Let $\mathcal{U}$ be an open cover, and $\mathcal{V}$ a point-countable open refinement. Let $D$ be countable with $\overline{D} = X$. For each $x \in D$, let $\mathcal{V}_x \subseteq \mathcal{V}$ be the finite set of elements containing $x$. Then $\mathcal{V}^* = \bigcup_{x \in X} \mathcal{V}_x \subseteq \mathcal{V}$ is countable. For each $V \in \mathcal{V}^*$, choose a $U_V \in \mathcal{U}$ such that $V \subseteq U$. Then $\mathcal{U}^* = \{ U_V \ : \ V \in \mathcal{V}^* \}$ is countable. Any $x \in X$ must have some $V \in \mathcal{V}$ as an open neighborhood, and since $D$ is dense, $V \in \mathcal{V}_d$ for some $d \in D$, which proves $x \in U_V$ and $\mathcal{U}^*$ is a subcover.

It suffices to show regularity. Let $A$ be closed and $x \notin A$. Let $\{V_n\}$ be a countable basis of $x$. Define the collection $\mathcal{U}_a(n) = \{ U \text{ nbd of } a \ : \ U \cap V_n = \emptyset \}$ and $\mathcal{U}(n) = \bigcup_{a \in A} \mathcal{U}_a(n)$. Clearly $\mathcal{U}_n \cap V_n = \emptyset$ for each $n$. $X$ is $T_2$, so $\{\mathcal{U}_n\}$ is an open cover of $A$. A closed subspace of a countably compact space is countably compact, so let $\{\mathcal{U}_1, \dots, \mathcal{U}_k\}$ be a finite subcover and

$$V = \bigcap_{n=1}^k V_n \qquad U = \bigcup_{n=1}^k \mathcal{U}_n \quad \implies \quad x \in V, A \subseteq U, \text{ and } U \cap V = \emptyset$$

Let $x \ne y$. Note that $\overline{\{x\}} \cap \overline{\{y\}}$ cannot be empty, so let $z$ be in their intersection and define the path $f : [0, 1] \to X$ where

$$f(t) = \left\{ \begin{array}{ll} x & t < 1/2 \\ z & t = 1/2 \\ y & t > 1/2 \end{array} \right.$$

Any nbd of $z$ must contain both $x$ and $y$, so if $V \subseteq f$ is any open set, $f^{-1}(V)$ is one of the following sets: $\{[0, 1/2), (1/2, 1], [0, 1]\}$, which are all open in $[0, 1]$.

Let $D$ be a countable dense subset. For each $x \in X$, define $\mathcal{D}_x = \{ A \subseteq D \ : \ x \in \overline{A} \}$, which is nonempty as it has $D$ itself. If $x \ne y$, being $T_2$ ensures nbds $U$ of $x$ and $V$ of $Y$ such that $U \cap V = \emptyset$. Clearly $D \cap V \in \mathcal{D}_y$, and $U \cap (D \cap V) = \emptyset$ implies $x \notin \overline{D \cap V}$, so $D \cap V \notin \mathcal{D}_x$. Therefore, the function $f : X \to 2^{2^D}$ defined as $f(x) = \mathcal{D}_x$ is injective, and $\text{card}(X) \le \text{card}(2^{2^D}) = 2^\mathfrak{c}$

Let $A \subseteq X$ such that $K \cap A$ is open in $K$ for every $K \subseteq X$ compact. We wish to show $A$ is open in $X$. For $x \in A$, let $x \in V \subseteq K$ where $V$ is open and $K$ is compact. As $A \cap K$ is open in $K$, there exists some open $W$ such that $x \in W \cap K \subseteq A \cap K$. Thus, choose $U = V \cap W$. Clearly it is open and $x \in U \subseteq W \cap K \subseteq A$.

Let $A = \{x_n\}$ be a countable, closed and discrete set. Then we could define $f : A \to \R$ as $f(x_n) = n$. This is clearly continuous ($A$ is discrete), and we can extend it to $F : X \to \R$ via the Tietze extension theorem. But then $F$ must be bounded, so $A$ has to be finite.

Let $\tau$ be the topology of $X$ and $\tau'$ the order topology of $X$. The open rays $(x, {\rightarrow})$ and $({\leftarrow}, y)$ are all open in $\tau$, and they form a subbasis for $\tau'$, so $\tau' \subseteq \tau$. Then the identity map $\text{id} : (X, \tau) \to (X, \tau')$ is continuous. But $\tau$ is compact and $\tau'$ is Hausdorff (LOTS spaces are $T_2$), so this is a homeomorphism and $\tau = \tau'$.

If the space is locally euclidean, this means the open sets which are embeddable in some $\R^n$ is an open cover, and being Lindelöf, we can take an countable subcover $\{V_n\}$, where each has a homeomorphism $f : V_n \to f(V_n) \subseteq \R^{k_n}$. For each $n$, consider

$$\mathcal{U}_n = \{ f^{-1}(B(r, 1/m)) \ : \ r \in \Q^{k_n} \ \text{ and } \ m \in \N \}$$

Because $\Q^{k_n}$ is dense in $\R^{k_n}$, the balls $B(r, 1/m)$ form a countable basis of $f(V_n)$, so $\mathcal{U}_n$ is a countable basis of $V_n$ (since $f$ is homeomorphic). This gives us a countable basis $\bigcup_{n < \omega} \mathcal{U}_n$ of $X$.

