# Bei's Study Notes

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### Topological space (definition dump)

Definition (Topological space, open set): A topological space $(X,\mathcal J)$ is a set $X$ with a collection $\mathcal J$ of subsets of $X$ that satisfies:

1. $X, \emptyset \in \mathcal J$;
2. If $O_1, O_2, ... \subseteq \mathcal J$, then $\bigcup_\alpha\,O_\alpha \in \mathcal J$;
3. If $n \in \mathbb Z^+, O_1, ... O_n \in \mathcal J$, then

$$\bigcap_{i=1}^n O_i \in \mathcal J\,.$$

Sets in $\mathcal J$ are called "open set"s.

An example of a topology on $\mathbb R$ contains all the open intervals in $\mathbb R$. Thus the name "open set"s.

Definition (Induced topology): If $(X,\mathcal J)$ is a topological space and $A$ is a subset of $X$, we may make $A$ into a topological space by defining the topology $\mathcal F = \{U\mid U=A\cap O, O\in\mathcal J\}$, then $(A, \mathcal F)$ forms a topology space. $\mathcal F$ is called induced (or relative) topology.

Definition (Product topology): If $(X_1,\mathcal J_1)$ and $(X_2,\mathcal J_2)$ are both topological spaces, the direct prodct of both naturally forms a topological space $(X_1\times X_2, \mathcal J)$. $\mathcal J$ is called the product topology.

NOTE This lifts the dimension of the topological space.

Open balls on $\mathbb R^n$ naturally form a topology.

Definition (Continuous mapping): If $(X,\mathcal J)$ and $(Y,\mathcal K)$ are topological spaces, a map $f:X\to Y$ is continuous if the inverse image $f^{-1}[O] \equiv \{x\in X \mid f(x) \in O\}$ maps every open set in $Y$ to an open set in $X$.

Definition (Homeomorphism): If $f$ is continuous, one-to-one, onto, and its inverse is continues, then $f$ is called a homeomorphism, and the spaces are said to be "homeomorphic".

NOTE Not to be confused with homomorphism and homomorphic.

Definition (Closed set): The complement of an open set is called a "closed set". Sets in a topology can be open, close, both, or neither.

Definition (Connected): The topology is said to be conneted if the only subsets that are both open and closed are $X$ and $\emptyset$. $\mathbb R^n$ is connected.

Definition (Closure): If $(X, \mathcal J)$ is a topological space, $\forall A \subseteq X$, the closure $\overline A$ is the intersection of all open sets that contains $A$.

Properties:

1. $\overline A$ is closed;
2. $A \subseteq \overline A$;
3. $\overline A = A \iff A$ is closed.

NOTE Meaning "to make a set closed". Closure of a set is unique and is necessarily in the topology.

Definition (Interior, Boundary): Interior of $A$ is defined as the union of all the open sets contained in $A$. The boundy of $A$, denoted $\dot A$ (or $\partial A$), is defined as elements in $\overline A$ but not the interior of $A$, $\equiv \mathrm{int}(A)$.

NOTE alternatively, $\partial A \equiv \overline A \cap \overline {X \setminus A}$.
NOTE alternatively, $\mathrm{int}(A) = X \setminus \overline {X \setminus A}$.
NOTE alternatively, $\partial A \equiv \{ p \mid p \in X. O \in \mathcal J. p \in O \to (\exists \, a, b \in O. a \in A \wedge b \notin A) \}$.

Definition (Hausdorff): A topological space is Hausdorff if any two distinct points can be included in two disjoint open sets.

$\mathbb R^n$ is Hausdorff.

### Compactness

One of the most powerful notions in topology is that of compactness, which is defined as follows.

Definition (Open cover): If $(X, \mathcal J)$ is a topological space and a collection of open sets $C=\{O_\alpha\}$ has $\bigcup_\alpha\,O_\alpha = X$, then $C$ is said to be an open cover of $X$, and $C$ "covers" $X$. Also if $Y$ is a subset of $X$, and $Y \subseteq \bigcup_\alpha\,O_\alpha$, then $C$ is said to be an open cover of $Y$ and $C$ "covers" $Y$. A subcollection of $C$ forms a subcover if it also covers $X$ (or $Y$).

Definition (Compact space): If every open cover can be written as finite subcover, then the topological space is compact.

Alternative definitions of compact space. The following are equivalent:

1. A topological space $X$ is compact.
2. Every open cover a $X$ has a finite subcover.
3. $X$ has a sub-base such that every cover of the space by members of the sub-base has a finite subcover (Alexander's sub-base theorem).
4. Any collection of closed subsets of $X$ with the finite intersection property has nonempty intersection.
5. Every net on X has a convergent subnet (see the article on nets for a proof).
6. Every filter on X has a convergent refinement.
7. Every ultrafilter on X converges to at least one point.
8. Every infinite subset of X has a complete accumulation point.

Definition (Open cover of a set, subcover of a set): If $(X, \mathcal J)$ is a topological space and $A$ is a subset of $X$. A open cover $U$ is a open cover of $A$ if $A \subset U$. A subcover that also covers $A$ is called a subcover of $A$.

Definition (Compact subset): $A$ is said to be compact if every open cover of A has a finite subcover.

The relation ship between compact space and compact subset is given by these two theorems:

1. Compact subset of a Hausdorff space is closed.
2. Closed subset of a compact space is compect.

Heine-Borel Theorem. A closed interval $[a, b]$ of $\mathbb R$ is compact.

