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.

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

If $\overline{U} \cap \overline{V} = \emptyset$, then $U \cap V = \emptyset$.

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 $b \in V$.

A path component is always a subset of a connected component: If $f : [0, 1] \to X$ is a path from $x$ to $y$, then $f([0, 1])$ is a connected set containing $x$ and $y$.

In a partition topology, every open set is clopen, and equals its own closure.

If $x \in U$ and $y \in V$ with $U \cap V = \emptyset$, then in particular, $y \notin \overline{U}$, which must be clopen.

Take $x \ne y$. Then there’s some $A$ clopen with $y \in A$ and $x \notin A$, so $y$ cannot be in the connected component of $x$.

Path connected components are subsets of connected components. So path connected components are singletons, which implies constant paths.

Let $x \in A$ and $y \notin A$ with $A$ clopen. Then $\chi_A$ (indicator function of $A$) is continuous, since any preimage is one of $\{\emptyset, A, A^C, X\}$ and all are open. $\chi_A(x) = 1$ and $\chi_A(y) = 0$.

Contrapositively, suppose every nbd of $x$ must contain $y$. Then $f(0) = x$ and $f(t) = y$ for $t \in (0, 1]$ is continuous.

Let $W$ be a nbd of $x$ which is pseudometrizable. Then the open balls $B(x, 1/n) \subseteq W$ form a countable local basis.

A path component is always a subset of a connected component (seen in T40).

Let $x \in X$. If $K \subseteq X$ is compact $T_2$, then $\{x\} \cap K$ is clearly closed in $K$. This proves $\{x\}$ is closed.

If $f : X \to \R$ is continuous and nonconstant, then for some $x \ne y$ there exists $z$ with $f(x) < z < f(y)$ and the sets $f^{-1}((-\infty, z))$ and $f^{-1}((z, \infty))$ are disjoint, open, and nonempty.

By definition, a discrete family must be disjoint, so in an ultraconnected space, it has at most one nonempty closed set $F$. Then just choose $X$ to be the open set with $F \subseteq X$.

Take $x \ne y$ and suppose $U$ is a nbd of $x$ not containing $y$. Let $x \in B \subseteq U$ with $B$ clopen, so $y \notin B$. Similarly, $y \in B^C$ is clopen and $x \notin B^C$.

If $C$ is a clopen set, then the indicator function $\chi_C$ is continuous, so it must be constant equal to zero or one (meaning $C = \emptyset$ or $C = X$).

Let $x \in V \subseteq K$ with $V$ open and $K$ compact (in particular, $K$ is closed). Then the sets of the form $U \cap V$ where $U$ is any open set is a local basis for $x$ and $\cl{U \cap V} \subseteq \cl{K} = K$ are all compact.

If $X$ is not discrete, then some path is non-constant, so some $f : [0, 1] \to X$ is injective, meaning $\text{card}(X) \ge \text{card}([0, 1]) = \mathfrak{c}$.

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.

A $T_4$ space is normal and $T_2$, in particular, $R_0$. So by (T37), it is completely regular.

The space is $T_2$, so given $x \ne y$, $\{y\}$ is closed. It’s also completely regular, so there is a continuous map $f : X \to [0, 1]$ with $f(x) = 0$ and $f(y) = 1$.

The space is $T_2$ and completely regular. By (T35), it is regular.

If $X$ is a countable union of nowhere dense sets, then either it is a union of zero sets (so $X$ is empty), or $X \ne \emptyset$ can’t have empty interior (so $X$ is not Baire).

Given an open cover $\mathcal{U}$, define $\mathcal{U}_n = \mathcal{U}$ for all $n$ and so $\bigcup_{n < \omega} \mathcal{F}_n$ is a countable cover, since each $\mathcal{F}_n \subseteq \mathcal{U}_n$ is finite.

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.

It suffices to show $T_2$. For $x \ne y$. The space is $T_0$, so by swapping $x$ and $y$ if needed, suppose $y$ has a nbd not of $x$, which means $x \notin \cl{\{y\}}$. Completely regular implies there’s a separation between $x$ and $\cl{\{y\}}$, which is a separation of $x$ and $y$.

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

Note that $T_0$ implies $\cl{\{x\}} \ne \cl{\{y\}}$ for any $x \ne y$. So contrapositively, if $A = \cl{\{x\}} = \cl{\{y\}}$, then the closed set $A$ is the closure of two distinct singletons (not unique), so the space is not sober.

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”.

Let $p$ be isolated and $\phi : X \to X$ with $\phi(p) = q$ a homeomorphism. Then $q$ is isolated. Do this for every $q$.

For $x \ne y$, $\{x, y\}$ is discrete, so some nbd of $y$ does not contain $x$ and vice-versa.

