Difficulty: 2

Let $X$ be infinite and $A \subseteq X$ countably infinite. If $X$ has no limit points, $A$ is closed and for each $x \in A$, choose a nbd $V_x$ with $V_x \cap X = \emptyset$. Then $\{V_n\} \cup \{A^C\}$ is a cover. A finite subcover would imply $A$ is finite.

For $f : X \to \R$ continuous, $\{f^{-1}(-n, n)\}$ is a countable cover.

A $k$-netork is a network: This was done in (T11).

Let $f : X \to [0, 1]$ continuous with $f(A) = \{0\}$ and $f(b) = 1$. Then $U = f^{-1}([0, 1/2))$ and $V = f^{-1}((1/2, 1])$ are disjoint with $A \subseteq U$ and $b \in V$.

The discrete metric $d(x, y) = 1$ iff $x \ne y$ is complete: If \abs{x_n - x_m} < 1/2, then $x_n = x_m$.

If no disjoint nonempty closed sets exist, one of the disjoint closed sets is empty, and the other is contained in $X$, which is clopen.

Contrapositively, if all paths were constant, then any basis of path connected sets must be made up of singletons, proving it would be discrete.

Contrapositively, assume $A,B$ are disjoint, nonempty, and closed. Then we could separate them with $U,V$ open sets, disproving hyperconnectivity.

Let $X$ be partitioned into two sets with at least 2 elements each. One of them contains $p$, and so the other is contained in $X \setminus \{p\}$, so it is disconnected.

Every open set is dense, so $\overline{V} = X$ is trivially clopen for any open $V$.

Let $V$ be open. Then $\overline{V}$ is clopen, so it is either empty or $V$ is dense (no other nonempty open sets are disjoint with $V$).

Let $x \ne y$. $T_1$ implies $\{x\}$ and $\{y\}$ are closed, and normal implies there are $\{x\} \subseteq U$ and $\{y\} \subseteq V$ open sets with $U \cap V = \emptyset$. So it is $T_2$ and normal.

Let $\{V_n\}$ be a countable local basis of $p$. Let $U_m = \bigcap_{n \le m} V_n$. Now just remove duplicates.

Rigorously, define $M(V) = \{ m \in \omega \ : \ V_m \subset V \}$ for each $V$ open. Then $W_0 = U_0$ and $W_{n+1} = U_{M(W_n)}$ is a valid definition with $\{W_n\}$ well ordered by reverse-inclusion, unless some $M(W_n)$ is empty, for which $\{W_n\}$ is finite, and that’s okay.

Take a local basis of $p$ which is well-ordered by reverse inclusion. Then it’s isomorphic to some ordinal $\alpha$ and it can be enumerated with $\{V_\beta\}_{\beta < \alpha}$. Since $p \in \overline{A}$, use the axiom of choice to construct $x_\beta \in V_\beta \cap A \setminus \{p\}$.

This is a similar proof to First countable $\implies$ Fréchet Urysohn (T183), just with transfinite sequences now (and using the full axiom of choice, not just countable).

If $x \ne y$ are indistinguishable, then $\overline{\{x\}} \cap \overline{\{y\}} = \emptyset$. So no two distinct points are distinguishable.

Take $x \ne y$. If $\{x\}$ is open, it’s a nbd of $x$ and not of $y$. If it is closed, $\{x\}^C$ is a nbd of $y$ not of $x$.

Let $x \ne y$ such that $x \in W$ and $y \notin W$ for some open $W$. Then $x \notin F = W^C$, so by regularity, choose disjoint open sets $U,V$ with $F \subseteq U$ and $x \in V$. Since $y \in U$, this proves being $T_2$. $T_2$ and regular is $T_3$ by definition.

Contrapositively, if $x \ne y$ were indistinguishable, then $\{x, y\}$ would have no isolated point.

Take a countable local basis $\{V_n\}$ of $p$. By constructing $W_n = \bigcup_{j=1}^n V_n$, $\{W_n\}$ is a shrinking local basis. Since $p \in \overline{A}$, select $x_n \in W_n \cap A \setminus \{p\}$ for each $n$ and $x_n \to p$.

If $\text{scl}(A)$ denotes the sequential closure, then Fréchet Urysohn means “$\overline{A} \subseteq \text{scl}(A)$ for all $A$” and sequential means “$\text{scl}(A) \subseteq A$ implies $A$ is closed for all $A$”. So if the former is true, then assuming $\text{scl}(A) \subseteq A$, we have $\overline{A} \subseteq \text{scl}(A) \subseteq A$, proving $A$ is closed.

Contrapositively, if $X$ is infinite, let $A = \{x_0, x_1, x_2, \dots\} \subseteq X$ be countable. Since no subsequence of $A$ is eventually constant, any subsequence must not converge.

Any bijective function is a homeomorphism in discrete space. Just take the permutation that swaps $a$ and $b$.

Essentially the same proof as Fréchet Urysohn $\implies$ Sequential (T184). Just swap “sequential closure” with “radial closure”.

Define $d(x, y) = 1$ if $x,y$ belong to the same set in the partition, and 0 otherwise. This is a pseudometric.

Any countable family is a countable union of finite sets. Finite sets are clearly locally finite. So any countable family is $\sigma$-locally finite.

It has a bijection $f: X \to n$ if finite, or $f : X \to \omega$ if infinite. It’s a homeomorphism either way.