1. Topological Spaces
Definition. Topology, Topological Space, and Open Sets.
A pair $\ts$ is called a topological space if $X$, as a set, is equipped with a collection $\T$ of subsets (called a topology) of $X$ that satisfies:
- $\emptyset\in\T$, $X\in\T$
- Arbitrary union is in $\T$. i.e.$$\{U_\alpha\}_{\alpha\in\Lambda}\subset\T\implies \bigcup_{\alpha\in\Lambda}U_{\alpha}\in\T $$
- Finite intersection is in $\T$, i.e.$$\{U_i\}_{i=1}^n\subset\T\implies \bigcap_{i=1}^nU_{i}\in\T $$
In this case, every $U$ in $\T$ is called an open set.
Definition. Metric topology
$\ms$ is a metric space, if $N_{\cdot}(\cdot)$ is the neighborhood introduced by $d$, let
then $\T$ is called a metric topology introduced by $d$.
Example. Several topologies. Given a set $X$.
- $\ms$ is a metric space, then $(X, \tau)$ is a topological space where $\tau$ is the metric topology of $X$ introduced by $d$.
- $\{\emptyset, X\}$ is called the trivial topology or indiscrete topology of $X$. (The reason that this space is called trivial will be shown in the part of Hausdorff Space)
- $2^X$ is called the discrete topology of $X$.
Definition. Comparisons of topologies
$\T_i$ are generated by basis $\B_i$ on $X$ ($i=1,2$), then we say
- $\T_1$ is finer (or larger) than $\T_2$ and $\T_2$ is coarser (or smaller) than $\T_1$ if $\T_1\supset \T_2$
- $\T_1$ is strictly finer than $\T_2$ and $\T_2$ is strictly coarser than $\T_1$ if $\T_1\supsetneq \T_2$
Definition. Subspace Topology and Subspace
$\ts$ is a topological space with $Y\subset X$. The the collection
is called a subspace topology of $\T$, and $(Y,\T_Y)$ is called a subspace of $\ts$ correspondingly.
When dealing with a space $X$ and a subspace $Y$ , one needs to be careful when one uses the term “open set”. Does one mean an element of the topology of $Y$ or an element of the topology of $X$? The following lemma is useful in this case.
Lemma. transitivity of openness
Let $Y$ be a subspace of $X$. If $U$ is open in $Y$ and $Y$ is open in $X$, then $U$ is open in $X$.
Pf. Since $U$ is open in $Y$, $U$ = $Y\cap V$ for some set $V$ open in $X$ . Since $Y$ and $V$ are both open in $X$ , so is $Y\cap V$.
Definition. Closedness
$\ts$ is a topological space, $S$ is called closed in $(X, \T)$ iff $X\setminus S\in \T$
Same in metric space, closedness is not contrary to openness.
Lemma. $\ts$ is a topological space, then the followings are true
- $\emptyset$ and $X$ are closed.
- arbitrary intersection of closed subsets are is closed.
- finite union of closed subsets is closed.
Lemma. Munkres Thm 17.2
$Y\subset X$ is a subspace of $X$, then $S$ is closed in $Y$ (with respect to the subspace topology) iff there exists a closed subset $C\subset X$ such that $S=C\cap Y$
One direction of this lemma is unexpectedly useful in proofs
Pf. If there exists a closed set $C$ in $X$ such that $S=C\cap Y$. Then $X\setminus C$ is open in $X$, so by definition of subspace topology, $(X\setminus C)\cap Y$ is open in $Y$. But $(X\setminus C)\cap Y=Y\setminus(C\cap Y)=Y\setminus S$ is open in $Y$, so $S$ is closed in $Y$.
Conversely, if $S$ is closed in $Y$, $Y\setminus S$ is open in $Y$, by definition of subspace topology, $\exists U\in\T_X$ such that $Y\setminus S=Y\cap U$ where $X\setminus U$ is closed in $X$, and $S=Y\setminus (Y\cap U)=Y\cap (Y\setminus U)=Y\cap (X\setminus U)$. So $X\setminus U$ is such a closed set.
Corollary Monkres Thm 17.3
$Y\subset X$ is closed. $S\subset Y$ is closed in $Y$ iff $S$ is closed in $X$.
