## Generalized inverse limits indexed by arbitrary total orders

#### UNC Charlotte

2016 March 11th

Presentation for the Spring Topology and Dynamics Conference 2016 Special Session on Continuum Theory at Baylor University

### Abstract

Call an idempotent uppersemicontinuous continuum-valued surjective relation on $$X^2$$ a CV-relation. The presenter and S. Varagona showed that an inverse limit of a linearly ordered compactum indexed by an ordinal and bonded with a single CV-relation is metrizable if and only if the ordinal is countable. This result may be generalized to any totally ordered index.To demonstrate this, the presenter will give a simple characterization for the inverse limit bonded by the simple CV-relation $$\gamma$$ in terms of the lexicographic product of the factor space and linearly ordered index.

## Background and Motivation

#### Bonding relation: $$f\subseteq_{cl} X^2$$

A.K.A. u.s.c. bonding map $$f:X\to C(X)$$

• $$f$$ is surjective.
• $$f$$ is full.

#### Inverse Limit: $$\varprojlim\{X,f,\omega\}$$

• $$\vec x(n)\in f(\vec x(n+1))$$

Assume $$X$$ is a Hausdorff continuum

#### Some properties of $$\varprojlim\{X,f,\omega\}$$: [Charatonik and Roe 2014]

• Nonempty
• Metrizable (if $$X$$ is)
• Compact
• Connected (if $$f$$ continuum-valued)

#### Totally ordered index: $$\varprojlim\{X,f,L\}\subseteq X^L$$

• $$\vec x(\alpha)\in f(\vec x(\beta))$$ for all $$\alpha\lt\beta$$ in $$L$$
• Assume $$f$$ is idempotent: $$f(x)=\{z:\exists y\in f(x)\text{ such that }z\in f(y)\}$$, i.e. $$f=f\circ f$$.

#### Some properties of $$\varprojlim\{X,f,L\}$$:

• Nonempty
• Compact
• Connected (if $$f$$ continuum-valued)
• Hausdorff
• Metrizable (?)

#### Silly example: $$L$$ is countable

$$\varprojlim\{X,f,L\}\subseteq X^L$$ is metrizable.

#### Silly example: the identity relation $$\iota$$

$$\varprojlim\{X,\iota,L\}\cong X$$ is metrizable

We now assume $$f\not=\iota$$.

#### Counterexample:

Varagona first showed that $$\varprojlim\{I,\gamma,\omega_1\}$$ is the closed long ray of length $$\omega_1$$, so not metrizable (or even Corson compact, $$W$$, Frechet-Urysohn, first-countable, etc.)

## The $$\Gamma$$ condition

There exist $$x,y\in I$$ such that $$\langle x,x\rangle,\langle x,y\rangle,\langle y,y\rangle$$ are all in $$f$$.

### The total orders $$\check L$$, $$\hat L$$

For any total order $$L$$, we may define $$\check L=\{A\subseteq L:a\in L,b\lt a\Rightarrow b\in L\}$$ and $$\hat L=\{A\in\check L:A\text{ is closed}\}$$, which are both totally ordered by $$\subseteq$$.

### The LOTS $$\check L$$, $$\hat L$$

Give $$\check L$$ its usual order topology generated by the sets $$(A,B)=\{C\in\check L:A\subsetneq C\subsetneq B\}$$. (Sim. for $$\hat L$$.)

Note $$\check L,\hat L$$ are always compact spaces.

### Metrizability of $$\varprojlim\{X,f,L\}$$

If $$f$$ has $$\Gamma$$, then $$\varprojlim\{X,f,L\}\supseteq\varprojlim\{2,\gamma,L\}=\check L$$

FACT: $$\check L$$ is metrizable iff $$\check L$$ is Corson compact iff $$\check L$$ is second-countable iff $$L$$ is countable

Therefore, $$\varprojlim\{X,f,L\}$$ cannot be metrizable or even Corson compact unless $$f$$ lacks $$\Gamma$$ or $$L$$ is countable.

## Characterizing $$\varprojlim\{X,\gamma,L\}$$

$$\check L$$ contains the lexicographic product $$L\times_{lex} 2$$, adding new points for leftward sets without a supremum.

### LOTS $$M$$

Suppose $$M$$ is a (compact) LOTS with minimum $$0$$ and maximum $$1$$.

A point in $$\varprojlim\{M,\gamma,L\}$$ is a thread which is valued $$1$$ on some closed leftward set $$A\in\hat L$$, except it may have any value of $$M$$ on its supremum (if it exists), and $$0$$ otherwise.

#### Temptation:

$\varprojlim\{M,\gamma,L\}\cong^?\hat L\times_{lex}M$ where $$\langle A,m\rangle$$ corresponds to the thread valued $$m$$ on $$\sup A$$, valued $$1$$ for points in $$A\setminus\{\sup A\}$$, and valued $$0$$ otherwise.

#### Problems:

• If $$l,l+1\in L$$ are successors, we cannot have separate points for the single thread where $$\vec x(l)=1$$ and $$\vec x(l+1)=0$$.
• If $$A\in\hat L$$ has no supremum, its corresponding thread is forced to be $$1$$ on $$A$$ and $$0$$ otherwise, so should correspond to a single point.

### Solution:

$$\varprojlim\{M,\gamma,L\}\cong(\hat L\times_{lex}M )/ \sim$$

• $$\langle(\leftarrow,l),1\rangle \sim \langle(\leftarrow,l],0\rangle$$
• When $$A\in\hat L$$ has no supremum or $$A=\emptyset$$, $$\langle A,m\rangle \sim \langle A,m'\rangle$$

## Future work

• It can be shown that $$\varprojlim\{M,\nu,L\} \cong (\check L\times M)/\sim$$ where $$\langle\emptyset,m\rangle\sim\langle A,0\rangle$$. Any other simple computations?
• Can it be shown that any idempotent $$f$$ satisfies condition $$\Gamma$$?
• Can similar techniques be used for a family of bonding relations $$\{f_{\alpha,\beta}:\alpha\lt\beta\in L\}$$?

### References

• Wlodzimierz J. Charatonik and Robert P. Roe, On Mahavier Products, Topology and its Applications, 166, (2014), 92-97.
• Steven Clontz, Characterizations of Generalized Inverse Limits Indexed by Total Orders, in preparation
• Steven Clontz and Scott Varagona, Destruction of Metrizability in Generalized Inverse Limits, Topology Proc. 48 (2016), 289-297.
• Sina Greenwood and Judy Kennedy, Connected generalized inverse limits, Topology and its Applications, 159 (2012), no. 1, 57-68.
• W. T. Ingram and William S. Mahavier, Inverse limits of upper semi-continuous set valued functions, Houston Journal of Mathematics, vol. 32 (2006) no. 1, 119-130.
• Van Nall, Connected inverse limits with a set-valued function, Topology Proc. 40 (2012), 167-177.
• Scott Varagona, Generalized Inverse Limits Indexed by Totally Ordered Sets, http://arxiv.org/abs/1511.00266
• Patrick Vernon, Inverse limits of set-valued functions indexed by the integers, Topology Applications 171 (2014), 35-40.

# Thank you!

Slides available at Clontz.org.