Equivalence of Definitions of Ultraconnected Space
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Theorem
The following definitions of the concept of Ultraconnected Space are equivalent:
Definition 1
A topological space $T = \struct {S, \tau}$ is ultraconnected if and only if no two non-empty closed sets are disjoint.
Definition 2
A topological space $T = \struct {S, \tau}$ is ultraconnected if and only if the closures of every distinct pair of elements of $S$ are not disjoint:
- $\forall x, y \in S: \set x^- \cap \set y^- \ne \O$
Definition 3
A topological space $T = \left({S, \tau}\right)$ is ultraconnected if and only if every closed set of $T$ is connected.
Proof
$(1)$ iff $(2)$
Let $T = \struct {S, \tau}$ be ultraconnected by Definition 1:
- no two non-empty closed sets of $T$ are disjoint.
Let $x, y \in S$.
By Topological Closure is Closed, both $\set x^-$ and $\set y^-$ are closed.
- $\set x^- \cap \set y^- \ne \O$
That is, $T = \struct {S, \tau}$ is ultraconnected by Definition 2.
$\Box$
Let $T = \struct {S, \tau}$ be ultraconnected by Definition 2:
- $\forall x, y \in S: \set x^- \cap \set y^- \ne \O$
Let $V_1, V_2$ be closed sets of $T$.
Let $x \in V_1, y \in V_2$.
Then:
- $\set x^- \cap \set y^- \ne \O$
But then from Topological Closure of Subset is Subset of Topological Closure we have that:
- $\set x^- \subseteq V_1^-$
- $\set y^- \subseteq V_2^-$
But from Closed Set Equals its Closure $V_1^- = V_1, V_2^- = V_2$.
So from Intersection is Subset:
- $\set x^- \cap \set y^- \subseteq V_1$
- $\set x^- \cap \set y^- \subseteq V_2$
from which:
- $V_1 \cap V_2 \ne \O$
As $V_1$ and $V_2$ are arbitrary, it follows that $T$ is ultraconnected by Definition 1.
$\blacksquare$
$(1)$ iff $(3)$
Closed Sets Intersect implies Closed Sets are Connected
Let $T = \struct {S, \tau}$ be ultraconnected in the sense that:
- no two non-empty closed sets of $T$ are disjoint.
Let $F \subseteq S$ be an arbitrary closed set of $T$.
Aiming for a contradiction, suppose $F$ is not connected.
Then there exist non-empty closed set $G, H$ in $F$ that are disjoint (and whose union is $F$).
By Closed Set in Closed Subspace, $G$ and $H$ are closed in $T$.
Because $G \cap H = \O$, $T$ is not ultraconnected.
This is a contradiction.
Thus $F$ is connected.
As $F$ is arbitrary, this applies to all closed set of $T$.
Thus $T = \struct {S, \tau}$ is ultraconnected in the sense that:
- every closed set of $T$ is connected..
$\Box$
Closed Sets are Connected implies Closed Sets Intersect
Let $T = \struct {S, \tau}$ be ultraconnected in the sense that:
- every closed set of $T$ is connected.
Let $G$ and $H$ be closed sets of $T$.
Then their union $G \cup H$ is closed in $T$.
By assumption, $G \cup H$ is connected.
By Closed Set in Closed Subspace, $G$ and $H$ are closed sets of $G \cup H$.
Because $G \cup H$ is connected, $G \cap H$ is non-empty.
Because $G$ and $H$ were arbitrary, this applies to intersection of all such closed sets
Thus $T = \struct {S, \tau}$ is ultraconnected in the sense that:
- no two non-empty closed sets of $T$ are disjoint.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness