Chapter 01 - Topological Spaces

TDA in pictures

Introduction

Detecting topological features with a filtration:

“Barcodes” of persistence homology:

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Topological spaces: basic notions

Theory - Spaces, subspaces, connectedness

Spaces, subspaces, connectedness

Topological space

A topological space is a set $X$ (the ‘space’) together with a topology structure.

A topology structure is a determination of open subsets of $X$.

Formally: open sets are collected together in a set of sets $\tau$, the topology.

Open sets (in $\tau$) must satisfy axioms:

• $\varnothing$ and $X$ are open
• All unions of opens are open
• Finite intersections of opens are open

A set $A\subset X$ is called closed when $X\backslash A$ is open.

Although these axioms for open sets seem rather obscure, a long history (leading up to and in consequence of the axioms) has shown that the concept of open set they determine accurately formalizes our intuitions about spatial connectedness, and the aspect of proximity that underwrites connectedness.

Many geometric sets have natural topology structures:

Some notes:

• The trivial topology has only the minimum of open sets, $\varnothing$ and $X$.
• The discrete topology has the maximum of open sets, $\tau =2^X$. (All subsets are open.)
• Sometimes a set is simultaneously closed and open, thus ‘clopen’. E.g. $\varnothing$ and $X$ are always clopen.

Subspace topology

For any subset $A\subset X$, there is a subspace topology structure on $A$.

The open sets of $A$ are the intersections $A\cap U$ where $U$ is open in $X$.

Beware that open sets $A\cap U$ in a subspace topology might not be open in $X$ itself.

Connectedness

A topological space $X$ is connected when it is impossible to decompose it as a disjoint union of opens:

$$X\ne U\bigsqcup V(\text{for any opens }U,V)$$

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Illustration

Simple topology - connectedness

Simple topology - connectedness

Three points: $X=\{A,B,C\}$, with two (nontrivial) open sets.
This is not a topology because the intersection $\{B\}$ is not open:

This is a topology, and it is connected:

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Continuum - real number interval

Continuum - real number interval

The standard topology structure for real numbers is generated by open intervals $(a,b)$.

(“Generated”: it is the smallest topology structure containing all such intervals.)

In $n$D, the open sets are generated by open boxes $(a,b)\times (c,d)$ or (equivalently) open balls ${\mathbf{B}}_{\epsilon }(\vec{c})=\{\vec{x}\in {\mathbb{R}}^n\mid |\vec{x}-\vec{c}|<\epsilon \}$.

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Open in S but not in R2

Open in $S$ but not in ${\mathbb{R}}^2$

Consider the subspace $S=[0,1{]}^2\subset {\mathbb{R}}^2$ equipped with the subspace topology inherited from the topology structure on ${\mathbb{R}}^2$:

An open set in $S$ may be given as the intersection with a ball in ${\mathbb{R}}^2$, but this set is not open in ${\mathbb{R}}^2$ because the top and right edges would be included. No open ball can be found around a point in this edge and contained in $S$.

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Theory - Metric, open balls, continuity

Metric, open balls, continuity

Metric space

A metric space is a set $X$ (the ‘space’) together with a metric structure.

A metric structure is a determination of distance between points in the set.

Formally: distance is measured by a function $d:X\times X\to \mathbb{R}$ satisfying:
Triangle inequality:

• $d(x,z)\le d(x,y)+d(y,z)$

  

and more basic axioms:

• $d(x,y)\ge 0$ for all $x,y\in X$
• $d(x,y)=d(y,x)$
• $d(x,y)=0⟺x=y$

In ${\mathbb{R}}^n$ the distance function is, of course, $d(\vec{x},\vec{y})=\sqrt{(x_1-y_1{)}^2+\cdots +(x_n-y_n{)}^2}$.

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Using any distance function, we can define open balls:

$${\mathbf{B}}_r(x)=\{y\in X\mid d(y,x)<r\}$$

Then we can generate a topology on $X$ from these open balls. We are, in effect, using the topology of $\mathbb{R}$ in the codomain of $d:X\times X\to \mathbb{R}$ to induce a topology back on the space $X$.

Metric space topology

Any metric space $(X,d)$ has a natural metric space topology structure.

The open sets are generated by open balls ${\mathbf{B}}_r(x)$. That is, open sets are all unions of open balls:

$$\bigcup _{x,r}{\mathbf{B}}_r(x)$$

Every structure of metric space automatically has a natural structure of topological space due to the above definition.

A topology structure formalizes only the notion of connectedness, or one might say “touching vs. separated”, while a metric structure adds a quantitative measure of distance. As such it implies notions like “double the distance” or “$100\times$ the distance.” But a metric structure does imply at least a topology structure, because any points that are separated by some distance are in fact separated.

