Math 4201 Exam 2 Review

Connectedness and compactness of metric spaces

Connectedness and separation

Definition of separation

Let $X=(X,\mathcal{T})$ be a topological space. A separation of $X$ is a pair of open sets $U,V\in \mathcal{T}$ that:

1. $U\neq \emptyset$ and $V\neq \emptyset$ (that also equivalent to $U\neq X$ and $V\neq X$)
2. $U\cap V=\emptyset$
3. $X=U\cup V$ ($\forall x\in X$, $x\in U$ or $x\in V$)

Some interesting corollary:

• Any non-trivial (not $\emptyset$ or $X$) clopen set can create a separation.
  • Proof: Let $U$ be a non-trivial clopen set. Then $U$ and $U^c$ are disjoint open sets whose union is $X$.
• For subspace $Y\subset X$, a separation of $Y$ is a pair of open sets $U,V\in \mathcal{T}_Y$ such that:
  1. $U\neq \emptyset$ and $V\neq \emptyset$ (that also equivalent to $U\neq Y$ and $V\neq Y$)
  2. $U\cap V=\emptyset$
  3. $Y=U\cup V$ ($\forall y\in Y$, $y\in U$ or $y\in V$)
  • If $\overline{A}$ is closure of $A$ in $X$, same for $\overline{B}$, then the closure of $A$ in $Y$ is $\overline{A}\cap Y$ and the closure of $B$ in $Y$ is $\overline{B}\cap Y$. Then for separation $U,V$ of $Y$, $\overline{A}\cap B=A\cap \overline{B}=\emptyset$.

Definition of connectedness

A topological space $X$ is connected if there is no separation of $X$.

Lemma of connected subspace

If $A,B$ is a separation of a topological space $X$, and $Y\subseteq X$ is a connected subspace with subspace topology, then $Y$ is either contained in $A$ or $B$.

Easy to prove by contradiction. Try to construct a separation of $Y$.

Theorem of connectedness of union of connected subsets

Let $\{A_\alpha\}_{\alpha\in I}$ be a collection of connected subsets of a topological space $X$ such that $\bigcap_{\alpha\in I} A_\alpha$ is non-empty. Then $\bigcup_{\alpha\in I} A_\alpha$ is connected.

Easy to prove by lemma of connected subspace.

Lemma of compressing connectedness

Let $A\subseteq X$ be a connected subspace of a topological space $X$ and $A\subseteq B\subseteq \overline{A}$. Then $B$ is connected.

Easy to prove by lemma of connected subspace. Suppose $C,D$ is a separation of $B$, then $A$ lies completely in either $C$ or $D$. Without loss of generality, assume $A\subseteq C$. Then $\overline{A}\subseteq\overline{C}$ and $\overline{A}\cap D=\emptyset$ (from $\overline{C}\cap D=\emptyset$ by closure of $A$). (contradiction that $D$ is nonempty) So $D$ is disjoint from $\overline{A}$, and hence from $B$. Therefore, $B$ is connected.

Theorem of connected product space

Any finite cartesian product of connected spaces is connected.

Prove using the union of connected subsets theorem. Using fiber bundle like structure union with non-empty intersection.

Application of connectedness in real numbers

Real numbers are connected.

Using the least upper bound and greatest lower bound property, we can prove that any interval in real numbers is connected.

Intermediate Value Theorem

Let $f:[a,b]\to \mathbb{R}$ be continuous. If $c\in\mathbb{R}$ is such that $f(a)<c<f(b)$, then there exists $x\in [a,b]$ such that $f(x)=c$.

If false, then we can use the disjoint interval with projective map to create a separation of $[a,b]$.

Definition of path-connected space

A topological space $X$ is path-connected if for any two points $x,x'\in X$, there is a continuous map $\gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=x'$. Any such continuous map is called a path from $x$ to $x'$.

• Every connected space is path-connected.
  • The converse may not be true, consider the topologists’ sine curve.

Compactness

Definition of compactness via open cover and finite subcover

Let $X=(X,\mathcal{T})$ be a topological space. An open cover of $X$ is $\mathcal{A}\subset \mathcal{T}$ such that $X=\bigcup_{A\in \mathcal{A}} A$. A finite subcover of $\mathcal{A}$ is a finite subset of $\mathcal{A}$ that covers $X$.

$X$ is compact if every open cover of $X$ has a finite subcover (i.e. $X=\bigcup_{A\in \mathcal{A}} A\implies \exists \mathcal{A}'\subset \mathcal{A}$ finite such that $X=\bigcup_{A\in \mathcal{A}'} A$).

Definition of compactness via finite intersection property

A collection $\{C_\alpha\}_{\alpha\in I}$ of subsets of a set $X$ has finite intersection property if for every finite subcollection $\{C_{\alpha_1}, ..., C_{\alpha_n}\}$ of $\{C_\alpha\}_{\alpha\in I}$, we have $\bigcap_{i=1}^n C_{\alpha_i}\neq \emptyset$.

Let $X=(X,\mathcal{T})$ be a topological space. $X$ is compact if every collection $\{Z_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ satisfies the finite intersection property has a non-empty intersection (i.e. $\forall \{Z_{\alpha_1}, ..., Z_{\alpha_n}\}\subset \{Z_\alpha\}_{\alpha\in I}, \bigcap_{i=1}^n Z_{\alpha_i} \neq \emptyset\implies \bigcap_{\alpha\in I} Z_\alpha \neq \emptyset$).

Compactness is a local property

Let $X$ be a topological space. A subset $Y\subseteq X$ is compact if and only if every open covering of $Y$ (set open in $X$) has a finite subcovering of $Y$.

• A space $X$ is compact but the subspace may not be compact.
  • Consider $X=[0,1]$ and $Y=[0,1/2)$. $Y$ is not compact because the open cover $\{(0,1/n):n\in \mathbb{N}\}$ does not have a finite subcover.
• A compact subspace may live in a space that is not compact.
  • Consider $X=\mathbb{R}$ and $Y=[0,1]$. $Y$ is compact but $X$ is not compact.

Closed subspaces of compact spaces

A closed subspace of a compact space is compact.

A compact subspace of Hausdorff space is closed.

Each point not in the closed set have disjoint open neighborhoods with the closed set in Hausdorff space.

Theorem of compact subspaces with Hausdorff property

If $Y$ is compact subspace of a Hausdorff space $X$, $x_0\in X-Y$, then there are disjoint open neighborhoods $U,V\subseteq X$ such that $x_0\in U$ and $Y\subseteq V$.

Image of compact space under continuous map is compact

Let $f:X\to Y$ be a continuous map and $X$ is compact. Then $f(X)$ is compact.

Tube lemma

Let $X,Y$ be topological spaces and $Y$ is compact. Let $N\subseteq X\times Y$ be an open set contains $X\times \{y_0\}$ for $y_0\in Y$. Then there exists an open set $W\subseteq Y$ is open containing $y_0$ such that $N$ contains $X\times W$.

Apply the finite intersection property of open sets in $X\times Y$. Projection map is continuous.

Product of compact spaces is compact

Let $X,Y$ be compact spaces, then $X\times Y$ is compact.

Any finite product of compact spaces is compact.

Compact subspaces of real numbers

Every closed and bounded subset of real numbers is compact

$[a,b]$ is compact in $\mathbb{R}$ with standard topology.

Good news for real numbers

Any of the three properties is equivalent for subsets of real numbers (product of real numbers):

1. $A\subseteq \mathbb{R}^n$ is closed and bounded (with respect to the standard metric or spherical metric on $\mathbb{R}^n$).
2. $A\subseteq \mathbb{R}^n$ is compact.

Extreme value theorem

If $f:X\to \mathbb{R}$ is continuous map with $X$ being compact. Then $f$ attains its minimum and maximum. (there exists $x_m,x_M\in X$ such that $f(x_m)\leq f(x)\leq f(x_M)$ for all $x\in X$)

Lebesgue number lemma

For a compact metric space $(X,d)$ and an open covering $\{U_\alpha\}_{\alpha\in I}$ of $X$. Then there is $\delta>0$ such that for every subset $A\subseteq X$ with diameter less than $\delta$, there is $\alpha\in I$ such that $A\subseteq U_\alpha$.

Apply the extreme value theorem over the mapping of the averaging function for distance of points to the $X-U_\alpha$. Find minimum radius of balls that have some $U_\alpha$ containing the ball.

Definition for uniform continuous function

$f$ is uniformly continuous if for any $\epsilon > 0$, there exists $\delta > 0$ such that for any $x_1,x_2\in X$, if $d(x_1,x_2)<\delta$, then $d(f(x_1),f(x_2))<\epsilon$.

Theorem of uniform continuous function

Let $f:X\to Y$ be a continuous map between two metric spaces. If $X$ is compact, then $f$ is uniformly continuous.

Definition of isolated point

A point $x\in X$ is an isolated point if $\{x\}$ is an open subset of $X$.

Theorem of isolated point in compact spaces

Let $X$ be a nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.

Proof using infinite nested closed intervals should be nonempty.

Variation of compactness

Limit point compactness

A topological space $X$ is limit point compact if every infinite subset of $X$ has a limit point in $X$.

• Every compact space is limit point compact.

Sequentially compact

A topological space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.

• Every compact space is sequentially compact.

Equivalence of three in metrizable spaces

If $X$ is a metrizable space, then the following are equivalent:

1. $X$ is compact.
2. $X$ is limit point compact.
3. $X$ is sequentially compact.