# Set Theory, Notes 1: Ordinals and Cardinals

Unless otherwise specified, we will assume ZF axioms.  Recall the following ZF axiom:

Axiom of Infinity:

$(\exists\omega)(\varnothing\in\omega\wedge(\forall x\in\omega)(x\cup\{x\}\in\omega)).$

We will call $\omega$ the set of natural numbers.  That is,

$\begin{array}{rcl}0&=&\varnothing\\1&=&0\cup\{0\}=\{0\}\\2&=&1\cup\{1\}=\{0,1\}\\&\vdots&\\n+1&=&n\cup\{n\}\\&\vdots&\end{array}$

We can then define $\omega+1=\omega\cup\{\omega\}$ and iterate as before, whence by applying the axiom of infinity again we can obtain $\omega\cdot 2:=\omega+\omega,$ and so on.  All such sets generated by this process are called ordinals.  This in turn gives us the canonical linear ordering of the ordinals where $n iff $n\in m.$

Definition 1.  An ordinal $\alpha$ is a successor ordinal iff $\alpha=\beta+1.$  A limit ordinal is an ordinal which is not a successor ordinal.

Proposition 2.  (Transfinite Induction)  Let $Ord$ denote the class of all ordinals (in accordance to VBG notion of class) and $C$ be a class.  If

1. $0\in C,$
2. $\alpha\in C\Longrightarrow\alpha+1\in C,$ and
3. if $\gamma$ is a limit ordinal and $\alpha\in C$ for all $\alpha<\gamma,$ then $\gamma\in C,$

then $C=Ord.$

Proof.  Since $<$ is a linear ordering on the ordinals, let $\alpha$ be the least ordinal such that $\alpha\notin C.$  But then $\alpha+1\in C$ which is a contradiction.  Hence $C=Ord.$

We can index ordinals with ordinals to generate the notion of a sequence of ordinals.  We define an nondecreasing (nonincreasing) sequence of ordinals an ordered set $\{\gamma_\alpha\}$ of ordinals where $\gamma_\alpha\leq\gamma_\beta$ iff $\alpha\leq\beta.$

Definition 3.  Let $\{\gamma_\alpha\}$ be an nondecreasing sequence of ordinals and $\xi$ be a limit ordinal and $\alpha<\xi.$  Then we define the limit of the sequence as

$\displaystyle\lim_{\alpha\to\xi}\gamma_\alpha=\sup_{\alpha<\xi}\{\gamma_\alpha\}.$

A dual definition can be defined for nonincreasing sequences, in which case the limits can be respectively distinguished as left and right limits.  A sequence $\{\gamma_\alpha\}$ is continuous if for every limit ordinal $\xi$ in the indexing subclass we have

$\displaystyle\lim_{\alpha\to\xi}\gamma_\alpha=\gamma_\xi.$

An example of a sequence which is not continuous may be one of the form $S=(...,\gamma_{\beta},\gamma_{\beta+1},...)$ where $\gamma_{\beta},\gamma_{\beta+1}$ are both limit ordinals.  So in this case

$\displaystyle\lim_{\alpha\to\beta+1}\gamma_\alpha=\sup_{\alpha<\beta+1}\{\gamma_\alpha\}=\gamma_{\beta}\neq\gamma_{\beta+1}$

(since the sup is actually a max in this case).

Definition 4.  (Ordinal Arithmetic)  We define

1. $\alpha+0=\alpha,$
2. $\alpha+(\beta+1)=(\alpha+\beta)+1,$
3. $\alpha+\beta=\lim_{\gamma\to\beta}\alpha+\gamma.$

Multiplication:

1. $\alpha\cdot 0=0,$
2. $\alpha\cdot(\beta+1)=\alpha\cdot\beta+\alpha,$
3. $\alpha\cdot\beta=\lim_{\gamma\to\beta}\alpha\cdot\gamma.$

Exponentiation:

1. $\alpha^0=1,$
2. $\alpha^{\beta+1}=\alpha^\beta\cdot\alpha,$
3. $\alpha^\beta=\lim_{\gamma\to\beta}\alpha^\gamma.$

It follows that addition and multiplication are both associative, but not commutative.  In particular one can see that

$1+\omega=\omega\neq\omega+1$

and

$2\cdot\omega=\omega\neq\omega\cdot 2=\omega+\omega.$

Definition 5.  For a set $X,$ we define its cardinality, denoted $|X|,$ as the unique ordinal with which the set has a bijection.  The corresponding subclass of ordinals is called the class of cardinals.