Both directions instantly come from $S=Y\cap{\text{some closed set in X}}$ thus closed in $X$
Note that when we talk about closedness and openness in topological space, we always should specify what space is we comparing to relatively
Definition. interior, closure
$\T$ is a topology of $X$
- the set $S^\circ$ of interior points of $S\subset X$ is defined as $S^\circ:=\bigcup\{U\in\T\mid U\subset S\}$
- the closure $\clo S$ of $S$ is defined as $\clo S:=\bigcap\{E\subset X\mid S\subset X, \text{E is closed}\}$
By definition we have $S^\circ\subset S\subset \clo S$
Definition. Neighborhood
$\ts$ is a topological space, any element of $\{U\in\T\mid p\in U\}$ is called a neighborhood of $p\in X$.
Here we generalize the concept of neighborhood from metric space, that is, any open set containing the specific point is called a neighborhood of this point.
Definition. Limit point
$\ts$ is a topological space, a point $p$ is said to be a limit point of a set $A\subset X$, if every punctured (去心) neighborhood $U\setminus\{p\}$ of $p$ satisfies
The set of all limit points of $A$ is denoted by $A'$, called the derivative set of $A$.
Theorem. Localization Property of Closure
$\ts$ is a topological space, $A\subset \ts$, then $p\in \clo A$ iff every $U\in\T$ that contains $p$ must satisfy $U\cap A\neq \emptyset$
Pf. Instantly comes from the definition of closure.
Theorem. Closure is the union of a set itself and its derivative set
$\ts$ is a topological space, $X\supset A\neq \emptyset$, then $\clo A=A\cup A'$
Pf. If $p\in \clo A \minus A$, $\forall U\in\T$ that contains $p$ satisfies $U\cap A\neq \emptyset$, but $p\not \in A$, so $(U\minus \{p\})\cap A\neq \emptyset$, which means $p\in A'$
Theorem. $\ts$ is a topological space, $A\subset X$, TTFE
- $A$ is closed
- $\clo A = A$
- $A'\subset A$
Pf.
1 $\implies$ 2. By definition $\clo A$ is the smallest closed set containing $A$, so $A=\clo A$
2 $\implies$ 3. Shown by the above theorem
3 $\implies$ 1. By the above theorem $\clo A=A'\cup A$, so $A=\clo A$. But $\clo A$ by definition is closed.
2. Continuity
Definition. Continuous function
$\ts$, $(Y, \T')$ are two topological spaces. A function $f:X\to Y$ is called continuous on $X$ if
Theorem (Munkres Thm 18.1). $f:X\to Y$ is a function, TTFE
- $f$ is continuous on $X$
- preimage of any closed set in $Y$ should be closed in $X$
- $\forall A\subset X$, $f(\overline A)\subset \overline {f(A)}$
- $\forall p\in X$, neighborhood $V$ of $f(p)$, there exists $U\subset X$ a neighborhood of $p$ such that $f(U)\subset V$.
Pf.
1 $\iff$ 2. Instantly comes from the fact that a set is closed iff it’s complement is open.
1 $\iff$ 4. Instantly comes from the fact that a neighborhood is simply an open set in a topological space. This is more generalized than that in metric space.
2 $\implies$ 3. Let $w\in f(\clo A)$, and let $V$ be a neighborhood of $w$. Taking $x\in \clo A$ such that $f(x)=w$, (applying 4.) there exists an neighborhood $U$ of $x$ such that $f(U)\subset V$. Then $U\cap A\neq \emptyset$ and $V\cap f(A)\neq \emptyset$,
Definition. homeomorphism
$X$, $Y$ are two topological spaces, a bijection $f:X\to Y$ is called a homeomorphism if $f$ and $f^{-1}:Y\to X$ are both continuous
Definition. Embedding
An injection $f:X\to Y$ with $Z=f(X)$ is called a topological embedding of $X$ in $Y$ if $f':X\to Z$ obtained by restricting the range of $f$ is a homeomorphism between $X$ and $Z$.
Theorem. (Munkres Thm 18.2) Rules for constructing continuous functions
Let $X, Y, Z$ be topological spaces,
- (Constant) If $f :X\to Y$ maps all of $X$ into the single point $\{y_0\}$ of $Y$, then $f$ is continuous.
- (Inclusion) If $A$ is a subspace of $X$, the inclusion function $j : A \to X$ is continuous.
- (Composite) of continuous functions are continuous. If $f: X\to Y$ and $g: Y \to Z$ are continuous, then the map $g\circ f: X \to Z$ is continuous.
- (Restriction of domain) If $f: X \to Y$ is continuous, and if $A$ is a subspace of $X$, then the restricted function $f|_A : A \to Y$ is continuous.
- (Restriction and expansion of codomain) Let $f:X\to Y$ be continuous.
- If $Z$ is a subspace of $Y$ containing the image set $f (X)$, then the function $g: X\to Z$ obtained by (slightly) restricting the range of $f$ is continuous.