If $K$ is infinite, let $A = \{x_n\} \subseteq K$ be a countable subset. But then $A$ is discrete and sets of the form $\{x_n\}$ form a cover for $A$ with no finite subcover, so $K$ can’t be compact.

Contrapositively, suppose there is no nbd of $x$ which doesn’t contain $y$ as well. Then for the constant sequence $x_n \coloneqq x$, $x_n \to x$ and $x_n \to y$.

If $f : K \to X$ continuous and $K$ compact, $f(K)$ is compact, so it is closed by KC.

$T_2$ implies $\Delta \subseteq X^2$ is closed. So $\Delta \cap K$ is trivially closed in $K$ for any $K \subseteq X^2$.

Let $\{U_n\}$ and $\{V_n\}$ be countable local bases of $x \ne y$. Contrapositively, suppose $U_n \cap V_n \ne \emptyset$ for all $n$, and choose $z_n \in U_n \cap V_n$. Then $z_n \to x$ and $z_n \to y$.

$\R$ is homeomorphic to $\R^1$.

Not sure why they expected locally here. If $f : X \to f(X) \subseteq \R$ is a homeomorphism and $X$ connected, then $f(X)$ is connected. Every connected set in $\R$ is path connected, so $f(X) \simeq X$ is path connected.

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

If no nonempty disjoint open sets exist, $R_1$ implies all points are indistinguishable.

Let $\mathcal{V}$ be all nbds of $x$. Then $\bigcap \mathcal{V}$ is an open set. For any $y \ne x$, there is some nbd $U_y$ of $x$ with $y \notin U_y$, so $\{x\} = \bigcap \mathcal{V}$.

Any network of open sets is a $k$-network. If $K \subseteq \bigcup \mathcal{N}^* \subseteq V$ for some $K$ compact, we can take a finite subcollection of $\mathcal{N}^*$. Any basis is a network. Therefore, a countable basis is a countable $k$-network.

If $A \subseteq X$ is discrete, it can be covered by nbds which contain one element each. So if $A$ were to be Lindelöf, it has to have countably many elements.

If $\mathcal{V}$ is the collection of all nbds of $x$, $W = \bigcap \mathcal{V}$ is the smallest nbd of $x$, so $\{W\}$ is a compact local basis of $x$: for any $\mathcal{W}$ open cover of $W$, there must be some $V \in \mathcal{W}$ with $x \in V$, and then $\{V\}$ is a subcover.

If $\mathcal{V}$ is the collection of all nbds of $x$, $W = \bigcap \mathcal{V}$ is the smallest nbd of $x$, so $\{W\}$ is a local basis of $x$.

If $X = \{x_1, x_2, \dots\}$ is countable, define $K_n = \{ x_1, x_2, \dots, x_n \}$. Then $X = \bigcup_{n=1}^\infty K_n$ and every compact/finite set is contained in some $K_n$.

Cover a compact set $K$ with finite nbds. It has a finite subcover of finite sets, so it must be finite.

Every point $p$ is associated with the constant map $x \mapsto p$, which is continuous. So $X$ has at most as many continuous self-maps.

Let $p$ be the isolated point with nbd $V$ so that $V \cap X = \{p\}$. Every other point $x \ne p$ has a nbd not containing $p$, so if $U$ is the union of all open sets not containing $p$, we have $U \cap V = \emptyset$ and $U \cap V = X$.

Every singleton is a retract: the constant function $f : X \to \{p\}$ is continuous.

If $p$ is isolated, any set containing $p$ cannot be nowhere empty, since it contains the open set $\{p\}$.

Let $W$ be the smallest nbd of $x$. Then $W$ must be path connected since it is indiscrete as a subspace topology, and so any function $f : [0, 1] \to W$ must be continuous.

In a $T_1$ space, finite sets are discrete, in particular, they are $T_2$.

$X$ is a countable union of singletons. For each $x$, $\cl{\{x\}} = \{x\}$ because $X$ is $T_1$, and $\text{int}(\{x\}) = \emptyset$ because no point is isolated.

Suppose that for every $K$ compact $T_2$ and $f : K \to X$ continuous, that $f^{-1}(A)$ is open. In particular, for each $K \subseteq X$ compact $T_2$, the inclusion $\iota : K \to X$ is continuous, and so $\iota^{-1}(A) = K \cap A$ is open. So $A$ is open.

Suppose that for every $K$ compact and $f : K \to X$ continuous, that $f^{-1}(A)$ is open. In particular, for each $K$ compact and $T_2$, $f^{-1}(A)$ is open. So $A$ is open.

