Non-metric topological continua

Question

What important results hold for non-metric continua, or where can I find a survey of such results?

There are three definitions of a continuum around: a non-empty topological space that is

(1) connected compact metric, or

(2) connected compact Hausdorff [e.g., General Topology by Willard], or

(3) connected compact [ProofWiki].

I am interested in non-trivial properties commonly known for definition (1) that have been found to also hold for definitions (2) or even (3).

Answer 1

To save space, I will say:

(1) a compact, connected, metric space = type I space

(2) compact, connected, Hausdorff = type II

(3) compact and connected = type III

A few results:

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Theorem 1. "existence of non-cut points"

Let $X$ be a connected topological space. Then $c \in X$ is a $\textit{non-cut point}$ of $X$ if $X\setminus\{c\}$ is connected.

It has been known for a long time that type II spaces have at least two non-cut points. However, this paper shows that "any topological space with no non-cut points is not compact". Hence the non-cut point existence theorem also holds for type III spaces.

Proof of non-cut point existence

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Theorem 2. "boundary bumping theorems"

Loosely speaking, the boundary bumping theorems (there are a few variations, hence the plural) state that connected components of certain sets must "bump" (intersect) the boundaries of those sets.

Theorem 5.6 from Nadler's $\textit{Continuum Theory - An Introduction}$: Let $E$ be a proper subset of a type I space $K$. Then the boundary of every connected component of $E$ intersects the boundary of $E$. Kuratowski gives a proof of this theorem for type II spaces in $\textit{Topology II}$.

Intuitively I feel it should also be true for type III spaces, but I haven't found a proof :)

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Theorem 3. "type III spaces are horrible"

Let $S = \{0, 1\}$, with the topology $\tau = \{\emptyset, \{0\},\{0,1\}\}$. This is called the Sierpinski space, and is an example of a type III space. (It is connected because it cannot be realised as the union of two disjoint open sets, and it is compact because it is finite.) However, we see that $\{1\}$ is a compact subset of $S$ which is not closed. Thus for a type III space:

singletons are not necessarily closed, and hence compact sets are not necessarily closed.

On the other hand, we can't talk about whether or not compact implies bounded, because we have no notion of boundedness.

Answer 2

My old theorem (i.e. by Włodzimierz Holsztyński) about maps of inverse limits of ANRs, which involves universal mappings, holds for general Hausdorff compact spaces. Since image domains of universal maps are connected, this means that this theorem is about general Hausdorff compact continua. (I very soon had another and a much more elegant proof, and I had presented it many times publicly but not in print; I didn't use Smith-Moore general sequences in that newer proof).

There are many results in the theory of fixed points, in dimension theory and universal mappings which hold for general Hausdorff continua, it's only natural. Basically, most every topological theorem about metric metric continua allows for a Hausdorff compact version, and in practice the formulation of the generalization is natural.