Sequences as Continuous Functions: A Novel Proof That Closed Sets Contain Their Limits

This essay will use the results in my previous blogs : "Beyond Sequences: A Topological Approach to Density Arguments" and "Proving Homeomorphism with Yoneda Lemma: The Unification of epsilon-delta and epsilon-N Formulation". to give an approach for the really basic fact:

Proposition.

Let $F\subseteq X$$F\subseteq X$ be a closed set, $(x_n{)}_{n\in \mathbb{N}}$$(x_n)_{n\in\mathbb N}$ be a convergence sequence in $X$$X$ with $\mathrm{\forall }n,\in \mathbb{N},x_n\in F$$\forall n,\in\mathbb N,x_n\in F$. then we have $\begin{array}{c}{x}_0=\underset{n\to \mathrm{\infty }}{lim}{x}_n\in F\end{array}$$x_0=\lim_{n\to\infty}x_n\in F\\$.

Proof. Notice a convergence sequence is just a continuous function from ${\mathbb{N}}^{\ast }:=\{0\}\cup \{1/n:n\in \mathbb{N}\}$$\mathbb N^*:=\{0\}\cup\{1/n:n\in\mathbb N\}$ with $f(1/n)=x_n,f(0)=x_0$$f(1/n)=x_n,f(0)=x_0$.

Then we have $f^{-1}(F)$$f^{-1}(F)$ is closed as well and $\{1/n:n\in \mathbb{N}\}\subseteq f^{-1}(F)$$\{1/n:n\in\mathbb N\}\subseteq f^{-1}(F)$. Hence the closure of $\{1/n:n\in \mathbb{N}\}={\mathbb{N}}^{\ast }\subseteq f^{-1}(F)◻$$\{1/n:n\in\mathbb N\}=\mathbb N^*\subseteq f^{-1}(F)\quad\square$

 

(Co)Group Objects and Their Representable Group-Valued Presheaf Equivalence in Category Theory

Group ObjectExample of group object and Group Object preserving functorThe equivalence between category of group objects in $\mathcal C$ and representable group valued presheaves on $\mathcal C$.Cogroup objectHigher Homotopy Group

 

Group Object

Definition.

Let $\mathcal{C}$$\mathcal{C}$ be a category admitting finite products and in particular, terminal object $1$$1$. A group object in $\mathcal{C}$$\mathcal{C}$ is a quadruple

$(G,m,i,e),$$\bigl(G,\;m,\;i,\;e\bigr),$ where $G\in \mathrm{Ob}(\mathcal{C})$$G\in\mathrm{Ob}(\mathcal{C})$ and

• $m:G\times G\to G$$m: G\times G \to G$ is the multiplication morphism,

• $e:1\to G$$e: 1 \to G$ is the unit (identity) morphism,

• $i:G\to G$$i: G \to G$ is the inverse morphism,

subject to the usual associativity, unit and inverse axioms encoded by commutative diagrams in $\mathcal{C}$$\mathcal{C}$.

Associative Law:

Unit Law

Inverse Law

The morphism $\phi$$\varphi$ between Group objects are

Example of group object and Group Object preserving functor

Example of group object.

In $\mathrm{Top}$$\mathrm{Top}$, the group objects are topological groups, in $\mathrm{Man}$$\mathrm{Man}$, they will be Lie groups, in $\mathrm{Set}$$\mathrm{Set}$, they are the usual groups, in $\mathrm{Grp}$$\mathrm{Grp}$, they are abelian groups by Eckmann-Hilton argument. Since for a group object $G$$G$ in $\mathrm{Grp}$$\mathrm{Grp}$,we have two untial operators (multiplications) on $G$$G$. One is the $\cdot$$\cdot$ where $\cdot$$\cdot$ is the multiplication of the group. Another is $m:G\times G\to G$$m:G\times G\to G$, denote as $\circ$$\circ$.

$$(a\circ b)\cdot (c\circ d)=m(a,b)m(c,d)=m(a\cdot c,b\cdot d)=(a\cdot c)\circ (b\cdot d)$$

Since $m$$m$ is a group homomorphism. Hence $\circ =\cdot$$\circ =\cdot$.

Remark.

Let $F:\mathcal{C}\to \mathcal{D}$$F:\mathcal C\to\mathcal D$ be a product preserving functor, then $(G,m,i,e)$$(G,m,i,e)$ is a group object implies $(F(G),F(m),F(i),F(e)$$(F(G),F(m),F(i),F(e)$ is a group object.