Open interval $(0, 1)$ is not compact (since the open cover $O_\alpha = (1/\alpha, 1)$ has no finite subcover).

A subset of $\mathbb R$ is compact iff it is closed and bounded.

NOTE A unbounded set can totally be closed. For example, $\mathbb Z$ is obviously unbounded and closed.

Let $(X, \mathcal J)$ and $(Y, \mathcal K)$ be topological spaces. Suppose $(X, \mathcal J)$ is compact and $f: X \to Y$ is continuous. Then $f[X] \equiv \{y\in Y \mid y = f(x)\}$ is compact.

NOTE This transfers compactness through homeomorphisms.

A continuous function from a compact topological space into $\mathbb R$ is bounded and attains its maximum and minimum values.

Tychonoff theorem: Product of compact topological spaces is compact. Given the axiom of choice, the number of such spaces can be infinite.

An application of these is that $S^n$ is compact, because 1) the sphere in $\mathbb R^{n+1}$ is closed and bounded, therefore compact; 2) there is a continuous function from $\mathbb R^{n+1}$ to $S^n$.

### Convergence of sequences

To extend the normal definition of sequence convergence, a sequence $\{x_n\}$ of points in a topological space $(X, \mathcal J)$ is said to converge to point $x$ if $\forall O \in \mathcal J .( x \in O \to \exists N \in \mathbb Z. \forall n > N. x_n \in O)$. $x$ is called the limit of the sequence.

A point $y \in X$ is said to be a accumulation point of $\{x_n\}$ if every open neiborhood of $y$ contains infinitely many points of the sequence.

NOTE The difference between a limit and an accumulation point is that the former requires a particular set of infinite points in $\{x_n\}$. For example, the alternating sequence $1, -1, 1, -1, ...$ has two accumulation points $1$ and $-1$, but it does not have a limit.

Definition (First countable): For every point $p$ in $X$, if there is a countable collection of open sets $\{O_\alpha\}$ that for every neiborhood $O$ of $p$, $O$ contains at least one element in $\{O_\alpha\}$.

Definition (Second countable): There is a countable collection of open sets that every open set can be written as the union of some of the sets in the collections. The sets in that collection are called basis.

NOTE The basis of a linear space is a collection of vectors, s.t. every vector in the space is a linear combination of the basis. The basis of a topological space is a collection of open sets, s.t. every opens set in the space is a union of the basis.

NOTE $\mathbb R^n$ is second countable. Open balls with rational radii centered on rational coordinates can form a countable collection of open sets.
NOTE Every second countable space is first countable.

The relationship between compactness and convergence of sequences is expressed by Bolzano-Weierstrass theorem:

Bolzano-Weierstrass theorem Let $(X, \mathcal J)$ be a topological space and let $A \subset X$

1. If $A$ is compact, then every infinite sequence $\{x_n\}$ of points in $A$ has a accumulation point lying in $A$;
2. Conversely, if $(X, \mathcal J)$ is second countable and every sequence in $A$ has an accumulation point in $A$, then $A$ is compact.

Thus, in particular, if $(X, \mathcal J)$ is second countable, $A$ is compact iff every sequence in $A$ has a convergent subsequence whose limit lies in $A$.

### Paracompactness

Definition (Neighborhood) Given $p$ in topological space $(X, \mathcal J)$, a neighborhood $V$ of $p$ is a subset of $X$ that includes an open set $U$ containing $p$:

$$V \subseteq X, \exists U \in \mathcal J, U \subseteq V, p \in U$$

NOTE: $V$ may not be open, but it contains a open set $U$ that contains $p$.

Definition (Refinement of an open cover): Open cover $\{V_\beta\}$ of $X$ is said to be a refinement of open cover $\{O_\alpha\}$ of $X$ if $\forall V_\beta. \exists O_\alpha. V_\beta \subseteq O_\alpha$.

NOTE Refinements forms a partially ordered set.
NOTE Subcover is always a refinement of a open cover. A refinement of a open cover is not always a subcover.

Definition (Locally finite): $\{V_\beta\}$ is locally finite if each $x \in X$ has an open neighborhood $W$ such that only finitely many $V_\beta$ satisfy $W \cap V_\beta \neq \emptyset$.

NOTE Compactness requires a finite subcover, locally finiteness only requires a finte refinement. It is a weaker requirement.

Definition (Paracompactness): A space is paracompact if every open cover has a locally finite refinement.

NOTE: Locally finiteness is weaker than finitenes of subcovers. Therefore every compact space is paracompact.

NOTE: (wiki) Every metric space is paracompact. A topological space is metrizable if and only if it is a paracompact and locally metrizable Hausdorff space. Paracompactness has little to do with the notion of compactness, but rather more to do with breaking up topological space entities into manageable pieces.

A paracomact manifold $M$ implies:

1. $M$ admits a Riemannian metric and
2. $M$ is second countable.

The most important implication is that a paracompact manifold $M$ will have a partition of unity.

Definition (Partition of unity): If $(X, \mathcal T)$ is a topological space, and $R$ is a set of continuous functions from $X$ to unit interval $[0, 1]$, such that for every point $x \in X$:

1. there is a neighborhood of $x$ where all but finite number of functions in $R$ are $0$, and
2. the sum of all the functions at $x$ is $1$:

$$\sum_{\rho\in R}\rho(x) = 1$$

This is for the ease of defining integrals on the manifold.

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