Teacher: Thomas Ehrhard

Recap: about Scott-continuity

Let $P,Q$ be two partially ordered sets.

$f:P⟶Q$ is Scott-continuous:

iff for every directed subset $D\subseteq P$ such that $\bigvee D\in P$, $\bigvee f(D)$ exists in $Q$ and $f(\bigvee D)=\bigvee f(D)$

The Scott topology on $P$:

is the topology whose opens are upper sets (sets upper-closed) that are inaccessible by directed joins (i.e. if $D$ is directed and $\bigvee D\in O$, then $D\cap O\ne \varnothing$)

Lemma: $f:P⟶Q$ is Scott-continuous iff it is continuous for the Scott-topology

$⟹$: Let $V\subseteq Q$ be an open set. Let’s show that $f^{-1}(V)\subseteq P$ is open.

• upper set: if $x\in f^{-1}(V)$ and $y\ge x$, then as $f$ is monotone:

  $$V\ni f(x)\le f(y)\stackrel{\text{as }V\text{ is an upper-set}}{\in }V$$

• inaccessible by directed joins: if $D$ is directed and $\bigvee D\in f^{-1}(V)$, then if there existed a $x\in f^{-1}(V)\cap D$, then we would have $f(x)\in V\cap f(D)=\varnothing$ as $f(D)$ is directed, $\bigvee f(D)=f(\bigvee D)\in V$, but $V$ is inaccessible by directed joins.

$⟸$: Let $D\subseteq P$ be a directed set such that $\bigvee D\in P$.

$f:Cl(E)⟶Cl(F)$ monotone is Scott-continuous iff $\forall x\in Cl(E),\forall b\in |F|,b\in f(x)⟹\exists x_0\in {𝒫}_f(x);b\in f(x_0)$

Exercise Sheet: Problem statement

1.

Let $f:Cl(E)⟶Cl(F)$

Assume $f$ is linear. Let $(x_i{)}_{i\in I}\in Cl(E{)}^I$ such that

• $I$ is at most countable
• $i\ne j⟹x_i\cap x_j=\varnothing$
• $\bigcup _{i\in I}{x}_i\in Cl(E)$

Let’s show that $f$ preserves disjointness, compatibility of the family, and preserves the sup of $(x_i)$.

• Disjointness preservation: If $i\ne j$

  $x_i\cup x_j\subseteq \bigcup x_k\in Cl(E)$, so $x_i\uparrow x_j$, and by conditional multiplicativity and linearity:

  $$\varnothing =f(\varnothing )=f(x_i\cap x_j)=f(x_i)\cap f(x_j)$$

• Preserves the sup:

  $$f(\bigcup _{i\in I}{x}_i)=f(\bigcup _{J{\subseteq }_{finite}I}\bigcup _{j\in J}{x}_j)\stackrel{\text{continuity}}{=}\bigcup _{J{\subseteq }_{finite}I}f(\bigcup _{j\in J}{x}_j)$$

  since $(\bigcup _{j\in J}{x}_j{)}_{J{\subseteq }_{f}I}$ is directed in $Cl(E)$, and

  $$\bigcup _{J{\subseteq }_{finite}I}f(\bigcup _{j\in J}{x}_j)\stackrel{\text{linearity}}{=}\bigcup _{J{\subseteq }_{finite}I}\bigcup _{j\in J}f(x_j)=\bigcup _{i\in I}f(x_i)$$

Conversely, suppose $f$ preserves disjointness, compatibility of the family, and preserves the sup of $(x_i)$.

$f$ is monotone

Let $x\subseteq y$. As $y\mathrm{\setminus }x$ is a clique of $E$ and $y\mathrm{\setminus }x$ and $x$ are disjoint:

Scott-continuity of $f$

Fact: $f:Cl(E)⟶Cl(F)$ monotone is Scott-continuous iff

$$\forall x\in Cl(E),\forall b\in |F|,b\in f(x)⟹\exists x_0\in {𝒫}_f(x);b\in f(x_0)$$

Scott-continuity: to have a finite approximation in the output, you only need a finite approximation in the input.

Let $x\in Cl(E),b\in |F|$ such that $b\in f(x)$.

Then

And there exists a singleton $\{a\}$ that is suitable.

Actually, we have shown:

Lemma:

$$\forall x\in Cl(E),\forall b\in f(x),\exists !a\in x;b\in f(\{a\})$$

Uniqueness comes from the fact that if you have two distinct suitable $a$ and $a^{\prime }$, then as $f$ preserves disjointness $b\in f(\{a\})\cap f(\{a^{\prime }\})=\varnothing$: contradiction.

Conditional multiplicativity

As $f$ is monotone, $f(x\cap y)\subseteq f(x)\cap f(y)$. So we need to show

Let $b\in f(x)\cap f(y)$, then there exists unique $a\in x$, $a^{\prime }\in y$ such that

So as $\{a,a^{\prime }\}\in Cl(E)$ (since $x\cup y\in Cl(E)$), $a=a^{\prime }$.

Other conditions

$f(\varnothing )=\varnothing$ for the empty family

Let $x,y\in Cl(E)$ st $x\cup y\in Cl(E)$. As $f$ is monotone:

And conversely:

2.

2.1.

Assume $f$ is bilinear, show that it is stable.

Monotone

Continuous

Let $D\stackrel{\scriptscriptstyle\mathrm{def}}{=}(x_i,y_i{)}_{i\in I}$ be a directed set.

Then

• $D_1\stackrel{\scriptscriptstyle\mathrm{def}}{=}(x_i{)}_{i\in I}$ directed in $Cl(E_1)$
• $D_2\stackrel{\scriptscriptstyle\mathrm{def}}{=}(y_i{)}_{i\in I}$ directed in $Cl(E_2)$

Stable

• $(x_1,x_2)$ and $(y_1,y_2)$ compatible
• i.e. $x_1,y_1$ compatible and $x_2,y_2$ compatible

Lemma: $f:Cl(E)⟶Cl(F)$ is stable iff $f$ is monotone and

$$\begin{array}{c}\forall x\in Cl(E),\forall b\in f(x),\exists x_0\in {𝒫}_f(x);b\in f(x_0)\\ \text{ and }{x}_0\text{ is minimal: }\forall x^{\prime }\subseteq x,(b\in f(x^{\prime })⟹x_0\subseteq x^{\prime })\end{array}$$

and

If $f:Cl(E_1)\times Cl(E_2)⟶Cl(F)$ is separately stable, then it is stable

Bilinear and linear

So suppose $f$ is linear and bilinear: then for all $x_1,x_2$

Example of a function which is bilinear but not linear:

• $E_1=E_2=F=\{\ast \}$
• $f((\varnothing ,\varnothing ))=\varnothing$
• $f((\varnothing ,\{\ast \}))=\varnothing$
• $f((\{\ast \},\varnothing ))=\varnothing$
• $f((\{\ast \},\{\ast \}))=\{\ast \}$

2.2.

You can straightforwardly use the characterisation of question 1, as intersection and union is taken pointwise.

2.3.

$f,g:Cl(E)⟶Cl(F)$ linear: $f=g⟺\forall a\in |E|,f(\{a\})=g(\{a\})$

$\tilde{f}$ defined on singletons:

Uniqueness: