# Abstract connectedness

Is there an abstract structure that characterizes connectedness, analogously to how topological spaces characterize continuity?

Here's one way to make this question more precise: if $(X,T_X)$ is a topological space with underlying set $X$ and topology $T_X$, then consider the pair $(X,C_X)$ where $C_X$ is the set of connected subsets of $X$. A continuous map $f:(X,T_X) \to (Y,T_Y)$ has the property that the direct image of a connected set is connected. Thus, we have a functor from Top to the category whose objects are pairs $(X,C_X)$ of a set equipped with a set of subsets, and whose maps $f:(X,C_X)\to (Y,C_Y)$ are functions $f:X\to Y$ such that $f(U) \in C_Y$ for all $U\in C_X$. Is there a naturally defined full subcategory of the latter category that is "close" to the image of Top?

There are undoubtedly other ways that one could make this question more precise; if you have a better suggestion feel free to make it.

• For an example of a non-obvious property that connected subsets of a topological space satisfy, see the theorem mentioned in this answer. Mar 6, 2015 at 7:30
• @JochenWengenroth: The collection of all connected sets contains far more information than just the partition into connected components, because not every subset of a component is still connected. Mar 6, 2015 at 7:47
• Do you want the empty set to be connected? :-) [It may seem a trivial point, but ruling it out might aid in getting a smooth axiomatics.] Mar 6, 2015 at 11:42
• @ToddTrimble - I generally do not consider the empty set to be connected, but I suppose I'm willing to be flexible if necessary. Mar 6, 2015 at 21:04
• The notion of "separoid" may be related: en.wikipedia.org/wiki/Separoid "Given a topological space, we can define a separoid saying that two subsets are separated if there exist two disjoint open sets which contains them (one for each of them)." But possibly it captures something slightly different from what you're looking for; I haven't thought about it. Mar 6, 2015 at 21:36

There seems to be some literature on this already: this paper introduces the notion of a "connective space", i.e., a set equipped with a "connectology", and develops some theory. There was some related Mathematics Stackexchange discussion here which connects up with the nontrivial property mentioned by Eric Wofsey in a comment below the question.

• Such literature actually goes back quite a few years. I remember once reading A. D. Wallace's Separation spaces (1941): "The paper gives a set of axioms sufficient to construct a theory of connectivity of sets in terms of a new undefined concept X|Y, which may be read "X is separated from Y"." Mar 7, 2015 at 18:13
• This is exactly what I was looking for, thanks! Mar 9, 2015 at 19:00

Here is an ad hoc attempt. A connectivity space is an ordered pair $\ \mathbf X:=(X\ \mathcal C)\$ such that the following two axioms hold:

1. $\ \left(A\ne\emptyset\ne B\,\ \wedge\ \,A\ B\in\mathcal C\right)\\ \quad \Rightarrow\quad \left( A\cup B\in \mathcal C\ \ \Leftrightarrow\quad\exists_{x\in A\cup B} \left(A\cup\{x\}\in\mathcal C\ \ \wedge\ \ B\cup\{x\}\in \mathcal C\right)\ \right)$
2. $\ \forall_{x\ y\,\in\,A}\ \exists_{S\in\mathcal C}\ (x\ y\in S\ \ \wedge\ \ S\subseteq A)\quad\Rightarrow\quad A\in \mathcal C$

for every $\ A\ B\ \subseteq X.\$ Next, given connectivity spaces $\ \mathbf X:=(X\ \mathcal C)\$ and $\ \mathbf Y:=(Y\ \mathcal D),\$ A connectivity map (or connectivity morphism) is any function $\ f:X\rightarrow Y\$ such that $\ \forall_{A\in\mathcal C}\ f(A)\in\mathcal D.$

Given a topological space $\ \mathbf X:=(X\ T),\$ we get the induced connectivity space $\ \mathbf X_c := (X\ \mathcal C_T),\$ where $\ C_T\$ is the family of all connected subsets of $\ \mathbf X$. Thus every continuous map between two topological spaces is a connectivity map between the induced connectivity spaces.

