen set in a topological space is a set which is both open and closed.

In any metric space $(X, d)$, the sets $X$ and $\emptyset$ are clopen.

Proof: $X$ is open by definition.

Also by definition, an open set $U$ in $X$ is a set where every point in $U$ has a certain $\epsilon$-neighborhood contained entirely within $U$.

That is, there is no point in $U$ who cannot find an $\epsilon$-neighborhood contained entirely within $U$.

$\emptyset$ has no point, so $\emptyset$ is open vacuously.

Then $X^c = \emptyset$ is closed; also $\emptyset^c = X$ is closed. Therefore they are both clopen. $\blacksquare$

## Open Sets & Function Continuity

Consider two metric spaces $(X, d_X)$ and $(Y, d_Y)$, a function $f: X \to Y$, and a point $a$.

Definition: ($\epsilon$-$\theta$ definition of continuity) The function $f$ is continuous at $a$ if $\forall \epsilon > 0$, $\exists \theta > 0$ such that

I.e. for any given $\epsilon$, I need to find a range of $x$ such that the range of $f(x)$ will be smaller than $2 \epsilon$; or from a different point of view, I need to find a range of $x$ around a point $a$ such that $f(x)$ will be in (or just) a predetermined range around $f(a)$:

Definition: A function $f$ is continuous if it is continuous at every $x$ in its domain $X$.

There is an alternative formulation of continuity that we present here as a theorem.

Theorem: A function $f$ is continuous $\iff \forall$ open subset $U$ of $Y$ (w.r.t. $d_Y$), the preimage $f^{−1}(U)$ is open in $X$ (w.r.t. $d_X$).

## Limit Points / Closure / Interior / Exterior / Boundary

Let $(X, d)$ be a metric space.

Definition: Suppose $A \subseteq X$. The point $x \in X$ is a limit point of $A$ if $\forall \epsilon$, $\Phi(x, \epsilon)$ contains points of $A$ distinct from $x$, i.e. $\Phi(x, \epsilon)$ meets $A$ in a point $\neq x$. This is equivalent to saying that each neighborhood of $x$ has an infinite number of members of $A$.

• E.g. $2$ is a limit point of interval $(2,3)$.

Definition: The closure of $A$, denoted by $\operatorname{cl}(A)$, is the union of $A$ and the set of limit points of $A$,

• If $\exists \epsilon > 0$ such that $\Phi(x, \epsilon) \subset A$, we call $x$ an interior point of $A$.
• $\operatorname{int}(A) = \lbrace \text{all interior points of } A \rbrace$ is the interior of $A$
• $\operatorname{int}(A) \equiv (\operatorname{cl}(A^c))^c$
• Note that $\operatorname{int}(A)$ is open.
• If $\exists \epsilon > 0$ such that $\Phi(x, \epsilon) \cap A = \emptyset$, we call $x$ an exterior point of $A$.
• $\operatorname{ext}(A) = \lbrace \text{all exterior points of } A \rbrace$ is the exterior of $A$
• $\operatorname{ext}(A) \equiv \operatorname{int}(A^c)$
• If $x$ is neither an interior nor exterior point of $A$, it’s a boundary point of $A$
• $\partial{A} = \lbrace \text{all boundary points of } A \rbrace$ is the boundary of $A$
• $\partial{A} \equiv X - (\operatorname{int}(A) \cup \operatorname{ext}(A))$
• $X = \operatorname{int}(A) \cup \operatorname{ext}(A) \cup \partial{A}$

Further, we have $\operatorname{cl}(A) = \operatorname{int}(A) \cup \partial{A} = X - \operatorname{ext}(A)$

Proposition: Set $A$ is open $\iff$ all points in $A$ are interior points.

• 注意：$S$ 全部由 interior 构成，不代表 exterior 和 boundary 不存在。可以想象 $A = \lbrace (x, y) \mid x^2 + y^2 < 1 \rbrace$ 是一个内部涂满、边界为虚线的圆
• $\operatorname{int}(A) = A = \lbrace (x, y) \mid x^2 + y^2 < 1 \rbrace$
• $\partial{A} = \lbrace (x, y) \mid x^2 + y^2 = 1 \rbrace$ 为圆周上的点
• $\operatorname{ext}(A) = \lbrace (x, y) \mid x^2 + y^2 > 1 \rbrace$
• $\operatorname{cl}(A) = \lbrace (x, y) \mid x^2 + y^2 \leq 1 \rbrace$ 是一个内部涂满、边界为实线的圆