Difficulty: 1

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.

Yeah.

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.

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

Singletons are clopen.

The space is not a singleton.

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.

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

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

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

Shrink the cover twice.

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.

Left as an exercise for the reader.

Left as an exercise for the reader.

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.

Every subspace is finite.

Any open set has more than one point.

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

In particular, singletons are discrete.

Compact sets are countably compact.

Globally implies locally.

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

$\mathcal{P}(X) = \{\emptyset, X\}$ 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.

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.

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.

Important result to solve the Riemann hypothesis.

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

A space is a subspace of itself.

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

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

Globally implies locally.

Every metric is a pseudometric.

By definition.

By definition.

By definition.

By definition.

By definition.

By definition.

Groups/monoids are nonempty by definition.

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

A topology is a coarser topology of itself.

We’re just dropping the “separable”.

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.

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

In general, a countable topology is second countable.

Can’t argue with that.

Can’t argue with that.

Can’t argue with that.

Literally by definition.

Literally by definition.

Literally by definition.

Literally by definition.

It’s an order topology.

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

Finite implies countable.

It’s almost… so not quite.

The homotopy cannot be empty.

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.

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

The complement of any set must be finite.