Evident.

Any closed set is compact, so any closure of a nbd is compact.

Take one nbd from the local basis. Its closure is compact.

By definition.

Indeed.

A subcover is a refinement. A finite subcover is star-finite.

If finitely many intersect a nbd around the point, finitely many will intersect the point.

If true for any covers, then true for countable ones.

If true for any covers, then true for countable ones.

A subcover is a refinement. A finite subcover is star-finite.

This is Paracompact $\implies$ Metacompact, just countable this time.

By definition.

By definition.

A space is a subspace of itself.

An injective path is a path.

If it has a dispersion point… it has… a point.

Singletons are clopen.

The space is not a singleton.

By definition.

By definition.

An injective path is a path.

This is obvious, so fun fact: The converse requires the continuum hypothesis.

Big brain stuff.

Singletons are compact.

The path between two distinct points has at least $\mathfrak c$ points.

A complete metric is… a metric.

By definition.

The continuous map with $f(a) = 0$ and $f(b) = 1$ is not constant.

By definition.

Take a path between two points. It’s not constant.

A countable family is a countable union of singletons, and a singleton is trivially locally finite.

If $f$ continuous, $f([0, 1/3])$ and $f([2/3, 1])$ are connected.

$T_4$ is $T_2$ (hence, $T_1$) by definition.

$T_5$ is $T_2$ (hence, $T_1$) by definition.

In particular, $X$ is $T_1$ and normal, which implies $T_2$ (shown in T99), and so it must be $T_5$ by definition.

By definition.

By definition.

Shrink the cover twice.

If only singletons are connected and $X$ has a basis of connected sets, then the basis must be all singletons.

I’m not sure why this is here. $T_5$ implies $T_1$ (T100) and completely normal (T336). Completely normal implies normal (T36). $T_1$ and normal implies $T_4$ (T99).

By definition.

By definition.

Clear from their definitions.

Clear from their definitions.

A single set is trivially a countable union.

If each compact has a finite subcover, a countable union of them will have a countable subcover.

Take any open cover. Apply Lindelöf, then apply countably paracompact.

Take any open cover. Apply Lindelöf, then apply countably metacompact.

By definition.

A subcover is trivially, a subcolection with dense union.

Left as an exercise for the reader.

Left as an exercise for the reader.

Every set is clopen.

By definition.

By definition.

By definition.

By definition.

By definition.

By definition.

$T_6$ is $T_1$ and perfectly normal. Then it is completely normal (T156). $T_1$ and completely normal is $T_5$ (T101).

Normal and $T_2$ is the definition of $T_4$.

Any compact set is relatively compact.

By definition.

Globally implies locally.

By definition.

That’s right, “countable” does not mean infinitely countable.

Any topology is a subset of $\mathcal{P}(X)$, so there are finitely many open sets.

By definition, $\aleph_1$ is the smallest cardinality greater than $\aleph_0$. Assuming the continuum hypothesis, $\aleph_1 = \mathfrak c$.

By definition, $\aleph_1$ is the smallest uncountable cardinal.

Globally implies locally.

By definition.

Every subspace is finite.

By definition.

By definition.

By definition.

Fréchet Urysohn is stronger because it asserts there is a sequence, a.k.a a transfinite sequence of length $\omega$.

Any radially closed set must be, in particular, sequentially closed.

Any open set has more than one point.

If each local basis $\mathcal{V}_x$ is countable and $X$ is countable, then $\mathcal{B} = \bigcup_{x \in X} \mathcal{V}_x$ is a countable basis.

$\{x\}$ is a finite neighborhood of $x$.

Compact sets are countably compact.

By definition: $\sigma$-compact $\iff$ countable union of compacts $\{K_n\}$. Hemicompact $\iff$ countable union of compacts $\{K_n\}$ such that any compact set lies inside some $K_m$.

Globally implies locally.

This is just “Has a countable $k$-network $\implies$ Has a countable network” (T11) with $T_3$ added.

This is just “Has a countable network $\implies$ Has a $\sigma$-locally finite network” (T29) with $T_3$ added.

This is just “Has a countable $k$-network $\implies$ Has a $\sigma$-locally finite $k$-network” (T352) with $T_3$ added.