- If $Z$ is a space having $Y$ as a subspace, then the function $h: X\to Z$ obtained by expanding the range of $f$ is continuous.
- (Localization) The map $f: X\to Y$ is continuous if $X$ can be written as the union of open sets $U_\alpha$ such that $f |_{U_\alpha}$ is continuous for each $\alpha$.
Theorem. pasting / gluing lemma.
Let $X=A\cup B$ where $A$ and $B$ are closed in $X$. Let $f:A\to Y$ and $g:B\to Y$ be continuous. If $f=g$ on every $x\in A\cap B$, then the function $h:X\to Y$ defined by
is continuous.
Pf. Let $C$ be closed in $Y$, $h^{-1}(C)=f^{-1}(C)\cup g^{-1}(C)$ is closed in $X$.
Theorem. maps into products.
Let $f:A\to X\times Y$ given by the equation $f(x)=(f_1(x),f_2(x))$. Then $f$ is continuous iff $f_1:A\to X$ and $f_2:A\to Y$ are both continuous.
Pf. If $f$ is continuous, $f_1=\pi_1\circ f$ is a composite of two continuous functions, thus continuous. Same for $f_2$. If $f_1$ and $f_2$ are continuous on $A$. For each basis element $U\times V\subset X\times Y$, $f^{-1}(U\times V)=f_1^{-1}(U)\cap f_2^{-1}(V)$ is open in $A$.
The above theorem indicates, when considering open sets in a product space, it’s probably easier to consider the basis of the space. In fact, the above theorem could be generalized to the codomain of infinite products. See next post.
Theorem. (Munkres Ex 19.10) “product” of continuous functions are continuous
$f:A\to B$ and $g:C\to D$ are two continuous functions, then $f\times g:=(a,c)\mapsto (f(a), g(c))$ is also continuous on $A\times C$.
The proof is quite similar to the preceding one.
Note. maps from products.
There’s no useful criterion for the continuity of the map $f:A\times B\to X$. Specifically, when $f_1:A\to X$ and $f_2:B\to X$ are continuous separately, $f$ is not necessarily continuous. A classic problem in calculus is
$F=\frac 1 2$ as it approaches to $(0,0)$ from $y=x$, but $F=0$ as it approaches to $(0,0)$ from $x=0$. Thus $F$ is not continuous at $(0,0)$. But it could be checked from definition that $F$ is continuous if $x$ or $y$ is fixed.
Definition. Sequencing convergence
$\{x_n\}_{n=1}^\infty\subset X$ is said to be convergent to a point $x_0\in X$ if every neighborhood $V$ of $x_0$, there exists $N\in\nat$ s.t.
We know that in any metric space, the sequential limit is unique. It’s natural to consider the uniqueness of sequential limits in an average topological space. In fact, this property doesn’t hold in a random topological space. But if we add some constraints to it, this property will hold.
Such spaces are called Hausdorff spaces
3. Hausdorff Spaces
Definition. Hausdorff Space
$\ts$ is called a Hausdorff Space if $\forall p,q\in X$, $p\neq q$ there exists open sets $U,V$ such that
- $p\in U$
- $q\in V$
- $U\cap V\neq \emptyset$
Observation.
- Any metric space is Hausdorff (proof)
- Trivial space is not Hausdorff
- Discrete space is Hausdorff
- A nontrivial non-Hausdorff space is $\R$ with the topology $\{[a,+\infty)\mid a\in \R\}$
Theorem. Munkres Thm 17.8
Every finite point set in a Hausdorff space $X$ is closed.
Pf. Only need to show any singleton is closed (because finite union of closed sets is still closed) Let $\{x_0\}$ be a singleton in $X$. Let $x\in X$ be any arbitrary point with $x\neq x_0$, then $x$ and $x_0$ have disjoint neighborhoods $U$ and $V$, respectively. Since $U$ does not intersect $\{x_0\}$, the point $x$ cannot belong to the closure of the set $\{x_0\}$. As a result, the closure of the set $\{x_0\}$ is $\{x_0\}$ itself, so that it is closed.
Theorem. Hausdorff spaces are generalized “metric space”
$X$ is Hausdorff, then any converging sequence has a unique limit. Moreover, $p$ is a limit point $A$ iff any neighborhood of $p$ contains infinitely many points of $A$.
Pf. suppose $\xn$ converges to $x$, if $y\neq x$, it’s easy to prove that $\xn$ does not converge to $y$ by definition.
Theorem. Munkres Ex 2.17.12
Any subspace of a Hausdorff space is Hausdorff.
Instantly comes from the definition of subspace topology.