Do not expect that there is a very close relation between continuous maps and connectivity maps. But the relation between them should be interesting (a source of new MO-questions :-) ).

REMARK   It follows from the above definition (two axioms) that $\ \emptyset\in\mathcal C.\$ (Thank you Eric for this point).

EXAMPLE   An intersection $\ \bigcap_{n=1}^\infty A_n\$of a monotone sequence of closed connected subspaces $\ A_n\$ doesn't have to be connected. For instance, consider the following subspaces of $\ \mathbb R^2$:

$$A_n\ :=\ \mathbb R^2\setminus (-1;1)\times(-n;n)$$

for $\ n=1\ 2\ \ldots$

Of course a small modification will give a similar example for open connected subsets $\ B_n,\$ say:

$$B_n\ :=\ \mathbb R^2\setminus \{0\}\times[-n;n]$$

A point-free definition:

A connectivity structure $\ \mathcal C\$ in set $\ X,\$ with $\ \mathcal C_0:=\mathcal C\setminus \{\emptyset\},\$ is defined by the following 3 axioms:

• $\forall_{A\ B\,\in\,\mathcal C_0}\ \left(A\cup B\in\mathcal C_0\quad\Leftrightarrow\quad\exists_{S\in\mathcal C_0}\left(S\subseteq A\cup B\ \ \wedge\ \ A\cup S\in \mathcal C_0\ \ \wedge\ \ B\cup S\in\mathcal C_0\right)\ \right)$
• $\forall_{R\ S\,\in\,\mathcal C_0}\, \left(\left(R\cup S\,\subseteq\, A\right)\ \Rightarrow\ \exists_{Q\in\mathcal C_0}\ R\cup S\subseteq Q\subseteq A\ \right)\quad\Rightarrow\quad A\in\mathcal C_0$
• $\emptyset\,\in\,\mathcal C$

for every $\ A\ B\,\in\,X$.

A categorical definition

TERMINOLOGY   A category $\ \mathbf C\$ is called vague $\ \Leftarrow:\Rightarrow\,\ \forall_{X\ Y\,\in\,Obj(\mathbf C)}\ |\,MOR(X\ Y)\,|\ \le\ 1$

EXAMPLE   The category of all sets and of the identity embeddings is vague.

DEFINITION 1  of a (connected) union   Let category $\ \mathbf C\$ be vague. An object $\ C\$ of objects $\ A\ B\$ is called a union $\ \Leftarrow:\Rightarrow\$ two conditions hold:

• $\ \exists_{S\,\in\,Obj(\mathbf C)}\\ \ \quad |\,MOR(S\ A)\,|\ =\ |\,MOR(S\ B)\,|\ =\ |\,MOR(A\ C)\,|\ =\ |\,MOR(B\ C\,|\ =\ 1$
• whenever $\ D\$ is like $\ C\$ above then $\ |\,MOR(C\ D)\,|\ =\ 1$.

Thus with every vague category $\ \mathbf C\$we associate a u-graph, where two objects $\ A\ B\$ are connected (i.e. form an edge of the connectivity graph) $\ \Leftarrow:\Rightarrow\$ there exists a union of $\ A\$ and $\ B.$

DEFINITION 2  of merger   Let category $\ \mathbf C\$ be vague. Objects $\ A\ B\$ merge into an object $\ C\ \Leftarrow:\Rightarrow\$ two conditions hold:

• $\ |\,MOR(A\ C)\,|\ =\ |\,MOR(B\ C)\,|\ =\ 1$
• whenever $\ D\$ is like $\ C\$ above, then $\ |\,|MOR(C\ D)\,|\ =\ 1$

With every vague category $\ \mathbf C\$ we associate the merging graph of objects of $\ C\$, where two objects form an edge $\ \Leftarrow:\Rightarrow\$ if they merge.