A metric space is Hausdorff: For $x \ne y$, let $r = d(x, y)/2$ and then $B(x, r) \cap B(y, r) = \emptyset$.

$\R^n$ is a normed space, so it is metrizable.

Let $x \ne y$ and $V$ a pseudometrizable nbd of $x$. If $y \notin V$, we’re done. Otherwise, if they are indistinguishable, we have $r = d(x, y) > 0$ and then $y \notin B(x, r)$.

Let $V$ be a nbd such that $V \simeq W \subseteq \R^n$. If $\R^n$ is locally compact, $W \simeq V$ is as well. But that holds since the balls $B(x, \delta)$ form a basis and $\cl{B(x, \delta)}$ is compact in finite-dimensional metric spaces.

Take a star-finite open cover. $x$ is in some element $U$, which is open. So for any nbd $V$ of $x$, $U \cap V$ is a nbd which intersects finitely many elements. So it is locally finite.

The map $\phi(x) = (ba^{-1})x$ satisfies $\phi(a) = b$, is bijective, continuous, and $\phi^{-1}(x) = (ab^{-1})x$ is also continuous.

Any bijective function is a homeomorphism. Just take $\Phi$ as the permutation that swaps $a$ and $b$.

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

Countably compact implies every sequence has an accumulation point, which would contradict being discrete.

If $A,B$ are closed and disjoint, just apply the theorem to $A \subseteq U = B^C$, since $A$ must be countable. Then $A \subseteq V \subseteq \cl{V} \subseteq U$ so that $B \subseteq W = \cl{V}^C$ and $V \cap W = \emptyset$.

Any converging sequence is eventually constant, and this constant has to be unique.

If $x \in U$ and $y \in V$ with $U \cap V = \emptyset$, in particular, $y \notin \cl{U}$. So $x \in W = \text{int}(\cl{U})$ is regular and $y \notin W$. (blaze it)

If $x \in U$ is a nbd with $y \notin U$, let $O$ be a regular basis element with $x \in O \subseteq U$.

If $X$ has one or two points, clearly $X \setminus \{p\}$ has either 1 point (trivially totally disconnected), or it’s empty (also totally disconnected, since there are no connected components).

$\{x\}$ is homeomorphic to $\R^0$.

Some nbd around the isolated point has to be homeomorphic to $\R^0$. But that implies every point has a nbd homeomorphic to $\R^0$, and so every point is isolated.

For each $A \subseteq X$, as it is separable subspace, there is a countable $D \subseteq A$ with $A \subseteq \cl{D}$. So then $\cl{A} \subseteq \cl{\cl{D}} = \cl{D}$.

If $g \ne 1$, then the map $x \mapsto gx$ is continuous yet has no fixed point.

If $p \ne q$ were to be topologically indistinguishable, let $f(x) = p$ for all $x \ne p$ and $f(p) = q$. Then $f$ is continuous yet has no fixed point.

If $\text{card}(X) \ge \mathfrak{c} = \text{card}([0, 1])$, then there exists an injective function $f : [0, 1] \to X$, which is clearly continuous in the trivial space, and a path from $f(0)$ to $f(1)$.

If $f : X \to Y \subseteq \R^n$ is a homeomorphism, the balls of the form $B(r, 1/n)$ for $r \in \Q$ and $n \in \N$ form a countable basis for $\R^n$, so $f^{-1}(B(r, 1/n))$ forms a basis for $X$.

If $X \simeq Y \subseteq \R^n$, note that $\R^n$ is metrizable because it’s a normed vector space, and being metrizable is hereditary, so $Y \simeq X$ is metrizable.

If $\{X_i\}_{i \in I}$ is the partition for the topology of $X$, then for each $x \in X$, the set $X_i$ such that $x \in X_i$ is the smallest nbd of $x$.

Any set of the partition is a clopen set. So if $X$ is the only nonempty clopen set, the partition is $\{X\}$.

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

$T_0$ implies $x \ne y \implies \cl{\{x\}} \ne \cl{\{y\}}$. Thus, if $A = \cl{\{x\}}$, then this $x$ must be unique.

If there’s a basis of compact open sets, the ones intersecting $x$ form a local basis for $x$.

The open sets are trivially closed under finite intersections and form a basis. Since they are all compact, including $X$ itself, the space is spectral.

Take a basis of open injectively path connected sets. There’s a point $p$ of which the only possible basis element containing $p$ is $X$. So $X$ is injectively path connected.

Let $\mathcal{N} = \{ \{x\} \ : \ x \in X \}$ be the network of singletons. $\mathcal{N}$ is locally finite: Every point has a finite nbd, so it obviously intersects finitely many elements of $\mathcal{N}$.

Every countably compact set is also Lindelöf, so it is compact (T106), hence it is closed.