Example 1.

Let $\mathcal{D}$$\mathcal D$ be a concrete category over $\mathrm{Set}$$\mathrm{Set}$, and suppose that we have free-forgetful adjoint.

Let $(G,m,i,e)$$(G,m,i,e)$ be a group object, then $(U(G),U(m),U(i),U(e))$$(U(G),U(m),U(i),U(e))$ is a group object in $\mathrm{Set}$$\mathrm{Set}$, i.e. a group. Here $U$$U$ is the forgetful functor. In general, a group object is not a group. We will see an example in the next section, the group object in $[{\mathcal{C}}^{\mathrm{op}},\mathrm{Set}]$$[\mathcal C^{\mathrm{op}},\mathrm{Set}]$ are group vaule functor.

For example, Let $U:\mathrm{Meas}\to \mathrm{Set}$$U:\mathrm{Meas}\to\mathrm{Set}$ be the forgetful functor, it admit both left and right adjoint.

$${\mathrm{Hom}}_{\mathrm{Meas}}(D(X),Y)\cong {\mathrm{Hom}}_{\mathrm{Set}}(X,U(Y))$$

Here the sigma algebra of $D(X)$$D(X)$ is just $P(X)$$P(X)$. The right adjoint is given trivial topology(form a sigma algebra directly).

Hence $U$$U$ preserve group object.

Example 2.

Consider the Borel set functor $B:\mathrm{Top}\to \mathrm{Meas}$$B:\mathrm{Top}\to\mathrm{Meas}$. Restrict $B$$B$ at the category of second conutable space, denote as ${}_2\mathrm{Top}$$_2\mathrm{Top}$. Then $B(X\times Y)\cong B(X)\times B(Y)$$B(X\times Y)\cong B(X)\times B(Y)$. Hence it will preserve group object.

Let us prove the proposition: $B{:}_2\mathrm{Top}\to \mathrm{Meas}$$B:_2\mathrm{Top}\to \mathrm{Meas}$ is a product preserving functor.

Proposition. Let $(X_i,{\tau }_i{)}_{i\in \mathbb{N}}$$(X_i,\tau_i)_{i\in\mathbb N}$ be a family of topological space, then $B(\prod _{{}_{i\in \mathbb{N}}}(X_i,{\tau }_i))=\prod _{i\in \mathbb{N}}B(X_i,{\tau }_i)$$B\left(\prod_{_{i\in\mathbb N}}(X_i,\tau_i)\right)=\prod_{i\in\mathbb N}B(X_i,\tau_i)$.

Proof. We only need to prove that $\sigma (C_1)=\sigma (C_2)$$\sigma(C_1)=\sigma(C_2)$, here $C_1$$C_1$ is the class of the open set in the product topology and $C_2$$C_2$ is the class of cylinder set of the form $X_1\times ...X_{n-1}\times B_k\times ...$$X_1\times...X_{n-1}\times B_k\times...$ with $B_k\in B(X_k,{\tau }_k)$$B_k\in B(X_k,\tau_k)$. Now let $C$$C$ be the sub class of $C_2$$C_2$ such that $B_k$$B_k$ is open set, hence $C$$C$ is the topology basis of the product topology. Every open set in the topology could be written as countable union of elements in $C$$C$ since each $X_i$$X_i$ is second countable.

Then $\sigma (C_1)=\sigma (C)$$\sigma(C_1)=\sigma(C)$ and $\sigma (C)=\sigma (C_2)$$\sigma(C)=\sigma(C_2)$. $◻$$\square$

Corollary.

Let $G$$G$ be a second countbale topological group (for example, ${\mathbb{R}}^n$$\mathbb R^n$), then $G$$G$ with Borel sigma algebra form a group object in category of measurable space.

The equivalence between category of group objects in $\mathcal{C}$$\mathcal C$ and representable group valued presheaves on $\mathcal{C}$$\mathcal C$.

Proposition. An object $G$$G$ in a locally small category $\mathcal{C}$$\mathcal C$ is a group object if and only if there is a functor $F:{\mathcal{C}}^{\mathrm{op}}\to \mathrm{Grp}$$F:\mathcal C^{\mathrm{op}}\to\mathrm{Grp}$ such that

$${\mathrm{Hom}}_{\mathcal{C}}(-,G)=U\circ F$$

Where $U$$U$ is the forgetful functor from $\mathrm{Grp}$$\mathrm{Grp}$ to $\mathrm{Set}$$\mathrm{Set}$.