DEFINITION 3  of a connectivity category:

A category $\ \mathbf C\$ is a connectivity graph $\ \Leftarrow:\Rightarrow\$ it satisfies the following four axioms:

1. $\ \mathbf C\$ is vague;

2. every union of objects of $\ \mathbf C\$ is a merger;

3. if $\ A\ B\$ merge, and if $\ |\,MOR(A\ A')\,|\ 1\$ then $\ A'\$ and $\ B\$ merge too;

4. if $\ \mathbf D\$ is a non-empty family of objects such that the full induced merger subgraph $\ \mathbf D\$ is connected then there exists an object $\ C\$ which is a merger of $\ \mathbf D,\$ meaning that the following two conditions hold:
• $\ \forall_{A\in\mathbf D}\ \ |\,MOR(A\ C)\,|\ =\ 1$
• whenever $\ D\$ is like $\ C\$ then $\ \,MOR(C\ D)\,|\ =\ 1$

INTERPRETATION: The objects of a connectivity category play the role of non-empty connected spaces.

In the case of a topological space $\ \mathbf X,\$ the category $\ \mathbf C :=\mathbf C_{\mathbf X}\$ consists of the non-empty connected subsets of $\ \mathbf X,\$ and of the identity embeddings.

• I was just about to fix this nasty typo. Thank you Eric! Mar 6, 2015 at 8:47
• Note that these axioms do not imply singletons are connected, nor do they rule out having only singletons and the whole space being connected (which is impossible for topological spaces with more than two points by this answer). Mar 6, 2015 at 8:51
• I always wondered what would happen if we took as definition of continuity, something like that: A function $f : A \rightarrow B$ is continuous iff its graph (or even its closure with product topology) $\Gamma(f)$ is connected in $A\times B$. This, in my opinion, captures the notion of continuity of "walking" of any ball of radius $\epsilon$, you stay inside your space (in the sense of metric space).
– sure
Mar 6, 2015 at 8:57
• @EricWofsey -- thank you again; I fixed the issue of the empty set. (Perhaps there are more things to fix), Mar 6, 2015 at 9:39
• (1) the same reasons that 1 is not a prime number. For instance, if $\emptyset$ is connected then the decomposition of a space into connected components is not unique since you can always add more copies of $\emptyset$. (2) What motivated you to write down these axioms? How did you think of them? Mar 8, 2015 at 5:32

Let me answer as to how well the connected subspaces of a space determine the topology. I claim that for a large class of spaces, the collection of all connected subsets of the space determines the topology on $X$. However, the functions between nice connected spaces that map connected sets to connected sets are generally not continuous.

The following lemma states that the closure of a connected subspace of a regular space $X$ is determined by the collection of all connected subspaces of $X$. In particular, the closed connected subspaces of a regular space $X$ are determined by the collection of all the connected subspaces of $X$.

$\mathbf{Lemma}$ Suppose that $X$ is a regular space and $A\subseteq X$ is a connected subset. Then $\{a\}\cup A$ is connected if and only if $a\in\overline{A}$. Furthermore, if $A\subseteq B\subseteq X$, then $B\subseteq\overline{A}$ if and only if whenever $A\subseteq C\subseteq B$ then $\overline{C}$ is connected.

The proof of the above lemma is straightforward.

We shall say that a point $x$ is a connected space $X$ is a nbd-cut-point if there is some open neighborhood $U$ of $x$ such that whenever $V$ is an open set with $x\in V\subseteq U$, then $X\setminus V$ is not connected. Then it is easy to show that a connected space $X$ has no nbd-cut-points if and only if every closed subspace of $X$ is the intersection of connected closed subspaces of $X$. I should also mention that I made up the notion of a nbd-cut-point in order to answer this question, but the notion of a nbd-cut-point is closely related to the notion of a cut-point since every cut point is a nbd-cut-point.

$\mathbf{Proposition}$ Suppose that $X,Y$ are connected regular spaces with no nbd-cut points. If there is a bijection $f:X\rightarrow Y$ so that if $C\subseteq X$ then $C$ is connected if and only if $f[C]$ is connected, then $X$ and $Y$ are homeomorphic.

