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.

A space is a subspace of itself.

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.

$T_4$ 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).

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.

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.

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

Globally implies locally.

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.

Every subspace is finite.

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 $x$.

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.

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.

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”.

Left as an exercise for the reader.

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.

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