Topological structures that do not come from metric structures can also be useful and interesting. (E.g. in algebraic geometry.) But the notion of “touching” that they express is rather abstracted from the geometrical one we are accustomed to.

“All unions of open balls”

The collection “all unions of open balls” satisfies the axioms of a topology. The first two axioms clearly hold. To check the third axiom, one would have to formalize the concept of ‘generates’, which is applicable to the concept of ‘subbase’. The essential verification is that for any point inside a finite intersection of open balls, there is another open ball around that point which is contained in the intersection. The triangle axiom is needed to confirm this.

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LEMMA: open subsets via open balls

In any metric space $(X,d)$, a subset $A\subset X$ is open if and only if:

$$\forall x\in A,\exists r,x\text{with:}x\in {\mathbf{B}}_r(x)\subset A$$

In words: $A$ is open IFF any $x$ in $A$ has some open ball around it which is contained in $A$.

This lemma provides a functional characterization of open sets in a metric space.

Proof of LEMMA

Assume the Lemma condition holds; try to prove $A$ is open.

• For each $x$, write $B_x$ for some open ball over $x$ contained in $A$. Then $A={\bigcup }_{x\in A}{B}_x$, and $A$ is a union of open sets, thus open.

Now assume $A$ is open; try to prove the condition holds.

1. $A$ is necessarily the union of some open balls. (Definition of topology structure given a metric.)
2. Select any $x\in A$. By 1., this $x$ is in some open ball ${\mathbf{B}}_{r^{\ast }}(x^{\ast })$: $x\in {\mathbf{B}}_{r^{\ast }}(x^{\ast })\subset A$
3. Find a ball centered at $x$ which is inside ${\mathbf{B}}_{r^{\ast }}(x^{\ast })$, thus inside $A$.
   • Set $m:=d(x,x^{\ast })$
   • Definition of ball and $x\in {\mathbf{B}}_{r^{\ast }}(x^{\ast })$ implies $m<r^{\ast }$
   • Set $r:=\frac{1}{2}(r^{\ast }-m)$
   • Triangle Inequality: Every $y\in {\mathbf{B}}_r(x)$ satisfies $d(y,x^{\ast })\le d(y,x)+d(x,x^{\ast })$
   • Therefore $d(y,x^{\ast })<r+m<r^{\ast }$

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Continuous functions

A function between topological spaces $f:X\to Y$ is continuous when:

$$V\subset Y\text{is open}⟹f^{-1}(V)\subset X\text{is open}$$

This definition implies that a connected subset cannot be mapped to a disconnected subset by a continuous $f$. Given any disjoint union of opens $U\bigsqcup V\supset \mathrm{Im}Y$ that ‘disconnects’ the image of $f$, the definition implies that $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint opens covering $X$ and disconnecting it. The fact that the definition of continuity appears to go “the wrong way” (implication from codomain $Y$ to domain $X$) is related to the logical negation inside “no possible disjoint union of opens” in the definition of connectedness.

For example, the graph of a function with a jump discontinuity:

The disconnected image of $f$ can be separated by open sets in the codomain (vertical axis). The domain is not separated by the preimages – the function is not continuous.

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An equivalent definition of continuity can be cast in terms of the metric structure, if the topology comes from a metric:

Continuity and distance

A map $f:(X,d)\to (Y,d^{\prime })$ of metric spaces is continuous at $x_0$ when:

For all $\epsilon >0$, there exists $\delta >0$ such that:

$$d(x,x_0)<\delta ⟹d(f(x),f(x_0))<\epsilon$$

Then $f$ itself is continuous if it is continuous at every point $x_0$.

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Illustration

Illustration - epsilon-delta continuity

In Calc I we learn that $f:\mathbb{R}\to \mathbb{R}$ is continuous at $x_0$ when:

$$\forall \epsilon ,\exists \delta \text{ such that:}|x-x_0|<\delta ⟹|f(x)-f(x_0)|<\epsilon$$

Or, in terms of set containment:

$$\forall \epsilon ,\exists \delta \text{ such that:}f((x_0-\delta ,x_0+\delta ))\subset (f(x_0)-\epsilon ,f(x_0)+\epsilon )$$

Let us compare this to the definition in terms of open sets:

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Open-set continuity agrees with epsilon-delta continuity

Open-set continuity agrees with $\epsilon$-$\delta$ continuity

Assume that $f$ is continuous according to the topological definition.