However, I must mention that even for connected regular spaces with no nbd-cut-points, there are non-continuous maps where the image of a connected space is connected. For example, define an equivalence relation on $\mathbb{R}$ where $x\simeq y$ iff $x-y\in\mathbb{Q}$. Then there is a surjective function $f:\mathbb{R}/\simeq\rightarrow\mathbb{R}^{2}$ since $|\mathbb{R}/\simeq|=|\mathbb{R}^{2}|$. Define a mapping $g:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ by letting $g(x,y)=f([x])$ and define $h:\mathbb{R}\rightarrow\mathbb{R}^{2}$ by letting $h(x)=f([x])$. Then $h[U]=\mathbb{R}^{2}$ for each nonempty open set $U$. I claim that $g$ maps connected subsets to connected subsets. Let $\pi_{1},\pi_{2}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the projection maps where $\pi_{1}(x,y)=x,\pi_{2}(x,y)=y$. Suppose that $C\subseteq\mathbb{R}^{2}$ is connected. Then $\pi_{1}[C]$ is also connected, so $\pi_{1}[C]$ is either a singleton or an interval. If $\pi_{1}[C]$ is a singleton, then $g[C]$ is a singleton in $\mathbb{R}^{2}$. However, if $\pi_{1}[C]$ is an interval, then $g[C]=\mathbb{R}^{2}$. In either case, the function $g$ maps connected sets to connected sets, but the function $g$ is very far from being continuous. It therefore seems as if the notion of continuity cannot be described in terms of connectedness unless one describes the open or closed sets in terms of connectedness and directly translates the notion of continuity into a notion of a connected mapping.

A relevant idea by @sure was relegated (by her/him-self!) to comments. Let me give it justice here. Actually, I'll provide a counter-example:

THEOREM   There exists a non-continuous map $\ f : X\ \rightarrow Y\$ of a metric 1-dimensional compact connected space $\ X\$ into a metric 1-dimensional separable complete space $\ Y\$ such that the graph $\ G(f)\subseteq X\times Y\$ is a closed connected subset.

PROOF   Let $\ X:=S^1\$ be the unit circle in the complex plane, with center $\ 0.\$ Let $\ Y:=\mathbb R.\$ Define

$$\forall_{t\,:\, 0\le t< 2\cdot\pi}\ \ f\left(e^{\imath\cdot t}\right)\ :=\ \tan\left(\frac t4\right)$$

That's it, END of Proof.

Its only a trivial observation, but too long for a comment:

We represent a topological space in terms of closet's $(X, \tau)$, then $\tau \subset \mathcal{P}(X)$ (where $\mathcal{P}(X)$ is the set of part's of $X$) and $\tau$ is closed for finite unions, and general intersection, $X\in\tau$ and $\emptyset=\bigcap \tau$.

I define a connect structure as a ordered couple $(X, \chi)$ where $\chi\subset \mathcal{P}(X)$ have mutually disjoint elements and $X=\bigcup \chi$, a morphism of connect structures $f: (X_1, \chi_1)\to (X_2, \chi_2)$ is a map of set's $f: X_1\to X_2$ such that $\forall C_1\in \chi_1\ \exists C_2\in \chi_2: f(C_1)\subset C_2$.

Of course we have a (I seem a topological) functor $F$ from the topological spaces to connected structures, where $F(X, \tau)$ is $(X, \chi)$ with $\chi$ the sets of topological connect component. We have also that $\chi$ is the minimum of the set of families $\mathcal{A}\subset \tau$ with mutually disjoint elements, and $X=\bigcup \mathcal{A}$, ordered as $\mathcal{A}\leq \mathcal{B}$ if $\forall A\in \mathcal{A}\ \exists B\in \mathcal{B}: A\subset B$. This definition don't use "points" may be can be generalized (by some additional hypothesis) to locales (in term of closed sublocales) but I dont know.

• It sounds like you're doing something similar to what Jochen proposed in a comment. But I definitely want to consider connected subsets that are not components. Mar 6, 2015 at 21:08