That is, you could lift ${\mathrm{Hom}}_{\mathcal{C}}(-,G)$$\mathrm{Hom}_{\mathcal C}(-,G)$ to $F$$F$.

Proof.

Let $G$$G$ be a group object, we define the group structure as follows:

Let $f,g:X\to G$$f,g:X\to G$ be two morphisms in $\mathcal{C}$$\mathcal C$, consider the following sequence:

$$X\stackrel{(f,g)}{\to }G\times G\stackrel{m}{\to }G$$

And define $f\cdot g:=m\circ (f,g)$$f\cdot g:= m\circ (f,g)$.

We are not unfamiliar with this. In calculus, let $\mathcal{C}=\mathrm{Top}$$\mathcal{C}=\mathrm{Top}$, consider the topological group $\mathbb{R}$$\mathbb R$. We define the addition of two continuous functions as

$$X\stackrel{(f,g)}{\to }\mathbb{R}\times \mathbb{R}\stackrel{+}{\to }\mathbb{R}$$

And see $f+g$$f+g$ is continuous as well since it is the composition of continuous functions.

Let $1:X\to 1$$1:X\to 1$ be the unique map.

The identity is given by $e_{\ast }:\{\ast \}\cong \mathrm{Hom}(X,1)\to \mathrm{Hom}(X,G),e_{\ast }(\ast )=e\circ 1$$e_*:\{*\}\cong\mathrm{Hom}(X,1)\to\mathrm{Hom}(X,G),e_*(*)=e\circ 1$.

The inverse for $\mathrm{Hom}(X,G)$$\mathrm{Hom}(X,G)$ is given by $i_{\ast }:\mathrm{Hom}(X,G)\to \mathrm{Hom}(X,G),i_{\ast }(f)=i\circ f$$i_*:\mathrm{Hom}(X,G)\to \mathrm{Hom}(X,G),i_*(f)=i\circ f$.

Using the diagram above it is easy to see that ${\mathrm{Hom}}_{\mathcal{C}}(X,G)$$\mathrm{Hom}_{\mathcal C}(X,G)$ is a group object in $\mathrm{Set}$$\mathrm{Set}$.

Now assume that ${\mathrm{Hom}}_{\mathcal{C}}(-,G)$$\mathrm{Hom}_{\mathcal C}(-,G)$ could be lifted to a functor $F:{\mathcal{C}}^{\mathrm{op}}\to \mathrm{Grp}$$F:\mathcal C^{\mathrm{op}}\to\mathrm{Grp}$, we claim that $F$$F$ is a group object in the functor category, and since it is represented by $G$$G$, using Yoneda Lemma we see that $G$$G$ is a group object.

Then we have the following natural transformation in $[{\mathcal{C}}^{\mathrm{op}},\mathrm{Grp}]$$[\mathcal C^{\mathrm{op}},\mathrm{Grp}]$:

• Multiplication:

  $$\mu :F\times F\to F$$

  whose component at $A$$A$ is the group multiplication ${\mu }_A:F(A)\times F(A)\to F(A)$$\mu_A : F(A) \times F(A) \to F(A)$. Naturality means that for every $h:B\to A$$h : B \to A$ the square

commutes.

• Unit:

  $$\eta :1\Rightarrow F$$

where $\ast$$\ast$ denotes the constant functor $A⟼1.$$A\longmapsto 1.$

The component ${\eta }_A(1)\in F(A)$$\eta_A(1) \in F(A)$ is the identity element of the group, and naturality means $F(h)({\eta }_A(\ast ))={\eta }_B(\ast )$$F(h)(\eta_A(\ast)) = \eta_B(\ast)$.

• Inverse:

$$i:F\Rightarrow F$$

whose components are the inverse map $i_A:F(A)\to F(A)$$i_A:F(A)\to F(A)$.

By Yoneda Lemma, $F\cong {\mathrm{Hom}}_{\mathcal{C}}(-,G)$$F\cong\mathrm{Hom}_{\mathcal C}(-,G)$ and $F$$F$ is a group object implies that $G$$G$ is a group object. $◻$$\square$

Remark.

