03 May 2019 - tsp

Last update 03 May 2019

8 mins

*Note: The following notes have been taken during various lectures I attended.*

Since we are mostly interested in understanding topology to understand general relativity and gravitation one can thing about a motivation for studying topology. Basically the aim is to find the most simple and least restrictive set of rules to describe the concepts without any unnecessary assumptions.

Most basically spacetime is a *set*. In classical physics we require
continuity of functions but set’s dont provide that feature - the
weakest method to impose on sets is to introduce a *topology*.

If $M$ is a set a topology $O$ is a subset $O \subset P(M)$ where $P(M)$ is the power-set of $M$ (i.e. the set containing all subsets of $M$, i.e. the topology is a certain choice of subsets of $M$).

A topology satisfies three axioms:

- $0 \in O$, the empty set has to be contained in the topology and $M \in O$, the set $M$ has to be contained in the topology
- $U, V \in O \implies U \cap V \in O$, the intersection of two sets from the topology has to be a member of the topology too
- $U_\alpha \in \implies \Cup_{\alpha \in A} U_\alpha \in O$ (with $U_\alpha$ being an index set)

*Example*: Let $M = \{ 1, 2, 3 \}$.

Choose subsets as topology: $O_1 = \{ 0, \{ 1, 2, 3 \} \}$. This is a topology because it satisfies all constraints:

- $0 \in O$ $\to$ ok
- $\{1,2,3\} \in O$ $\to$ ok
- $0 \cup 0 \in O$, $0 \cup \{1,2,3\} \in O$, $\{1,2,3\} \cup 0 \in O$, $\{1,2,3\} \cup \{1,2,3\} \in O$ $\to$ ok

*Example*: Let $M = \{ 1, 2, 3 \}$, $O_2 = \{0, \{1\}, \{2\}, \{1,2,3\}\}$ is **not** a topology because
$\{1\} \cap \{2\} = \{1,2\} \not\in O$

*Example*: $M$ is any set. The topology $O$ can always be choosen as $O := \{ 0, M \}$ which
is called the *chaotic topology* which is always a topology.

*Example*: $M$ is any set. $O := P(M)$ is always a topology which is called the *discrete topology*.
Note that the power set consists of all possible subsets of $M$.

Note that the *chaotic topology* and *discrete topology* represent the most extreme corner
cases (with least and most members). They are not really useful but allow checking corner
cases.

*Example*: The standard topology can be defined on $\mathfrak{R}^d$ (i.e. the sets of d-tuples of
real numbers). One can (arbitrarily) define $O_{standard} \subset P(\mathfrak{R}^d)$. Note that
$O_{standard}$ contains noncountable many elements (so there is no explicit way to write down all
elements, one has to use an implicit definition).

- Define
*soft ball*: $B_{r \in \mathfrak{R^{+}}}(p \in \mathfrak{R}^d) := \{ (q_1, …, q_d) \mid \sum_{i=1}^{d} (q_i - p_i)^2 < r^2 \}$. This might remind someone of the euclidian norm but since there is no vector space structure it’s not a norm. $r$ might be seen like the radius of the soft ball, $p$ like the center - but since there is no vector space it’s not really a sphere. - $U \in O_{standard} \iff \forall p \in U : \exists r \in \mathfrak{R}^{+} : B_r(p) \subset U$. One can imagine that whatever point $U$ one can pick inside a given are without it’s boundary there is a point and a radius that lies around the choosen point. This is not possible on the boundary since the ball would not always be contained inside the set.

We assume $M$ is a *set*, $O$ is a *topology* (a set of subsets of open sets of $M$ with certain properties).
A topology with it’s set $(M, O)$ is a *topological space*.

$U \in O \iff U \subset M$ is called an *open set*, $M \backslash O \iff A \subset M$ is called an *closed set*. Note
that sets can be opened and closed at the same time - these are not contradictions. Sets may also be neither
open not closed.

A map $f : M \to N$ takes every point from the domain $M$ to any point in the set $N$. There may be points in $N$ that may not be hit by the map, points from $M$ may map to the same point in $N$

- If a map $f$ maps different points from $M$ onto the same point in $N$ the map $f$ is
*not injective*. - If points in $N$ are not hit by the map the map $f$ is
*not surjective*. - If a map is surjective and injective it’s
*bijective*

Note that a map is only defined on two sets - there is no further structure implied.