1. Select any $x_0\in \mathbb{R}$. Let $\epsilon >0$ be given.
2. The interval $V=(f(x_0)-\epsilon ,f(x_0)+\epsilon )$ is open.
3. By the topological definition of continuity, the preimage $f^{-1}(V)$ is open. It contains $x_0$.
4. By the Lemma for metric space continuity, find an open ball around $x_0$ inside $f^{-1}(V)$.
5. This ball is an interval $(x_0-\delta ,x_0+\delta )$.
6. All elements of this interval are mapped into $V$, confirming the $\epsilon$-$\delta$ condition.

Assume that $f$ is continuous according to the $\epsilon$-$\delta$ definition.

1. Select any $V\subset \mathbb{R}$ open set. We must show $f^{-1}(V)$ is open.
2. Select any $x_0\in f^{-1}(V)$. Define $y=f(x_0)$.
3. Then $y_0\in V$, and so $y_0\in {\mathbf{B}}_{\epsilon }(y_0)\subset V$ for some $\epsilon >0$ by metric space continuity.
4. This ${\mathbf{B}}_{\epsilon }(y_0)$ is an interval: $y\in {\mathbf{B}}_{\epsilon }(y_0)⟺|y-y_0|<\epsilon$
5. By the $\epsilon$-$\delta$ definition of continuity, produce some $\delta >0$ so that: $|x-x_0|<\delta ⟹|f(x)-y_0|<\epsilon$
6. Therefore, whenever $x\in {\mathbf{B}}_{\delta }(x_0)$, we know $f(x)\in V$.
7. So ${\mathbf{B}}_{\delta }(x_0)$ is open (by definition), is contained in $f^{-1}(V)$, and contains $x_0$, confirming the topological continuity condition.

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Theory - Compactness, homeomorphism, continuity, homotopy

Compactness, homeomorphism, continuity, homotopy

Compactness

A topological space is compact when every open cover has a finite subcover.

The entire real line $\mathbb{R}$ is not compact because the open cover $\{(n-1,n+1)\mid n\in \mathbb{Z}\}$ has no finite subset which also covers $\mathbb{R}$.

The closed interval $[0,1]$ is compact.

1. Let $\mathcal{U}=\{U_{\alpha }{\}}_{\alpha }$ be any open cover of $[0,1]$.
2. Define $A\subset [0,1]$:

$$A=\{x\in [0,1]|[0,x]\text{ can be covered finitely out of }\mathcal{U}\}$$

3. Let $\ell$ be the supremum of $A$.
4. We show $\ell =1$.
   • Suppose $\ell <1$ and seek a contradiction.
   • Take some open set $U^{\ast }$ in $\mathcal{U}$ containing $\ell$, and an open ball ${\mathbf{B}}_{\epsilon }(\ell )\subset U^{\ast }$ around $\ell$ and contained in $U^{\ast }$.
   • Move leftwards to $\ell -\frac{1}{2}\epsilon$. There is a finite cover of $[0,\ell -\frac{1}{2}\epsilon ]$ from $\mathcal{U}$.
   • Add $U^{\ast }$ to this cover, and get a finite cover of $[0,\ell +\frac{1}{2}\epsilon ]$.
   • This contradicts the supremum property of $\ell$. So we know $\ell =1$.
5. Take some open set $U^{†}$ in $\mathcal{U}$ containing $1$, and an open ball ${\mathbf{B}}_{{\epsilon }^{†}}(1)\subset U^{†}$.
6. There is a finite cover of $[0,1-\frac{1}{2}{\epsilon }^{†}]$ because $\ell =1$.
7. Add $U^{†}$ to this cover to get a finite cover of $[0,1]$. QED

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Homeomorphism

A homeomorphism is a map between topological spaces, $f:X\to Y$, satisfying:

• $f$ is bijective
• $f$ is continuous
• $f^{-1}$ is continuous

PROP: ‘Homeomorphism’ simplifies for compact metric spaces

If $f:X\to Y$ is a continuous bijection between metric spaces, and $X$ and $Y$ are both compact, then $f$ is necessarily a homeomorphism.

$$\text{metric space}+\text{compact}⟹\text{don’t check }{f}^{-1}$$

Homotopy

A homotopy is a morphism between continuous maps.

Given $f_0,f_1:X\to Y$ continuous maps, a homotopy is a continuous function:

$$F:X\times [0,1]\to Y,F(x,0)=f_0\text{and}F(x,1)=f_1$$

We say $f_0$ and $f_1$ are homotopic if any homotopy exists between them.

A homotopy is a continuous deformation of $f_0$ into $f_1$. Define $f_t(x)=F(x,t)$ to view it this way.

For example, suppose we have two maps $f_0,f_1:{\mathbb{S}}^1\to {\mathbb{S}}^1$:

The “winding number” is preserved by $F$, meaning that each $f_t$ has the same winding number.