This is just “Has a $\sigma$-locally finite $k$-network $\implies$ Has a $\sigma$-locally finite network” (T34) with $T_3$ added.

Just take one element from the local basis.

$\mathcal{P}(X) = \{\emptyset, X\}$ if and only if it has 0 or 1 point.

There’s only one possible topology.

There’s only one possible topology.

If you have an infinite amount of apples, then you have at least 2 apples.

Any open cover must contain $X$ as an element.

Some point has a neighborhood not containing another point.

A space is a subspace of itself.

Singletons form a network.

Every metric is a pseudometric.

By contraposition, if $d(x, y) = 0$, then they both have the same neighborhoods.

Globally implies locally.

A countable basis is a local basis for every point.

$T_2$ is $T_0$ with $R_1$.

Any two distinct points are distinguishable in a $T_0$ space.

Immediate by definition.

By definition, $T_1$ is $T_0$ and $R_0$.

$T_0$ ensures any two points are distinguishable. This is an if and only if.

Important result to solve the Riemann hypothesis.

Indiscrete spaces are connected and being indiscrete is hereditary.

It has finitely many self-maps (continuous or not).

A space is a subspace of itself.

The space is a countable union of finite sets.

If $X$ itself is not dense, then some point is isolated.

The only family of disjoint nonempty open sets is empty, which is countable.

$X = \bigcup \emptyset$ where $\emptyset$ is countable and each $A \in \emptyset$ is nowhere dense.

The space cannot be a partition of more than one nonempty closed set.

If $K \cap A$ is closed for every compact $K$, in particular $K_n \cap A$ is closed for each $\{K_n\}$ stipulated by the $k_{\omega,1}$-space property, proving $A$ is closed.

Globally implies locally.

Every metric is a pseudometric.

This is just “(Pseudometrizable ∧ $T_0$) $\implies$ Metrizable” (T265) but for local nbds.

By definition.

By definition.

By definition.

By definition.

By definition.

By definition.

By definition.

If an open refinement is a partition, its elements are pairwise-disjoint, so trivially star-finite.

Groups/monoids are nonempty by definition.

Any intersection of open sets is open, including countable ones.

$\kappa < 2^\kappa$ is true for any cardinal.

$\leq$ means $\lt$ or $=$.

A topology is a coarser topology of itself.

The space has a coarser metrizable topology, and any coarser topology of a separable space is separable as well (a dense set remains dense in coarser topologies).

We’re just dropping the “separable”.

A regular open neighborhood is… an open neighborhood.

If you have at least 3 apples, then you have at least 2 apples.

If you have at least 4 apples, then you have at least 3 apples.

If you have an infinite amount of apples, then you have at least 4 apples.

Vacuously true: there are no disjoint subsets with at least two points each.

A space is a subspace of itself.

If it has a fixed point… it has… a point.

The partition would be $\{X\}$ itself, the trivial partition.

In general, a countable topology is second countable.

By definition.

Can’t argue with that.

Can’t argue with that.

Can’t argue with that.

By definition.

As the partition forms a basis for $X$, it must refine any open cover, and each element of the partition is a clopen set.

A space is a subspace of itself (and is homeomorphic to itself).

Being a W-space is hereditary.

By definition.

By definition.

By definition.

By definition.

By one of the equivalent definitions.

It’s an order topology.

We can cover the space with countable sets, and take a countable subcover.

By definition (quasi-sober just drops uniqueness).

In order for $X \setminus \{p\}$ to be disconnected, it needs to have at least 2 points.

The space is a closed subspace of itself.

Finite implies countable.

It’s almost… so not quite.

The homotopy cannot be empty.

Separable means Density$= \aleph_0$.

$\cl{X} = X$

If it has a closed point… it has… a point.

A space is a subspace of itself.

Any set in any local basis is compact.

An arc is, by definition, an embedding (hence, injective).

An arc is, by definition, an embedding (hence, injective).

Every $p \in \emptyset$ is homeomorphic to $\R$.

The complement of any set must be finite.

By definition.

It’s impossible to have an infinite strictly decreasing sequence of open sets if there are only finitely many possible open sets.

Globally implies locally.