It cannot be determined from the map $f : M \to N$ alone if a map is continuous. This depends on the choosen topologies on $M$ and $N$ (by definition of continuity).

Let $(M, O_M)$ and $(N, O_N)$ be topological spaces (i.e. sets with the additional
structure of choosen topologies). Then a map $f : M \to N$ (i.e. $m \to f(m)$ with
$f(m)$ being the image of $m$ in the target space $N$) is **continuous** with respect
to the topologies $O_M$ and $O_N$ if $\forall V \in O_N : \text{preim}_f(V) \in O_M$
i.e. the *preimage* of $V$, which is an open set in $N$, is an open set in $M$.

The preimage is defined as being $f : M \to V$ with $\text{preim}_f(V) := \{ m \in M \mid f(m) \in V \}$

Note that the preimage is a weaker description than an inverse map or an inverse function because the preimage is also defined for elements that have multiple sets in $M$ that are mapped to the same set in $N$. Then the preimage is the union of all the sets from $M$ that map onto the same set in $N$.

*A map is continuous if and only if the preimages of all open sets in the
target space are open sets in the domain*

*Example*: $M = \{1,2\}$, $N = \{1,2\}$ (i.e. domain and target are the same sets)

Define an arbitrary map. For example $f : M \to N$ with $f(1) = 2$ and $f(2) = 1$

Choose $O_M := \{ 0, \{1\}, \{2\}, \{1, 2\} \}$ as topology on $M$ and $O_N := \{ 0, \{1,2\} \}$ as arbitrary topologies.

Checking if $f$ is continuous with respect to the choosen topologies:

- Check if $\text{preim}_f(0) = 0 \in O_M$
- Check if $\text{preim}_f(\{1,2\}) = M \in O_M$

Thus $f$ is continuous.

*Example*: $g : N \to M$ uses the same maps and same topology definitions.

- Check if $\text{preim}_g(0) = 0 \in O_N$
- Check if $\text{preim}_g(\{1,2\}) = M \in O_N$
- Check if $\text{preim}_g(\{1\}) = \{2\} \not\in O_N$ which violates the axioms
- Check if $\text{preim}_g(\{2\}) = \{1\} \not\in O_N$ which (too) violates the axioms

So $g$ is **not** continuous

If $f : M \to N$ and $g : N \to P$ one can define $g \circ f : M \to P$.

The composition is defined as $m \to (g \circ f)(m) := g(f(m))$.

*Theorem*: If $f$ and $g$ are continuous, then $g \circ f$ is also continuous.

*Proof*: Let $V \in O_p$ then

$ \begin{align} \text{preim}_{g \circ f}(V) := \{ m \in M \mid (g \circ f)(m) \in V \} \\ := \{ m \in M \mid f(m) \in \text{preim}_g(V) \} \\ := \underbrace{\text{preim}_f(\underbrace{\text{preim}_g(V)}_{\in O_N})}_{\in O_M} \end{align} $ q.e.d.

Note that there is an arbitrary number of ways to inherit a topology from some given topological space(s).

Assume $(M, O_M)$ where one wants to consider $S \subset M$. A topology may
always be invented on $S$ but it may be desireable to inherit one from
$O_M$ via using the *subset topology*.

$ \begin{align} O\mid_S \subset P(S) \\ O\mid_S := { U \cap S \mid U \in O_M } \end{align} $

- $0 = 0 \in O_M \cap S \to 0 \in O\mid_S$
- $S = M \in O_M \cap S \to S \in O\mid_S$

$ \begin{align} A \in S, B \in S \to \exists \hat{A} \in O_M, \hat{B} \in O_M \\ A = \hat{A} \cap S, B = \hat{B} \cap S \\ A \cap B = (\hat{A} \cap S) \cap (\hat{B} \cap S) \\ = (\hat{A} \cap \hat{B}) \cap S \in O\mid_S \end{align} $

$f : (M,O_M) \to (N,O_N)$ is continuous (easy to check). Now one can easily determine if $f\mid_S : (S \subset M) \to N$ is continuous. Even if $f\mid_S$ is complex to check whenever one inherits a topology $(S, O\mid_S)$ via the inheritance then it’s guaranteed that the restriction of the map $f\mid_S : S \to N$ is continuous. In case one had choosen the topology arbitrarily this would not be the case.

Dipl.-Ing. Thomas Spielauer, Wien (webcomplains389t48957@tspi.at)

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