Theory

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:

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

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 $x_0$.

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:

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.