In mathematics, the particular point topology (or included point topology) is a topology where sets are considered open if they are empty or contain a particular, arbitrarily chosen, point of the topological space. Formally, let X be any set and p ∈ X. The collection
T = {
S ⊆
X:
p ∈
S or
S = ∅}
of subsets of X is then the particular point topology on X. There are a variety of cases which are individually named:
If X = {0,1} we call X the Sierpiński space. This case is somewhat special and is handled separately.If X is finite (with at least 3 points) we call the topology on X the finite particular point topology.If X is countably infinite we call the topology on X the countable particular point topology.If X is uncountable we call the topology on X the uncountable particular point topology.A generalization of the particular point topology is the closed extension topology. In the case when X {p} has the discrete topology, the closed extension topology is the same as the particular point topology.
This topology is used to provide interesting examples and counterexamples.
Closed sets have empty interiorGiven an open set
A ⊂ X every
x ≠ p is a limit point of A. So the
closure of any open set other than
∅ is
X . No
closed set other than
X contains p so the interior of every closed set other than
X is
∅ .
Path and locally connected but not arc connected f ( t ) = { x t = 0 p t ∈ ( 0 , 1 ) y t = 1 f is a path for all
x,
y ∈
X. However since
p is open, the preimage of
p under a continuous injection from [0,1] would be an open single point of [0,1], which is a contradiction.
Dispersion point, example of a set withp is a
dispersion point for
X. That is
X{p} is totally disconnected.
Hyperconnected but not ultraconnectedEvery open set contains
p hence
X is hyperconnected. But if
a and
b are in
X such that
p,
a, and
b are three distinct points, then
{a} and
{b} are
disjoint closed sets and thus
X is not ultraconnected. Note that if
X is the Sierpinski space then no such
a and
b exist and
X is in fact ultraconnected.
Closure of compact not compactThe set
{p} is compact. However its
closure (the closure of a compact set) is the entire space
X and if
X is infinite this is not compact (since any set
{t,p} is open). For similar reasons if
X is uncountable then we have an example where the closure of a compact set is not a
Lindelöf space.
Pseudocompact but not weakly countably compactFirst there are no disjoint non-empty open sets (since all open sets contain 'p'). Hence every continuous
function to the real line must be
constant, and hence bounded, proving that
X is a
pseudocompact space. Any set not containing
p does not have a limit point thus if
X if infinite it is not weakly countably compact.
Locally compact but not strongly locally compact. Both possibilities regarding global compactness.If
x ∈ X then the set
{ x , p } is a compact neighborhood of
x. However the closure of this neighborhood is all of
X and hence
X is not strongly locally compact.In terms of global compactness,
X finite if and only if
X is compact. The first implication is immediate, the reverse implication follows from noting that
⋃ x ∈ X { p , x } is an open cover with no finite subcover.
Accumulation point but not a ω-accumulation pointIf
Y is some subset containing
p then any
x different from
p is an accumulation point of
Y. However
x is not an
ω-accumulation point as {
x,
p} is one neighbourhood of
x which does not contain infinitely many points from
Y. Because this makes no use of properties of
Y it leads to often cited counter examples.
Accumulation point as a set but not as a sequenceTake a sequence {
ai} of distinct elements that also contains
p. As in the example above, the underlying set has any
x different from
p as an accumulation point. However the sequence itself cannot possess accumulation point
y for its neighbourhood {
y,
p} must contain infinite number of the distinct
ai.
T0X is T
0 (since {
x,
p} is open for each
x) but satisfies no higher
separation axioms (because all open sets must contain
p).
Not regularSince every nonempty open set contains
p, no closed set not containing
p (such as
X{
p}) can be separated by neighbourhoods from {
p}, and thus
X is not regular. Since complete regularity implies regularity,
X is not completely regular.
Not normalSince every nonempty open set contains
p, no nonempty closed sets can be separated by neighbourhoods from each other, and thus
X is not normal. Exception: the Sierpinski topology is normal, and even completely normal, since it contains no nontrivial separated sets.
Separability{p} is
dense and hence
X is a
separable space. However if
X is uncountable then
X{p} is not separable. This is an example of a
subspace of a separable space not being separable.
Countability (first but not second)If
X is uncountable then
X is
first countable but not
second countable.
Comparable ( Homeomorphic topology on the same set that is not comparable)Let
p , q ∈ X with
p ≠ q . Let
t p = { S ⊂ X | p ∈ S } and
t q = { S ⊂ X | q ∈ S } . That is
tq is the particular point topology on
X with
q being the distinguished point. Then
(X,tp) and
(X,tq) are homeomorphic incomparable topologies on the same set.
Density (no nonempty subsets dense in themselves)Let
S be a subset of
X. If
S contains
p then S has no limit points (see limit point section). If
S does not contain
p then
p is not a limit point of
S. Hence
S is not dense if
S is nonempty.
Not first categoryAny set containing
p is dense in
X. Hence
X is not a
union of nowhere dense subsets.
SubspacesEvery subspace of a set given the particular point topology that doesn't contain the particular point, inherits the discrete topology.
