Generalized Inverse Limits

A generalized inverse limit of arcs is given by \(\varprojlim \{I, f_{\alpha \beta}, \kappa\}\) where

• \(I=[0,1]\)
• \(\kappa\) is an ordinal (often a cardinal)
• For each \(\alpha<\beta<\kappa\), \(f_{\alpha\beta}:I\to 2^I\) is an upper semicontinuous (u.s.c.) function which maps points to closed subspaces of \(I\).
  • Equivalently, a relation in \(I^2\) which is a closed subset, and for each \(x\), \(xf_{\alpha\beta}y_x\) for some \(y_x\).

\(\varprojlim \{I, f_{\alpha \beta}, \kappa\}\) is the collection of sequences \(\mathbf x\) in \(I^\kappa\) such that \(\mathbf x(\alpha)\in f_{\alpha\beta}(\mathbf x(\beta))\).

Necessarily, we must assume that \(f_{\alpha \gamma}=f_{\alpha \beta}\circ f_{\beta \gamma}\) for all \(\alpha<\beta<\gamma\).

One way to guarantee this is to require \(f_{\alpha \beta} = f\) for all \(\alpha<\beta\); such a function must be idempotent, as then \(f = f\circ f = f^2\).

Examples of idempotent u.s.c. functions

Example of \(\varprojlim \{I, f, \kappa\}\)

\[ f(x) = \begin{cases} [0,1] & x= 0 \\ \{1\} & x\not= 0 \end{cases} \]

It follows then that each sequence in \(\varprojlim \{I, f, \kappa\}\) must be of the form \[ \mathbf x_{\gamma,z}(\alpha) = \langle 1,1,\dots,z,0,0,\dots\rangle = \begin{cases} 0 & \alpha>\gamma \\ z & \alpha=\gamma \\ 1 & \alpha<\gamma \end{cases} \] for some \(\gamma\leq\kappa,z\in [0,1)\)

\[ \mathbf x_{\gamma,z}(\alpha) = \langle 1,1,\dots,z,0,0,\dots\rangle = \begin{cases} 0 & \alpha>\gamma \\ z & \alpha=\gamma \\ 1 & \alpha<\gamma \end{cases} \]

Then \(h(x_{\gamma,z})=\langle \gamma,z\rangle\) defines a homeomorphism from \(\varprojlim \{I, f, \kappa\}\) to the lexicographic order on \(\kappa\times [0,1)\cup\{\langle\kappa,0\rangle\}\); a.k.a. the long compactified ray of length \(\kappa\).

Metrizability in \(\varprojlim \{I, f_{\alpha \beta}, \kappa\}\)

Fact: provided the bonding maps are surjective u.s.c., \(\varprojlim \{I, f_{\alpha \beta}, \kappa\}\) is a closed and therefore compact subspace of \(I^\kappa\); if the bonding maps are continuum-valued, then \(\varprojlim \{I, f_{\alpha \beta}, \kappa\}\) is connected.

Of course, this guarantees our result is a Hausdorff continuum, but what about metrizability? No problem provided \(\kappa<\omega_1\). No problem provided \(f\) is trivial.

Corson Compactness

Say a space is Corson compact in the case that it may be embedded as a compact subspace of \(\Sigma\mathbb R^\kappa\) for some cardinal \(\kappa\).

• \(\Sigma\mathbb R^\kappa\) denotes all sequences of \(\mathbb R^\kappa\) with only countably many non-zero coordinates
• Obviously, closed subspaces of Corson compacts are Corson compact.
• Metric continuua may be embedded in \(I^\kappa\), and are therefore Corson compact.
• Compactified long rays and compactified linearly ordered uncountable cardinals (\(\kappa +1\)) are not Corson compact.

Say a bonding map satisfies condition \(\Gamma\) if there exist \(x,y\in I\) such that \(\langle x,x\rangle\), \(\langle y,y\rangle\), and \(\langle x,y\rangle\) all belong to the graph of the map.

Then by restricting to \(\{x,y\}\), we may easily find the points \[ \mathbf x_{\gamma}(\alpha) = \langle y,y,\dots,x,x,x,\dots\rangle = \begin{cases} x & \alpha\geq\gamma \\ y & \alpha<\gamma \end{cases} \] for \(\gamma\leq\kappa\). This is a copy of \(\kappa+1\) where \(\mathbf x_\gamma \leftrightarrow \gamma\) and \( U[\mathbf x_\gamma,\{\beta,\gamma\},\frac{1}{2}] \leftrightarrow (\beta,\gamma] \).

Condition \(\Gamma\) seems to pop up a lot.

From our doodles, we couldn't find non-trivial examples where \(\Gamma\) doesn't hold.

Showing \(\Gamma\)

Following illustrations show why \(\Gamma\) holds if \(f\) is nontrivial, idempotent, continuum-valued, and u.s.c.:

Note: most cases exploit the fact that if \(f(x)=[a,b]\), then \(f([a,b])=[a,b]\).

If \(\Delta\subseteq G(f)\)...

\(\Gamma\) is easily found to hold as seen above.

Can we avoid \(\Delta\)?

The graph of \(f\) cannot be connected.

Can we force exactly one point on \(\Delta\)?

Two remaining cases...

Case 1: two points on \(\Delta\) with nothing between

Case 2: exactly one incomplete fragment of \(\Delta\)

In conclusion: any nontrivial, idempotent, continuum-valued, and u.s.c. bonding map has condition \(\Gamma\).

As a result, a copy of \(\kappa+1\) may be found within \(\varprojlim \{I, f, \kappa\}\cap\{x,y\}^\kappa\).

Since \(\kappa\geq\omega_1\) implies \(\kappa+1\) is not Corson compact, it follows that \(\varprojlim \{I, f, \kappa\}\) cannot be Corson compact, much less metrizable!