Proposition 6.  If $|X|=\kappa,$ then $|P(X)|=2^\kappa.$

Proof.  For every $A\subseteq X,$ define

$\displaystyle\chi_A(x)=\left\{\begin{array}{ll}1&x\in A\\ 0&x\in X-A\end{array}\right..$

Hence the mapping $f:A\mapsto\chi_A(X)$ is a bijection between $P(X)$ and $\{0,1\}^X.$

Hence in this context, Cantor’s theorem immediately follows: $|X|<|P(X)|.$

Proposition 7.  Let $|A|=\kappa,$ $|B|=\lambda,$ and $A\cap B=\varnothing.$  Then

1. $|A\cup B|=\kappa+\lambda,$
2. $|A\times B|=\kappa\cdot\lambda,$
3. $\left|A^B\right|=\kappa^\lambda.$

Proof.  We have

$|A\cup B|=|A\sqcup B|=|\kappa\sqcup\lambda|=\kappa+\lambda.$

Also if $f:A\to\kappa$ and $B\to\lambda$ are bijections, then we can define $h:A\times B\to\kappa\cdot\lambda$ by $h:(a,b)\mapsto f(a)\cdot g(b)\in\kappa\cdot\lambda,$ and this is easily seen to be a bijection.

And if $k\in A^B$ then we can define $h:k\mapsto k(b)^b\in\kappa^\lambda,$ which is also seen to be a bijection.

It thus follows that addition and multiplication of cardinals are commutative (that is, ordinal operations are commutative on this subclass).

Above we said $1+\omega=\omega\neq\omega +1$ and $2\cdot\omega=\omega\neq\omega\cdot 2.$  But

$\omega=|\omega|=|1+\omega|=|\omega+1|$

and

$\omega=|n\cdot\omega|=|\omega\cdot n|=\left|\omega\uparrow\uparrow n\right|$

for any finite ordinal $n$ with Knuth notation

$\displaystyle\omega\uparrow\uparrow n=\omega^{\omega^{\cdot^{\cdot^{\cdot^\omega}}}}$

having $n$ raisings of $\omega.$  Consider the following convention for defining countable infinite ordinals:

$\displaystyle \omega,\omega+1,...,\omega\cdot 2,...,\omega^2,...,\omega^\omega,...,\omega\uparrow\uparrow\omega=\varepsilon_0,...,\varepsilon_1=\varepsilon_0\uparrow\uparrow\varepsilon_0,...,\\\varepsilon_2=\varepsilon_1\uparrow\uparrow\varepsilon_1,...,\varepsilon_\omega,...,\varepsilon_{\varepsilon_0},...,\varepsilon_{\varepsilon_{\ddots}},...$

They are all called countable infinite ordinals since for any one of them $\alpha,$ $|\alpha|=\omega.$  To clarify when we are talking about cardinal numbers versus ordinal numbers, we will use aleph notation: $\aleph_0=\omega.$  From Cantor’s theorem above, we know that if $|X|=\aleph_0,$ then $|P(X)|=2^{\aleph_0}>\aleph_0.$  That is, the ordinal corresponding to $2^{\aleph_0}$ must be greater than all of the countable ordinals above, otherwise its cardinality would be $\aleph_0.$  This necessitates the notion of uncountable ordinals and corresponding uncountable cardinals.  We use subscripts to characterize these: $\aleph_1=\omega_1,\aleph_2=\omega_2,$ etc.

Definition 8.  An infinite cardinal $\aleph_\alpha$ is a successor cardinal iff $\alpha$ is a successor ordinal, and it is a limit cardinal iff $\alpha$ is a limit ordinal.