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Topological spaces: further notions

Theory - Quotients, gluing

Quotients, gluing

Quotient topology

For any partition of $\mathcal{P}$ of $X$ into cells $Y_{\alpha }$, $\alpha \in \mathcal{A}$, there is a quotient topology structure on the set of cells $Y=\{Y_{\alpha }{\}}_{\mathcal{A}}$.

The open sets of $Y$ are the collections of cells whose unions are open in $X$.

A quotient space is therefore given by ‘collapsing’ some sets of points in $X$ into single objects $Y_{\alpha }$, and the open sets come from open sets in $X$ that are compatible with the partition, meaning they do not cross partition boundaries, or that they contain only whole cells.

There are many equivalent formulations of this notion, and these two are most common:

• Using an equivalence relation $\sim$ on $X$, whereby the equivalence classes $[x]$ determine cells of the partition. Then the quotient space is written $X/\sim$.
  • Recall that partitions and equivalence relations are interchangeable notions.
• Using a surjective function $f:X\twoheadrightarrow Y$, whereby the preimage fibers $f^{-1}(y)$ determine cells of the partition. Then $Y$ has the quotient topology when open sets are given by $V$ where $f^{-1}(V)$ is open in $X$.

Gluing

Two spaces $X_1$ and $X_2$ can be glued along specified subsets $A_1\subset X_1$ and $A_2\subset X_2$ using a bijection $\phi :A_1\to A_2$ by defining the quotient topology $X_1\bigsqcup X_2/\sim$ where $\sim$ identifies $a_1\in A_1$ with $\phi (a_1)\in A_2$.

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Illustration

Gluing shapes

Gluing shapes

• Glue a pair of triangles along an edge.
• Glue a line segment to a triangle, endpoint to vertex.

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Real projective space RPn

Real projective space ${\mathbb{RP}}^n$

Consider the unit sphere ${\mathbb{S}}^2\subset {\mathbb{R}}^3$. Antipodal points of ${\mathbb{S}}^2$ are those determining a straight line passing through the origin.

Define ${\mathbb{RP}}^2$ as the quotient ${\mathbb{S}}^2/\sim$ where equivalence classes of $\sim$ are the pairs of antipodal points.

Define ${\mathbb{RP}}^n$ similarly by gluing together antipodal points in ${\mathbb{S}}^n$.

We can also describe ${\mathbb{RP}}^2$ using a disk with antipodally glued boundary:

This surface is also derivable by gluing a disk to a Mobius band:

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Theory - Embedding, limit points, closure, boundary

Embedding, limit points, closure, boundary

Embedding

An embedding $f:A\hookrightarrow Y$ is a continuous injective map that induces a homeomorphism $f:A\to f(A)$.

It is important to emphasize that $f(A)$ should have the subspace topology inherited from $Y$.

For example, a knot in ${\mathbb{R}}^3$ is an embedding ${\mathbb{S}}^1\hookrightarrow {\mathbb{R}}^3$.

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01

Exercise - Touching upon

An open interval is mapped into ${\mathbb{R}}^2$ by $f$ such that one end has a limit point on the image of the middle.

The map $f$ is continuous and bijective. Show that $f^{-1}$ is not continuous.

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Embedding, limit points, closure, boundary (part 2)

Limit points

Given a subspace $S\subset X$, a point $p\in S$ is a limit point of $S$ when every open set around $p$ touches $S$ somewhere besides $p$.

$$p\in U\text{open}⟹(U\cap S)\backslash \{p\}\ne \varnothing$$

Notes:

• Limit points are sometimes called accumulation points.
• The limit point $p$ might be outside $S$!
• In a metric space, this can be reformulated: $p$ is a limit point of $S$ when there is some sequence $x_n\in S$ such that $d(x_n,p)\to 0$ as $n\to \infty$. This is written $x_n\to p$.
  • In other words, $p$ is “approached” from inside $S$.

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Interior and exterior

The interior of a set $S\subset X$ is the set of points $p\in S$ with an open neighborhood contained in $S$.

The exterior of a set $S\subset X$ is the interior of the complement of $S$.

Closure of a set

The closure $\overline{S}$ of a set $S\subset X$ is the union of $S$ with all limit points of $S$.

• Equivalently: the closure $\overline{S}$ is the complement of the union of all open sets in $X$ that don’t touch $S$.
• Equivalently: the closure $\overline{S}$ is the intersection of all closed sets containing $S$.

Boundary of a set

The boundary $\mathrm{Bd}(S)$ of a set $S\subset X$ is the set of points in both $\overline{S}$ and $\overline{(X\backslash S)}$.

• Equivalently: the boundary $\mathrm{Bd}(S)$ is the set of points for which every open neighborhood touches both the interior and the exterior of $S$.
• Boundary points are not always in $S$!

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