Particular point topology

Path and locally connected but not arc connected

For any x, y ∈ X, the function f: [0, 1] → X given by

${\displaystyle f(t)={\begin{cases}x&t=0\\p&t\in (0,1)\\y&t=1\end{cases}}}$

is a path. 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 with

p is a dispersion point for X. That is X \ {p} is totally disconnected.

Hyperconnected but not ultraconnected

Every non-empty open set contains p, and 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 Sierpiński space then no such a and b exist and X is in fact ultraconnected.

Compact only if finite. Lindelöf only if countable.

If X is finite, it is compact; and if X is infinite, it is not compact, since the family of all open sets ${\displaystyle \{p,x\}\;(x\in X)}$ forms an open cover with no finite subcover.

For similar reasons, if X is countable, it is a Lindelöf space; and if X is uncountable, it is not Lindelöf.

Closure of compact not compact

The 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. 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 compact

First 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 locally relatively compact.

If ${\displaystyle x\in X}$, then the set ${\displaystyle \{x,p\}}$ is a compact neighborhood of x. However the closure of this neighborhood is all of X, and hence if X is infinite, x does not have a closed compact neighborhood, and X is not locally relatively compact.

Accumulation points of sets

If ${\displaystyle Y\subseteq X}$ does not contain p, Y has no accumulation point (because Y is closed in X and discrete in the subspace topology).

If ${\displaystyle Y\subseteq X}$ contains p, every point ${\displaystyle x\neq p}$ is an accumulation point of Y, since ${\displaystyle \{x,p\}}$ (the smallest neighborhood of ${\displaystyle x}$) meets Y. Y has no ω-accumulation point. Note that p is never an accumulation point of any set, as it is isolated in X.

Accumulation point as a set but not as a sequence

Take a sequence ${\displaystyle (a_{n})_{n}}$ of distinct elements that also contains p. The underlying set ${\displaystyle \{a_{n}\}}$ has any ${\displaystyle x\neq p}$ as an accumulation point. However the sequence itself has no accumulation point as a sequence, as the neighbourhood ${\displaystyle \{y,p\}}$ of any y cannot contain infinitely many of the distinct ${\displaystyle a_{n}}$.

T₀

X is T₀ (since {x, p} is open for each x) but satisfies no higher separation axioms (because all non-empty open sets must contain p).

Not regular

Since every non-empty 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 normal

Since every non-empty open set contains p, no non-empty closed sets can be separated by neighbourhoods from each other, and thus X is not normal. Exception: the Sierpiński 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 topologies on the same set that are not comparable)

Let ${\displaystyle p,q\in X}$ with ${\displaystyle p\neq q}$. Let ${\displaystyle t_{p}=\{S\subset X\mid p\in S\}}$ and ${\displaystyle t_{q}=\{S\subset X\mid q\in S\}}$. That is t_q is the particular point topology on X with q being the distinguished point. Then (X,t_p) and (X,t_q) are homeomorphic incomparable topologies on the same set.

No nonempty subset dense-in-itself

Let S be a nonempty subset of X. If S contains p, then p is isolated in S (since it is an isolated point of X). If S does not contain p, any x in S is isolated in S.

Not first category

Any set containing p is dense in X. Hence X is not a union of nowhere dense subsets.

Subspaces

Every subspace of a set given the particular point topology that doesn't contain the particular point, inherits the discrete topology.