Definition 9.  Let $\alpha$ be a limit ordinal.  An increasing $\delta$sequence $(\beta_\gamma)_{\gamma<\delta}$ with $\delta$ a limit ordinal is cofinal in $\alpha$ if $\lim_{\gamma\to\delta}\beta_\gamma=\alpha.$  And if $\alpha$ is an ordinal, then we define its cofinality as

$\displaystyle\text{cf}\,\alpha=\inf\left\{\delta:\lim_{\gamma\to\delta}\beta_\gamma=\alpha\right\}.$

It is easy to verify that $\text{cf}\,\alpha=1$ iff $\alpha$ is a successor ordinal.  Also $\text{cf}\,0=0,\text{cf}\,\omega=\omega,$ and $\text{cf}\,\omega_\alpha=\omega_\alpha$ for any finite ordinal $\alpha.$

Proposition 10.  $\mbox{cf}\,\mbox{cf}\,\alpha=\mbox{cf}\,\alpha.$

Proof.  Let $\mbox{cf}\,\alpha=c.$  Then $\lim_{\gamma\to c}\beta_\gamma=\alpha$ (in particular it is the smallest such $c$).  Now if $\mbox{cf}\,\mbox{cf}\,\alpha=\mbox{cf}\,c=d,$ then certainly $d\leq c.$  Now since $(\beta_\gamma)$ is cofinal in $\alpha,$ a subsequence of indices $(\gamma_\delta)$ is cofinal in $c$ (where cofinality can be chosen to be $d$).  So

$\displaystyle\lim_{\delta\to d}\gamma_\delta=c.$

But then $\left(\beta_{\gamma_\delta}\right)$ is cofinal in $\alpha.$  That is,

$\displaystyle\lim_{\delta\to d}\beta_{\gamma_\delta}=\alpha,$

whence $d\geq c.$

Definition 11.  An ordinal $\alpha$ is regular if $\mbox{cf}\,\alpha=\alpha.$  It is singular if it is not regular.

Corollary 12.  If $\alpha$ is a limit ordinal, then $\mbox{cf}\,\alpha$ is a regular cardinal.

Theorem 13.  If $\kappa$ is an infinite cardinal, then $\kappa<\kappa^{\mbox{cf}\,\kappa}.$

[1]  Jech, Thomas.  Set Theory.  3rd Edition.  Springer Monographs in Mathematics.  Springer-Verlag.  2000.

# Fractional Sobolev Spaces

Recall for a function $f\in L^1(\Omega)$ for $\Omega\subseteq\mathbb{R}^n$ we define the Fourier transform of $f$ by

$\displaystyle\hat{f}(x)=F(f)(x)=\int_\Omega e^{-2\pi ix\cdot\xi}f(\xi)\,d\xi$

and the inverse Fourier transform by

$\displaystyle \check{f}(x)=F^{-1}(f)(x)=\int_\Omega e^{2\pi i x\cdot\xi}f(\xi)\,d\xi.$

If $f,\hat{f}\in L^1(\Omega),$ then $\check{\hat{f}}=\hat{\check{f}}=f$ a.e..

Now if $f\in L^1$ and $D^\alpha (f)\in L^1$ with $\alpha\in\mathbb{Z}^n,$ we say $D^\alpha f$ is the weak derivative of $f$ provided

$\displaystyle\int_\Omega f D^\alpha\phi\,dx=(-1)^{|\alpha|}\int_\Omega D^\alpha(f)\phi\,dx$

with $\phi\in C_C^\infty(\Omega),$ $|\alpha|=\sum_i\alpha_i,$ and

$\displaystyle D^\alpha\phi=\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n}}\phi.$

The Sobolev space $W^{k,p}(\Omega)$ is then defined as the set of all functions $f\in L^p(\Omega)$ such that $D^\alpha(f)\in L^p(\Omega)$ for all $|\alpha|\leq k.$  These become Banach spaces under the norm

$\displaystyle\|f\|_{k,p}=\left(\sum_{|\alpha|\leq k}\int_\Omega \left|D^\alpha(f)\right|^p\,dx\right)^{1/p}$

and Hilbert when $p=2,$ whence we denote $H^k(\Omega)=W^{k,2}(\Omega).$

For $1 we can say that $W^{k,p}(\mathbb{R}^n)$ consists of $f\in L^p(\mathbb{R}^n)$ such that

$\displaystyle F^{-1}\circ(1+|2\pi x|^2)^{k/2}F\circ f=(1-\Delta)^{k/2}f(x)\in L^p(\mathbb{R}^n)$

(using Fourier multipliers).

Bessel Potential Spaces:

We can generalize with $r\in\mathbb{R}$ and $1 and define the Bessel potential space $W^{r,p}(\mathbb{R}^n)$ as all $f\in L^p$ such that $\|f\|_{r,p}:=\|(1-\Delta)^{r/2}f\|_p<\infty.$  Note the fractional power of the Laplacian is defined by virtue of the fact that $L^p\cap L^1$ is dense in $L^p$, so we say

$\displaystyle(1-\Delta)^{r/2}(f):=(F^{-1}\circ (1+|2\pi x|^2)\circ F)(f).$

These spaces satisfy our desire of being Banach (and Hilbert when $p=2$).

Sobolev-Slobodeckij Spaces:

There is an alternative approach.  Recall that the Holder space $C^{k,\gamma}(\Omega)$ is defined as all functions $f\in C^k(\Omega)$ such that

$\displaystyle\|f\|_{C^{k,\gamma}}=\sum_{|\alpha|\leq k}\sup_{x\in\Omega}|D^\alpha|+\sum_{|\alpha|=k}\sup_{x,y\in\Omega}\frac{|D^\alpha f(x)-D^\alpha f(y)|}{|x-y|^\gamma}<\infty.$

That is, it is the set of functions on $\Omega$ which are $C^k$ and whose $k$-th partial derivatives are bounded and Holder continuous of degree $\gamma.$  These spaces are Banach under the above norm.  We can generalize the Sobolev spaces to incorporate similar properties.  Let us define the Slobodeckij norm for $f\in L^p(\Omega)$ with $1\leq p<\infty$ and $\theta\in (0,1)$ by

$\displaystyle [f]_{\theta,p}=\int_\Omega\int_\Omega\frac{|f(x)-f(y)|^p}{|x-y|^{\theta p+n}}\,dxdy.$

The corresponding Sobolev-Slobodeckij space $W^{s,p}(\Omega)$ is defined as all functions $f\in W^{\lfloor s\rfloor,p}(\Omega)$ such that

$\displaystyle\sup_{|\alpha|=\lfloor s\rfloor}[D^\alpha f]_{\theta,p}<\infty$

where $\theta=s-\lfloor s\rfloor\in(0,1).$  This becomes a Banach space under the norm

$\displaystyle\|f\|_{s,p}=\|f\|_{W^{\lfloor s\rfloor,p}}+\sup_{|\alpha|=\lfloor s\rfloor}[D^\alpha f]_{\theta,p}.$

[1] http://en.wikipedia.org/wiki/Sobolev_space (unclear text references)

[2]  Lieb, Elliot and Michael Loss.  Analysis.  2nd Edition.  Graduate Studies in Mathematics.  Vol. 14.  American Mathematical Society.  2001.

[3]  Evans, Lawrence.  Partial Differential Equations.  Graduate Studies in Mathematics.  Vol. 19.  American Mathematical Society.  1998.

# Cohomology of Pointed Sets

Let $M$ be a monoid.  Then $M$ is a pointed set with point $1.$  Let $\varphi:(M,0)\times(X,x_0)\to (X,x_0)$ be a map satisfying

$\varphi(1,x)=x,$ and

$\varphi(mn,x)=\varphi(m,\varphi(n,x)).$

Such a map is called a monoidal action, and we correspondingly call $X$ an $M$-pointed set.  Note that since $\varphi(1,x_0)=x_0,$ it is a pointed morphism.  We will also write $mx:=\varphi(m,x)$ for short.

