1. Compactness
Definition. covering
A collection $\A$ of subsets of a space $X$ is said to cover $X$ or be an covering of $X$ if $\bigcup\A=X$. It’s furthermore said to be an open covering if every element of $\A$ is open in $X$.
Definition. compactness
A topological space $X$ is said to be compact if any open covering admits a finite subcovering.
The definition is same as in metric spaces, with the openness defined by the topology.
Lemma. subspace compactness
If $X$ is a topological space, $Y$ is its subspace. $Y$ is compact iff every open (w/ $X$) covering of $Y$ admits a finite subcovering.
Note the openness is defined in $X$. This could also be taken as the definition of subspace compactness.
Lemma. closed subspace of a compact space is compact, i.e.
If $X$ is a compact space, $K\subset X$ is a closed subspace, then $K$ is compact.
Pf. Let $\mathcal O$ be an open covering of $K$, $X\setminus K$ is open, then $\mathcal O\cup \{X\setminus K\}$ covers $X$. By assumption, $\mathcal O$ has a finite subcollection $\mathcal O^*$ such that $\mathcal O^*\cup\{X\setminus K\}$ covers $K$.
This might seem natural, since we’ve learnt metric spaces before. But intuitive things in metric spaces are not necessarily true in compact spaces. For example, compactness $\implies$ closedness is no longer true generally. See the theorem below.
Theorem. compact subspace of a Hausdorff space is closed, i.e.
If $X$ is a Hausdorff space, $K\subset X$ is compact, then $K$ is closed in $X$.
Pf. Only need to show $X\setminus K$ is open. $\forall p\in X\setminus K$, there exists a neighborhood $U_p$ of $p$ such that
(since $X$ is Hausdorff) for every $x\in X$, there is a neighborhood $V_x\subset X$ of $x$ and $V_x\cap U_p = \emptyset$, then $V=\{V_x\mid x\in K\}$ is an open covering of $K$. Let $V^*$ be the corresponding finite subcovering of $K$, then we know
so $X\setminus K$ is open (any point is interior)
Theorem. continuous functions preserve compactness.
$X$ is a compact topological space, $f:X\to Y$ is continuous, then $f(X)$ is compact (w/ $\T_Y$). Or we can say $f(X)$ is a compact subspace of $Y$.
Pf. Let $\A$ be an open covering of $f(X)$, then $\{f^{-1}(A)\mid A\in\A\}$ is an open covering of $X$. Since $X$ is compact, it could then be reduced to a finite subcovering $\{f^{-1}(A_i)\}_{i=1}^n$, then $f(X)\subset\cup_{i=1}^n A_i$.
Corollary. verifying a map if a homeomorphism
If $X$ is compact, $Y$ is Hausdorff, $f:X\to Y$ is a continuous bijection, then $f$ is a homeomorphism
Pf. It suffices to show that $f^{-1}:Y\to X$ is continuous. Let $A$ be closed in $X$, since $X$ is compact, $A$ is then compact. Since $f=(f^{-1})^{-1}$ is continuous, the preimage of $A$ under $f^{-1}$, i.e. $f(A)$ is then compact in $Y$. Since $Y$ is Hausdorff, $f(A)$ is closed in $Y$. So $f^{-1}$ is continuous by definition.
Corollary. construct quotient map from a continuous surjection
If $p:X\to Y$ is a continuous surjection, $X$ is compact, $Y$ is Hausdorff, then $p$ is a quotient map.
Just not that the natural projection map is strongly continuous.
Lemma. Tube Lemma.
$X$ and $Y$ are two topological spaces. $Y$ is compact. If an open set $O$ contains $x_0\times Y:=\{x_0\}\times Y$, then there exists a neighborhood $W\subset X$ of $x$ such that
Pf. Since $Y$ is compact, $x_0\times Y$ is homeomorphic to $Y$ thus compact. Let $\{U_i\times V_i\}_{i=1}^n$ be the reduced finite collection of open sets that covers $x_0\times Y$, where $U_i$ and $V_i$ are open in $X$ and $Y$ respectively. Let $W:=\cap_{i=1}^n U_i\neq \emptyset$. By definition $W$ is a neighborhood of $W$ and $W\times Y\subset O$.
Theorem. Tychonoff’s Theorem
An arbitrary product of compact spaces is compact.
The proof of this theorem of complicated enough that all Munkres’ Chapter 5 is about this. We’ll give the proof of the baby version of this theorem as follows.
Theorem. (baby version) Tychonoff’s Theorem
The product of finitely many compact spaces is compact.
Pf. It suffices to show that if $X$ and $Y$ are compact, then $X\times Y$ is also compact. Take $x\in X$. Let $\A$ be an open covering of $x\times Y:=\{(x,y)\mid y\in Y\}$. Since $Y$ is compact, consider the continuous map into product $y\mapsto (x, y)$, we know $x\times Y$ is compact since it’s the image of $Y$ under a continuous map. $\A$ can be reduced to a subcovering, denoted by $\{A_i\}_{i=1}^n$. Their union $N$ is an open set containing $x\times Y$. By the tube lemma, there exists a neighborhood $W_x\subset X$ of $x$ such that $x\times Y\subset W_x\times Y\subset N$. Since $X$ is compact, we can find finitely many open sets in $X$, denoted by $\{W_i\}_{i=1}^m$, where for each $i$, $W_i\times Y$ can be covered by finitely many elements of $\A$.