Easy to see that the group objects in $[{\mathcal{C}}^{\mathrm{op}},\mathrm{Set}]$$[\mathcal C^{\mathrm{op}},\mathrm{Set}]$ are the objects in $[{\mathcal{C}}^{\mathrm{op}},\mathrm{Grp}]$$[\mathcal{C^{\mathrm{op}}},\mathrm{Grp}]$

We already see that functors $F:{\mathcal{C}}^{\mathrm{op}}\to \mathrm{Grp}$$F:\mathcal{C^{\mathrm{op}}}\to\mathrm{Grp}$ are group objects. Conversely, let $K$$K$ be a group object in $[{\mathcal{C}}^{\mathrm{op}},\mathrm{Set}]$$[\mathcal C^{\mathrm{op}},\mathrm{Set}]$, then the natural transformation at each object will give $K(X)$$K(X)$ a group structure.

Cogroup object

The cogroup object in $\mathcal{C}$$\mathcal C$ is just a group object in ${\mathcal{C}}^{\mathrm{op}}$$\mathcal C^{\mathrm{op}}$. Hence, $X$$X$ is a cogroup object if and only if ${\mathrm{Hom}}_{\mathcal{C}}(X,-)$$\mathrm{Hom}_{\mathcal C}(X,-)$ could be lifted to a group valued functor $F$$F$.

Example. Let us give $S^n:=\{x=(x_1,...,x_{n+1})\in {\mathbb{R}}^{n+1}:\parallel x{\parallel }_2=1\}$$S^n:=\{x=(x_1,...,x_{n+1})\in\mathbb R^{n+1}:\|x\|_2=1\}$ a cogroup structure in ${\mathrm{hTop}}_{\ast }$$\mathrm{hTop}_*$.

Collapse the equator $S^{n-1}:=\{x\in S^n:x_{n+1}=0\}$$S^{n-1}:=\{x\in S^n:x_{n+1}=0\}$ to a point. We get a quotient map

$$q:S^n\to S^n/S^{n-1}\cong S^n\vee S^n$$

Remark. Image the $\mathrm{O}$$\mathrm{O}$ and $\mathrm{\infty }$$\infty$...

Define the comultiplication

$$\mathrm{\nabla }:S^n\to S^n\vee S^n$$

by

$$\begin{array}{c}\mathrm{\nabla }(x)=\{\begin{array}{ll}{i}_1(x),& x_{n+1}>0,\\ \ast ,& x_{n+1}=0,\\ i_2(x),& x_{n+1}<0,\end{array}\end{array}$$

and counit

$$ϵ:S^n\to \{\ast \}$$

collapsing the entire sphere to the base point.

The coinverse is the antipodal map

$$\chi :S^n\to S^n,\chi (x)=-x$$

 

Higher Homotopy Group

Definition.

Consider ${\mathrm{Hom}}_{{\mathrm{hTop}}_{\ast }}(S^n,-)$$\mathrm{Hom_{hTop_*}}(S^n,-)$, since $S^n$$S^n$ is a cogroup, we could induce a group strcture on ${\mathrm{Hom}}_{{\mathrm{hTop}}_{\ast }}(S^n,X)$$\mathrm{Hom}_{\mathrm{hTop}_*}(S^n,X)$.

Define the multiplication as the composition of

$$S^n\stackrel{\mathrm{\nabla }}{\to }{S}^n\vee S^n\stackrel{{\gamma }_1\vee {\gamma }_2}{\to }X$$

The identity element is $\ast \circ ϵ$$*\circ\epsilon$, the inverse of $[f]$$[f]$ is given by $[f{]}^{-1}=[f\circ \chi ]$$[f]^{-1}=[f\circ\chi]$.

But in practice we usually define the higher homotopy group in the following way.

Consider the unit cube $[0,1{]}^n$$[0,1]^n$ and the quotient map $\pi :[0,1{]}^n\to [0,1{]}^n/\partial ([0,1{]}^n)\cong S^n$$\pi:[0,1]^n\to [0,1]^n/\partial([0,1]^n)\cong S^n$.

Then instead of consider $f:S^n\to X$$f:S^n\to X$, we consider $f\circ \pi :[0,1{]}^n\to X$$f\circ \pi:[0,1]^n\to X$ in ${\pi }_n(X)$$\pi_n(X)$.

That is, all the homotopy classes of maps

$$f:[0,1{]}^n\to X$$

for which $\partial ([0,1{]}^n)$$\partial([0,1]^n)$ is mapped on to the base point $x$$x$.

Let us define the product in ${\pi }_n(X)$$\pi_n(X)$.

Definition.