Suppose $M$ acts on $X$ and that $X$ is a differential pointed set (so $d^2(x)=x_0$).  Now define the pointed set

$\displaystyle Hom_M(X,M)=\{f:X\to M:f(x_0)=1\}$

with point $1:X\to M$ defined by $1(x)=1$ for all $x\in X.$  Then this is a differential pointed set with differential defined by

$(df)(x)=f(dx)$

since

$(d^2f)(x)=d\left(f(dx)\right)=(df)(dx)=f(d^2x)=f(x_0)=1$

for all $x\in X,$ so $d^2f=1.$  We then define the cohomology of $X$ with coefficients in $M$ as

$\dot{H}(M,X)=[1]-(dHom(X,M)-\{1\})$

where of course $[1]$ is the equivalence class of maps under the relation $f\sim g$ iff $df=dg=1.$

Let $X$ be an $M$-pointed set.  Consider the two sets

$X^M=\{x\in X:mx=x\}$

$X_M=X/\sim$

where $x\sim y$ iff $mx=y$ or $x=my$ for some nonzero $m\in M.$  (Note the “or” gives us symmetry).  $x_0$ belongs to both of these (as a class of elements in the latter case such that $mx=x_0$ (since $mx_0=x$ is never true for nonzero $m$ and $x\neq x_0$)).  The point (or trivial class) in $X_M$ is called the torsion subpointed set of $X$ with respect to $M.$

Proposition 1.  Let $X$ and the pointed natural numbers $(\mathbb{N},0)$ be $M$-pointed monoids.  Then

$X^M=Hom_M(\mathbb{N},X).$

Proof.  Let $m\in M$ and $n\in\mathbb{N}$ and define $\phi:X^M\to Hom_M(\mathbb{N},X)$ be defined by $m\phi(x)(n)=\sum_{i=1}^nx:=nx$ (since $mx=x$).  The map is injective since if $\phi(x)=\phi(y),$ then $nx=ny$ for all $n,$ but this just implies $x=y.$  As $\mathbb{N}$ is a monoid with one generator, any morphism $\varphi:\mathbb{N}\to X$ is determined by the image of $1.$  But such a map induces a preimage under $\phi:$ $\phi^{-1}(\varphi)=\varphi(1)\in X_M.$  And we have

$\phi(0_X)(n)=n0_X=0_X.$

So $\phi$ is a morphism.

We don’t however have that $X_M=\mathbb{N}\times X$ without some way of moving natural numbers between components of $\mathbb{N}\times X$ (i.e. a tensor product).  We can define a tensor product of two $M$-pointed sets $X$ and $Y$ which are monoids, denoted $X\odot_M Y,$ as $X\times Y/\sim$ with the usual bilinear relations.  We will denote simple elements by $x\odot y.$  It’s easy to verify that this is a monoid.  We then have the result:

Proposition 2.  Let the hypotheses of Proposition 1 hold, then

$X_M=\mathbb{N}\odot_{\mathbb{N}[M]} X.$

Proof.     Define $\phi:\mathbb{N}\odot_M X\to X_M$ by $\phi(n\odot x)=nx.$  Suppose $nx=n'y.$  Then

$\begin{array}{lcl}1\odot nx=1\odot n'y&\Rightarrow&1\odot n(1_Mx)=1\odot n'(1_My)\\&\Rightarrow&1\odot (n1_M)x=1\odot (n'1_M)y\\&\Rightarrow&n1_M\odot x=n'1_M\odot y\\&\Rightarrow&n\odot x=n'\odot y.\end{array}$

Also if $x\in X_M,$ then since action $mx=x$ for all $m\in M,$ we have a preimage $\phi^{-1}(x)=1\odot x.$  And lastly $\phi(0,0_X)=00_X=0_X,$ so it is a morphism.

If the functors $\cdot^M$ and $\cdot_M$ are left (right) exact functors in the category of pointed sets and $X$ has an injective (projective) resolution, then we can define the $M$-pointed cohomology (homology) via the right (left) derived functors $D^n(F,M)=H^n(F(I)^M)$ ($D_n(F,M)=H_n(F(P)_M)$).

# Homology on Pointed Sets 2

We wish to address four of the Eilenberg-Steenrod axioms for this category.  First note that for dimension we have $H(x_0)=x_0$ where $x_0$ denotes the trivial pointed set.

