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

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

NeighborhoodPermalink

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

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

We call $\mathrm{\Phi }(x_i,ϵ)$ the $ϵ$-neighborhood of $x_i$ in space $X$. Obviously, $x_i\in \mathrm{\Phi }(x_i,ϵ)$.

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

Open and Closed SetsPermalink

Definition: A set $U$ in space $X$ is open if $\mathrm{\forall }{x}_i\in U,\mathrm{\exists }{ϵ}_i>0$ such that $\mathrm{\Phi }(x_i,{ϵ}_i)\subseteq U$.

• 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 $\mathrm{\Phi }(x_i,ϵ)=\{x_j\in X|d(x_i,x_j)<ϵ\}$ and a point $x_j\in \mathrm{\Phi }(x_i,ϵ)$, $\mathrm{\forall }{ϵ}^{\prime }$ we can define $\mathrm{\Phi }(x_j,{ϵ}^{\prime })=\{x_k\in X|d(x_j,x_k)<{ϵ}^{\prime }\}$. However, according to triangle inequality, we only know that $d(x_i,x_k)\le ϵ+{ϵ}^{\prime }$. It’s not guaranteed that $d(x_i,x_k)<ϵ$, i.e. not guaranteed that $x_k\in \mathrm{\Phi }(x_i,ϵ)$, i.e. not guaranteed that $\mathrm{\Phi }(x_j,{ϵ}^{\prime })\subseteq \mathrm{\Phi }(x_i,ϵ)$. $◼$

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 $U$. $◼$

Proposition: Let $\{U_i|i\in I\}$ 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 $\{1,2,3\}$ 就是 $\{x_1,x_2,x_3\}$ 的 index set。
• 可以是 infinite union

Proposition: Let $\{U_i|i=1,\dots ,N\}$ be a finite collection of open sets in space $X$. Then the set $\stackrel{N}{\underset{i=1}{\cap }}{U}_i$ is open.

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

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 $\mathrm{\varnothing }$ 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 $ϵ$-neighborhood contained entirely within $U$.

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

$\mathrm{\varnothing }$ has no point, so $\mathrm{\varnothing }$ is open vacuously.

Then $X^c=\mathrm{\varnothing }$ is closed; also ${\mathrm{\varnothing }}^c=X$ is closed. Therefore they are both clopen. $◼$

Open Sets & Function ContinuityPermalink

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

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

I.e. for any given $ϵ$, I need to find a range of $x$ such that the range of $f(x)$ will be smaller than $2ϵ$; 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 $X$.

考虑函数 $f(x)=\{\begin{array}{ll}x-3& x\le 0\\ x+4& x>0\end{array}$。取 $a=0,ϵ<7$。对任意 $x>0$，$f(x)-f(a)>7>ϵ$，所以你不可能找到一个符合条件的 $\theta$ 值，所以这个函数不连续。

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

Theorem: A function $f$ is continuous $⟺\mathrm{\forall }$ 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 / BoundaryPermalink

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

Definition: Suppose $A\subseteq X$. The point $x\in X$ is a limit point of $A$ if $\mathrm{\forall }ϵ$, $\mathrm{\Phi }(x,ϵ)$ contains points of $A$ distinct from $x$, i.e. $\mathrm{\Phi }(x,ϵ)$ meets $A$ in a point $\ne x$. 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 $\mathrm{cl}(A)$, is the union of $A$ and the set of limit points of $A$,

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

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

Proposition: Set $A$ is open $⟺$ 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 $A=\{(x,y)\mid x^2+y^2<1\}$ 是一个内部涂满、边界为虚线的圆
  • $\mathrm{int}(A)=A=\{(x,y)\mid x^2+y^2<1\}$
  • $\partial A=\{(x,y)\mid x^2+y^2=1\}$ 为圆周上的点
  • $\mathrm{ext}(A)=\{(x,y)\mid x^2+y^2>1\}$
  • $\mathrm{cl}(A)=\{(x,y)\mid x^2+y^2\le 1\}$ 是一个内部涂满、边界为实线的圆