2. Connected Spaces
Definition. separation.
A separation of a set $X$ is given by two disjoint, nonempty, open subsets $A,B$ such that $A\cup B=X$.
Definition. connected space.
A set $X$ is said to be connected if $X$ has (admits) no separations.
Lemma. subspace separation / connected subspace.
If $Y$ is a subspace of $X$, a pair of disjoint, nonempty sets $A$ and $B$ in $Y$ are said to separate $Y$ if
- $A\cup B=Y$
- $A\cap \clo B = \emptyset$ (w/ $X$)
- $B\cap \clo A = \emptyset$ (w/ $X$)
Similarly, $Y$ is called connected if there exists no such pairs of sets to separate it.
I’ll show that the “separate” here indeed corresponds to a separation of $Y$.
- If $A$ and $B$ forms a separation of $Y$. By subspace topology, we know the closure of $A$ in $Y$ is $Y\cap\clo A$, then $B\cap\clo A = B~\cap$ the closure of $A$ in $Y$. Since $B$ is open in $Y$, $A$ is then closed in $Y$, so the closure of $A$ in $Y=A$. So $B\cap\clo A=\emptyset$. Similarly $A\cap\clo B=\emptyset$.
- If two sets $A$ and $B$ separate $Y$, then $\clo A\cap B=\clo B\cap A=\emptyset$. Then $A=\clo A\cap Y$, $B=\clo B\cap Y$, then $A$ and $B$ are both closed in $Y$ by definition. So $B=Y\setminus A$ is open in $Y$, and $A=Y\setminus B$ is open in $Y$.
This lemma seems confusing.
Lemma.
A topological space $X$ is connected iff the only subsets that are both open and closed are $X$ and $\emptyset$.
Pf. If $Y\subset X$ is both open and closed, then $Y$ and $X\setminus Y$ forms a separation of $X$.
Lemma. connected subspace is not separable.
If $\{C, D\}$ is a separation of $X$, and $Y\subset X$ is a connected subspace, then $Y\subset C$ or $Y \subset D$
Pf. Otherwise $Y\cap C$ and $Y\cap D$ is a separation of $Y$.
Theorem. continuous map preserves connectivity.
Let $X$ be a connected space. If $f:X\to Y$ is continuous, then $f(X)$ is connected.
Pf. Otherwise let $U$ and $V$ be a separation of $f(X)$. by definition $f^{-1}(U)$ and $f^{-1}(V)$ is a separation of $X$.
Theorem. the union of a collection of connected subspaces of a space $X$ is connected.
Pf. easy to see contradiction from the fact that a connected subspace lies entirely in one component of the separation.
Theorem. Let $A$ be a connected subspace of $X$. If $A\subset B\subset \clo A$, then $B$ is connected.
A more intuitive version of this theorem is, adding its several limit points to a connected subspace wouldn’t change the connectivity of the space. I’ll skip the proof because it’s just a check of definition.
Theorem. A finite product of connected spaces is connected.
Pf. Suppose $X$ and $Y$ are two connected topological spaces. Let $(a,b)$ be a “base” element in the product $X\times Y$. Note that $X\times b:=\{(x,b)\mid x\in X\}$ is connected as it’s homeomorphism to $X$. For each $x\in X$, $x\times Y:=\{(x,y)\mid y\in Y\}$ is connected for the same reason. $T_x:=X\times b\cup x\times Y$ is connected because $(x,b)\in X\times b\cap x\times Y$. $X\times Y=\bigcup_{x\in X}T_x$ is connected because $\forall x\in X.(a,b)\in T_x$.
Definition. path and path connectivity
$X$ is a topological space. Given two points $x$ and $y$ of $X$, a path in $X$ from $x$ to $y$ is a continuous map $f:[0,1]\to X$ such that $f(0)=x$ and $f(1)=y$. The space $X$ is said to be path connected if every pair of points of $X$ can be joined by a path in $X$.
Note. If a space if path connected, it’s then connected. This is because continuous maps preserves connectivity.
Example. There are connected spaces that are not path connected.
Let $\Sigma:=\{(x,\sin\frac 1 x)\mid x\in (0,1]\}$. $\clo\Sigma:=\Sigma\cup 0\times[0,1]$ is called the topologist’s sine curve. $\clo\Sigma$ is connected but not path connected.
$\clo\Sigma$ is connected because $\Sigma$ is connected. $\Sigma$ is connected because it’s the product of two connected spaces. $\sin \frac 1 X$ is connected because it’s the image of a connected domain of a continuous function. I’ll show that $\clo\Sigma$ is not path connected. Suppose there’s a path $f:[0,1]\to \Sigma$ from $(0,0)$ to $(x_0,\sin\frac 1 {x_0})$, $f$ is a continuous map into products, so this in fact requires $f$ is continuous from $0$ to $\sin\frac 1 {x_0}$, but the latter does not even converge to $0$ as $x_0$ approaches to $0$.