If $(X,x_0)$ and $(Y,y_0)$ are pointed sets, then we define the intuitive product pointed set as the pointed set $(X\times Y,(x_0,y_0)).$  If $X$ and $Y$ are also differential sets with differentials $d_1$ and $d_2,$ then we define the product differential component-wise: $d_1\times d_2 (x,y)=(d_1(x),d_2(y)).$  This is clearly a differential on $X\times Y,$ which we in turn call the product differential set.

If we then define the relation $\sim$ on $X\times Y$ where $(x,y)\sim (x',y')$ iff $(d_1(x),d_2(y))=(d_1(x'),d_2(y')),$ then we have $[(x,y)]_{d_1\times d_2}=[x]_{d_1}\times [y]_{d_2}.$  Hence we have

$\begin{array}{lcl}H(X\times Y)&=&[(x,y)]-(d(X\times Y)-(x_0,y_0))\\&=&[x]\times[y]-\left((d_1(X),d_2(Y))-(x_0,y_0)\right)\\&=&\left([x]-(d_1(X)-\{x_0\}),[y]-(d_2(Y)-\{y_0\})\right)\\&=&H(X)\times H(Y).\end{array}$

We can similarly define a formal sum of pointed spaces $\sqcup_i(X_i,x_i)$ whose corresponding homology is just defined as

$H\left(\bigsqcup_i(X_i,x_i)\right)=\bigsqcup_iH(X_i).$

So we have two forms of additivity.  We also have excision, for if $(X,A)$ is a pair of sets with $A$ a sub-differential set of $X$ and $U\subseteq A,$ then

$\begin{array}{lcl}H(X-U,A-U)&=&[x_0]\cap\left(X-U-\left((A-U)-\{x_0\}\right)\right)-d\left(X-U-(A-U)\right)\\&=&[x_0]\cap\left(X-(A-\{x_0\})\right)-d(X-A)\\&=&H(X,A).\end{array}$

We lastly address homotopy and omit the long exact sequence as we have not developed a dimensional concept (although recall we constructed a short exact sequence under an assumption on $A$).  Let $f:(X,A)\to (Y,B)$ be a map between paired sets.  We say this is a paired set morphism if $f(A)\subseteq B.$  If we furthermore have that these are paired pointed sets $(X,A,x_0)$ and $(Y,B,y_0)$ with $A$ and $B$ sub-pointed sets, a paired set morphism $f$ between them is a paired pointed morphism if we also have that $f(x_0)=y_0.$  If both are differential sets as well, we can define a paired differential morphism if we further have $f\circ d_1=d_2\circ f.$  Let $f$ be a paired differential morphism between such sets.  Then we would like for its restriction $f:H(X,A)\to H(Y,B)$ to be well-defined.  Hopefully some combination of requirements like normality and being a paired differential morphism will work, but I have not yet convinced myself.  If it worked, we would in turn just define two maps $f,g:(X,A)\to (Y,B)$ to be homotopic mod $A$ provided they agreed on $H(X,A).$

# Homology on Pointed Sets

Let $(X,x_0)$ be a pointed set (i.e. there is a nullary operation $x_0:\varnothing\to X$ where $x_0(\,)=x_0$).

Definition 1.  A pointed set is a differential set if there is an endomorphism $d:X\to X$ (i.e. $d(x_0)=x_0$) such that $d^2(x)=x_0$ for all $x\in X.$  We say two elements $x,y\in X$ are homologous, denoted $x\sim y,$ if $d(x)=d(y).$

Proposition 2.  The homologous relation is an equivalence relation.

Definition 3.  Elements of the equivalence class $[x_0]$ are called cycles.  Elements of the set $d(X)$ are called boundaries.  The homology of $X$ is defined to be the set

$H(X)=[x_0]-\left(d(X)-\{x_0\}\right).$

Note that all boundaries are cycles, and $x_0\in H(X),$ so $H(X)$ can have a pointed set structure with point $x_0.$  Now let the quotient set $X/\sim$ be pointed as $(X/\sim,[x_0]),$ and note that $(d(X),x_0)$ can be pointed since $d(x_0)=x_0.$

Proposition 4.  $(X/\sim,[x_0])\approx (d(X),x_0)$ as pointed sets.

