Introduction to topological spaces
We want now to extend the idea of continuity presented in limits, which is a function $f : E^{n} \to E^{n}$ is continuous if given $x$ then $\forall\varepsilon > 0$ $\exists \delta$ such that $\forall y : \lVert y -x \rVert < \delta \implies \lVert f(y) - f(x) \rVert < \varepsilon$. But we want to get rid of the idea of distance, and base our definition on the idea of neighborhoods, which in $E^{n}$ are just spherical radius centered around a point.
Definition of Topological space
We define a set $X$ to be a topological space and for each $x \in X$ we define a set of subsets of $X$ called $N_{x} : \forall S \in N_{x}, S \subseteq X$ such that they satisfy the following axioms:
- $\forall S \in N_{x} : x \in S$
- if $S, T \in N_{x} \implies S \cap T \in N_{x}$
- if $S \in N_{x}, U \subseteq X, S \subseteq U \implies U \in N_{x}$, this implies that the union of $S_{1}, S_{2} \in N_{x}$ is in $N_{x}$.
- If $S \in N_{x}$ and $T = \left\{ z \in S \mid S \in N_{z} \right\}$ then we have that $T \in N_{x}$. We call $T$ the interior of $S$
The first three conditions seems to be reasonable, and coherent to our intuitive idea of neighborhoods. Let’s use our geometrical interpretation of a neighborhood as circles in a 2D plane. 1. is trivial to check, the second condition is just the smallest radius of the two neighborhoods, the third condition is a little bit more tricky, as $U$ is not guaranteed to be a circle with some radius, but it satisfies the first two conditions (we see here a relaxing of the strict ball radius condition). The fourth condition is just the circle without the boundary.
In brief, if we have a set $X$ and an assignment of neighborhoods to each point $x \in X$ then we have a topology over the set $X$.
This is good resource for topological spaces. Another definition is this (this should be a preferred definition because it’s easier to remember):
We say that a couple $(X, \mathcal{T})$ is a topological space on a set $X$ when $\mathcal{T}$ is a set of subsets of $X$ such that:
- $\varnothing, X \in \mathcal{T}$
- $G_{i} \in \mathcal{T}, \forall i \in A \implies \bigcup_{i}G_{i} \in \mathcal{T}$
- $G_{i} \in \mathcal{T}, \forall i \in \left\{ 1, \dots, n \right\} \implies \bigcap_{i}G_{i} \in \mathcal{T}$.
This behaves well for infinite unions and finite intersections. But it is equivalent, but now I don’t know exactly why. Still the most important thing here is that we have a natural notion of neighbourhood .
It’s easy to see that every set $S \subseteq X$ is a subspace-topology. An easy way to convince ourselves about why we need finite intersection bound for closeness is the classical example of accumulation point studied in Limiti, e.g. the succession of sets $\left\{(0, \frac{1}{n}] \mid n \in \mathbb{R} \right\}$.
Product Topologies
One interesting observation is that if $(X, T_{x})$ and $(Y, T_{y})$ are topologies, then there exists a product topology $X \times Y$. We just take all the possible open sets in $T_{x}$ and product that with $T_{y}$ and one can observe that the axioms of topology are good also in this case. One can observe that this kind of proof is somehow similar to the proofs presented in Grammatiche Regolari when we try to prove things joining other things together.
Continuity of topological spaces
Let’s take, say $X$ and $Y$ to be two topologies, then the function $f : X \to Y$ is continuous if $\forall x \in X, \forall S \in N_{f(x)}, S \subseteq Y$ we have that $f^{-1}(S) \in N_{x}$. This is just a difficult way to say that if the function maps to a open set in the co-domain, then it should start from an open set in the domain. We can also write the above in the following manner: Given an open set $U \subseteq Y$ we want the pre-image $f^{-1}(U)$ to be an open set for the domain topology, which seems a lot easier to understand.
This is just a different characterization of the ideas of continuity we developed in #Motivation.
Path-connected spaces
We say that a topology is path connected if for every $a, b \in \mathcal{X}$ exists a continuous function $f$ from $f: \left[ 0, 1 \right] \to \mathcal{X}$ such that $f(a) = 0$ and $f(b) = 1$ this function describes a path in the topology that connects the two points. This is usually considered a stronger condition on the connected condition of the topological space and it’s very useful for initial exploration of the mathematical space.
Th: path-connected space implies connected.