Let $[f],[g]\in {\pi }_n(X)$$[f],[g]\in\pi_n(X)$, then define $[f]\cdot [g]$$[f]\cdot [g]$ by

$$\begin{array}{c}[f]\cdot [g](t_1,...,t_n)=\{\begin{array}{l}f(2t_1,t_2,...,t_n)t_1\in [0,1/2]\\ g(2t_1-1,t_2,...,t_n)t_1\in [1/2,1]\end{array}\end{array}$$

This $\cdot$$\cdot$ is untial. The identity element is the constant map.

As you can see, this multiplication is defined at the first component. it seems not canonical, but we will show that

This $\cdot$$\cdot$ is unital, with the identity element given by the constant map.

As you will see, the multiplication is defined on the first component. It may seem non-canonical, but, by the Eckmann–Hilton argument, no matter which component you choose to define the product, they all coincide.

Proposition.

For $n\ge 2,{\pi }_n(-)$$n\ge 2,\pi_n(-)$ is an $\mathrm{Ab}$$\mathrm{Ab}$ valued functor.

Proof. We will use Eckmann–Hilton argument, let us define $[f]\star [g]$$[f]\star [g]$ by

$$\begin{array}{c}[f]\star [g](t_1,...,t_n)=\{\begin{array}{l}f(t_1,2t_2,...,t_n)t_2\in [0,1/2]\\ g(t_1,2t_2-1,...,t_n)t_2\in [1/2,1]\end{array}\end{array}$$

We need to prove

$$([f]\cdot [g])\star ([h]\cdot [k])=([f]\star [h])\cdot ([g]\star [k])$$

Then by the Eckmann–Hilton argument, $\cdot =\star$$\cdot=\star$ and they are commutative.

The left hand side is

$$\begin{array}{c}[f]\cdot [g](t_1,...,t_n)=\{\begin{array}{l}f(2t_1,t_2,...,t_n)t_1\in [0,1/2]\\ g(2t_1-1,t_2,...,t_n)t_1\in [1/2,1]\end{array}\end{array}$$

$\star$$\star$ with

$$\begin{array}{c}[h]\cdot [k](t_1,...,t_n)=\{\begin{array}{l}h(2t_1,t_2,...,t_n)t_1\in [0,1/2]\\ k(2t_1-1,t_2,...,t_n)t_1\in [1/2,1]\end{array}\end{array}$$

i.e.

$$\begin{array}{c}([f]\cdot [g])\star ([h]\cdot [k])=\{\begin{array}{l}f(2t_1,2t_2,...,t_n)t_1\in [0,1/2],t_2\in [0,1/2]\\ g(2t_1-1,2t_2,...,t_n),t_1\in [1/2,1],t_2\in [0,1/2]\\ h(2t_1,2t_2-1,...,t_n),t_1\in [0,1/2],t_2\in [1/2,1]\\ k(2t_1-1,2t_2-1,...,t_n),t_1\in [1/2,1],t_2\in [1/2,1]\end{array}\end{array}$$

The right hand sides is

$$\begin{array}{c}[f]\star [h](t_1,...,t_n)=\{\begin{array}{l}f(t_1,2t_2,...,t_n)t_2\in [0,1/2]\\ h(t_1,2t_2-1,...,t_n)t_2\in [1/2,1]\end{array}\end{array}$$

$\cdot$$\cdot$ with

 

$$\begin{array}{c}[g]\star [k](t_1,...,t_n)=\{\begin{array}{l}g(t_1,2t_2,...,t_n)t_2\in [0,1/2]\\ k(t_1,2t_2-1,...,t_n)t_2\in [1/2,1]\end{array}\end{array}$$

Hence

$$\begin{array}{c}([f]\star [h])\cdot ([g]\star [k])=\{\begin{array}{l}f(2t_1,2t_2,...,t_n),t_1\in [0,1/2],t_2\in [0,1/2]\\ h(2t_1,2t_2-1,...,t_n),t_1\in [0,1/2],t_2\in [1/2,1]\\ g(2t_1-1,2t_2,...,t_n),t_1\in [1/2,1],t_2\in [0,1/2]\\ k(2t_1-1,2t_2-1,...,t_n),t_1\in [1/2,1],t_2\in [1/2,1]\end{array}\end{array}$$

Compare the LHS and RS we can see that these are not only homotopic, but exactly the same map.

$$([f]\cdot [g])\star ([h]\cdot [k])=([f]\star [h])\cdot ([g]\star [k])◻$$