Proof.     Since the equivalence class $[x]=\{y:d(x)=d(y)\},$ it follows that the map $[x]\mapsto d(x)$ is injective.  This map is also clearly surjective since $d$ is defined on $X,$ which is partitioned via $\sim.$  And of course we also have $[x_0]\mapsto x_0.$

Now suppose we have a sequence $\{X_i,x_i\}$ of pointed sets and a sequence $\{d_{i+1}:X_{i+1}\to X_i\}$ of pointed set homomorphisms such that $d_i\circ d_{i+1}(x)=x_{i-1}$ for all $x\in X_{i+1}.$  We will denote such a collection by $\{(X_i,x_i),d_i\}$ and call it a pointed set complex.  We can also define the $i$th homology of the complex as

$H_i(X)=[x_i]-(d_{i+1}(X_{i+1})-\{x_i\})$

where we have the intuitive equivalence relation on each set $X_i$ with $x\sim y\Leftrightarrow d_i(x)=d_i(y).$  We will also call a pointed set complex exact if $H_i(X)=(\{x_i\},x_i)=:0_i$ for all $i.$

Let $(A,x_0)$ be a sub-differential set of a differential set $(X,x_0).$  That is, $d(A)\subseteq A.$  We will call $A$ a normal sub-differential set if for $x\in X$ we have $d(x)\in A\Rightarrow x\in A.$

Definition 5.  We define the relative homology of $X$ with respect to $A$ as the set

$H(X,A)=[x_0]\cap(X-(A-\{x_0\}))-d(X-A).$

Theorem 6.  If $A$ is a normal sub-differential set of $X,$ then there are homomorphisms $i,j$ such that

$x_0\longrightarrow H(A)\stackrel{i}{\longrightarrow}H(X)\stackrel{j}{\longrightarrow}H(X,A)\to x_0$

is an exact complex, where $x_0$ denotes the trivial pointed set $(\{x_0\},x_0)$.

Proof.     Let $x\in H(A),$ then $x\in A,$ $d(x)=x_0,$ and $x$ is not the boundary of any element in $A$ (excluding $x_0$).  Since $x\in A,$ let us just define $i(x)=x.$  This map is fine provided $x$ is not the boundary of any element in $X-A.$  But this is not possible since $A$ is normal, so the map is well-defined.

Now let $x\in H(X).$  Then $x\in X,$ $d(x)=x_0,$ and $x$ is not the boundary of another element in $X.$  We want to send it to a cycle in $X-(A-\{x_0\})$ which is not the boundary of an element in $X-A.$  We already have that $x$ can’t be the boundary of something in $X-A$ since it’s not the boundary of anything in $X.$  So we will define $j$ by

$j(x)=\left\{\begin{array}{lcl}x&\mbox{if}&x\in X-A\\x_0&\mbox{if}&x\in A\end{array}\right.$

The image points are of course boundaries since the final map sends everything to $x_0.$  It follows that $j\circ i(x)=x_0$ for $x\in H(A),$ so we have a differential complex of homology pointed sets.  To prevent confusion, we will denote the homology sets of this complex of homology sets by $H_2,H_1$ and $H_0.$  Since the homology pointed sets will all have $x_0$ as their point, we will simply find the underlying sets.  $H_2=x_0$ since $i$ is just the identity map, that is, $[x_0]_i=\{x_0\}.$  For $H_1$ note that $[x_0]_j=H(A),$ so we have that

$H_1=H(A)-(i(H(A))-\{x_0\})=H(A)-(H(A)-\{x_0\})=\{x_0\}.$

Lastly we have $[x_0]_j=H(X,A)$ and

$j(H(X))=[x_0]\cap(X-(A-\{x_0\}))-(d(X)-\{x_0\}).$

So

$\begin{array}{lcl}H_0&=&H(X,A)-\left([x_0]\cap(X-(A-\{x_0\}))-(d(X)-\{x_0\})\right)\\&=&H(X,A)-\left([x_0]\cap(X-(A-\{x_0\}))-(d(X-A)-\{x_0\})\right)\\&=&H(X,A)-(H(X,A)-\{x_0\})\\&=&\{x_0\}.\end{array}$

$\square$