Proof: we assume path-connected and disconnected and try to prove a contradiction. By hypothesis of disconnected space, we have two disjoint open sets $A, B$ that make up the original space $\mathcal{X}$, let’s consider now two points in these sets. We know that we have a continuous function $f$ that connects these two points. We would like to use the disconnected hypothesis to prove that the function is not continuous, getting a contradiction. Let’s consider the sets $f^{-1}(A)$ and $f^{-1}(B)$ these two sets are disjoint in $\left[ 0, 1 \right]$ and their union is $[0, 1]$ because $f$ is continuous and $[0, 1] = f^{-1}(\mathcal{X}) = f^{-1}(A) \cup f^{-1}(B)$. But these two sets are non empty. Which contradicts the connectivity of $[0,1]$ proved in [[Metric Spaces#Connectedness of $ left[ 0, 1 right]$]]. Thus we conclude that the original set is connected.
One Wrong proof This is a wrong proof, some wrong proofs are educational, so I’ll include this here. Consider a ball topology (strict inequality) on the real interval with the extremes $[0], [1]$ included. We can observe that $f^{-1}(\mathcal{X}) = f^{-1}(A) \cup f^{-1}(B) = [0, a) \cup (b, 1]$, we can conclude that $a = b$, these sets are open sets for construction, but their union is not $[0, 1]$ which gives the contradiction.
Some definitions
Def: Interior and boundaries
Given a topology $(X, \mathcal{T})$ we call a set $S \subseteq X$ open if $S \in \mathcal{T}$. A set is closed if its complement is open. This is a easier definition of open and closeness compared to the definition present in real analysis. In this case the concept of neighbourhood is built in.
We cal interior set of $A$ the union of all open sets of $A$. We call boundary the difference between the closure of the set $A$ and its interior set.
Def: connected and disconnected spaces
Intuitively, we say that a topology is disconnected if we can split the original set in at least two parts. Which means: given a topology $\mathcal{X}, \mathcal{T}$ we can find $A, B \in T : A \cap B = \varnothing \land A \cup B = \mathcal{X}$. If the above condition is false, then we say that the topology is connected.
Def: Homeomorphism
A function $f: X \to Y$ where $X, Y$ are topological spaces is said to be a homeomorphism if
- Bijective
- Continuous
- Inverse is continuous This definition allows us to define a concept of equivalence of two topologies. If two spaces are homeomorphic we write $X \simeq Y$.
Def: Isotopy
$A$ is the ambient space, the space where the topologies live. We define a isotopy connecting $X \subseteq A$ and $Y \subseteq A$ to be a continuous map $\phi: X \times [0, 1] \to A$ such that $\phi(X, 0) = X$ and $\phi(X, 1) = Y$ and $\forall t \in [0, 1]$ we have that $\phi( \cdot, t)$ is a homeomorphism between $X$ and its image. We say that two spaces are isotopic if there is an isotopy connecting them.
Def: Surface
A surface is a topological space in which each point has a neighbourhood homeomorphic to the plane, and for which any two distinct points possess disjoint neighborhoods.
Classification Theorem
Any closed surface is homeomorphic either to the sphere, or to the sphere with a finite number of handles added, or to the sphere with a finite number of discs removed and replaced by Möbius strips. No two of these surfaces are homeomorphic.
This is one of the most important theorems in topology, but for a student that just started to understand this field, it will be probably a little difficult to do so.
Furstenberg’s proof of infinite prime numbers
There is a nice proof explained here about the infiniteness of prime numbers by just assuming basic properties of arithmetic and successions.
Statement
Let’s consider the set $\mathbb{Z}$ we define the set $S(a, b) := \left\{ an + b \mid n \in \mathbb{Z} \right\}, a, b \in \mathbb{Z}$. Then we consider open a set $U \subseteq \mathbb{Z}$ if we have that $x \in U \implies \exists a, b \in \mathbb{Z} : x \in S(a, b) \land S(a, b) \subseteq U$ which means that the set $U$ is composed by unions (that could also be infinite) of $S(a, b)$ successions.
Proof
We prove that with this definition of openness we have a topology:
- $\varnothing, \mathbb{Z}$ are contained in this definition, check.
- We have infinite unions, because by definition if we can unite $S(a, b)$ together as much as we want.
- We have finite unions, because it’s just taking the MCD. This is not a formal mathematical proof. but you can be convinced that this is right. Then we note that if a set is finite and non-empty it’s complementary set is not closed. We notice that the set $S(a, b)$ is both closed and open for all $a, b \in \mathbb{Z}$. This is quite easy to be convinced with. We notice that if we take $UU = \bigcup_{p \text{ is prime}} S(p, 0)$ this is equal to $\mathbb{Z} - \left\{ -1, 1 \right\}$. So we know that this set is only open and cannot be closed, by the observation of finite sets. We note now that $$ \left( \bigcup_{p \text{ is prime}} S(p, 0) \right) ^{c} = \bigcap_{p \text{ is prime}} S(p, 0)^{c} $$ We know that $S(p, 0)^{c}$ is open, and now conclude that if the number of $p$ was finite, then by property of topology the set should be open, which implies that the set $UU$ should be closed. Contradiction, which means the number of primes if infinite.