Suppose a set $A$ is sequentially closed. Let $p \in X$ and $(x_\alpha)_{\alpha < \lambda} \subseteq A$. Take $V$ a countable nbd of $p$. If $x_\alpha \to p$, then there is an ordinal $\mu$ for which $x_\alpha \in V$ for all $\alpha \ge \mu$. The set of ordinals $\mu \le \alpha < \lambda$ is isomorphic to some $\lambda'$. So we have $(y_\alpha)_{\alpha < \lambda'} \subseteq V$, still with $y_\alpha \to V$.

From a converging transfinite sequence within a countable set, we can extract a $\omega$-sequence that still converges to the same value, this was done in (T211). So we construct a sequence $(z_n)$ with $z_n \to p$, and so $p \in A$ as we wished to prove.

Let $A$ be the set of points which have a countable nbd. Then $A$ is locally countable and Lindelöf, so it is countable (T499). This must imply $B = A^C$ is empty, as otherwise it would need to have an isolated point $p \in B$ which has no countable nbd, so $p \in U$ and $U \cap B = \{p\}$ would imply $U \setminus \{p\} \subseteq A$ is uncountable (thanks to Almanzoris, I had proved it on my own but it was far more complicated than this nice argument).

We prove it is normal and $G_\delta$ (T257). For the latter, let $A$ be any closed set, and define $V_n = \bigcup_{x \in A} B(x, 1/n)$. Let $p \in \bigcap_{n=1}^\infty V_n$. For each $n$ there is an $x_n \in A$ such that $p \in B(x_n, 1/n)$ and therefore, $x_n \to p$. Since $A$ is closed, $p \in A$. This proves $A = \bigcap_{n=1}^\infty V_n$ is $G_\delta$.

To prove normality, we use the fact that for a closed set $Y$, $d(Y, x) = 0 \implies x \in Y$. For any disjoint closed sets $A$ and $B$, define $r_x = d(B, x)/2$ for each $x \in A$ and $s_y = d(A, y)/2$ for each $y \in B$. Define $U = \bigcup_{x \in A} B(x, r_x)$ and $V = \bigcup_{y \in B} B(y, s_y)$. It’s clear $A \subseteq U$ and $B \subseteq V$. If $p \in U \cap V$, then $d(x, p) < r_x$ and $d(y, p) < s_y$ for some $x \in A$ and $y \in B$. But then $d(x, y) < r_x + s_y = d(B, x)/2 + d(A, y)/2 \le d(x, y)$ is a contradiction.

We use the axiom of choice (specifically the well-ordering principle) to denote a bijection $f : X \to \alpha$ where $\alpha$ is some initial ordinal. If $\alpha < \omega$, the order topology on $\alpha$ is discrete, so $f$ induces an order in $X$ which is discrete. If $\kappa = \card{\alpha}$ is infinite, then there is a bijection $g : \alpha \to \alpha \times \Z$ (the well-ordering of $\alpha$ allows us to construct a bijection from $\alpha \times \Z$ to $\alpha$ and then apply Schröder–Bernstein). The dictionary order on $\alpha \times \Z$ is discrete since the element $(\beta, n)$ is between $(\beta, n-1)$ and $(\beta, n+1)$, so the bijection $g \circ f$ yields a discrete order in $X$.

Let $A$ be countable and dense. Let $\mathcal{B}$ be the finite intersections of open rays of the form $(x, {\rightarrow})$ or $({\leftarrow}, x)$ for $x \in A$. For any $z \in X$, assume it is not a minimum or maximum of $X$ and suppose $z \in (a, b)$. The proof is similar for those cases by taking $[z, b)$ or $(a, z]$. Then $(a, z) \ne \emptyset$, as otherwise $X = ({\leftarrow}, z) \cup (a, {\rightarrow})$ would be disconnected. Furthermore, there must be some $x \in A$ in this interval, otherwise some $w \in (a, z)$ with $(a, z) \cap A = \emptyset$ would imply $w \notin \cl{A}$. Similarly, let $y \in (z, b)$ and so $z \in (x, y) \subseteq (a, b)$. Since $(x, y) \in \mathcal{B}$, this is a countable basis.

Let $\{K'_n\}$ be the compact sets of which every compact $K \subseteq K_n$ for some $n$, and define $K_n = \bigcup_{n \le k} K'_k$, which has the same property. Let $\{U_n\}$ be a local basis for some $x$ and $V_n = \bigcap_{k \le n} U_k$ is also a countable local basis. We wish to prove $x \in V_n \subseteq K_n$ for some $n$. By contradiction, assume there is no such $n$. That means we can extract a sequence $x_n \in V_n \setminus K_n$. Then $x_n \to x$ and so $K = \{x_n \ : \ n < \omega\} \cup \{x\}$ is compact. But then $K \subseteq K_n$ for some $n$, which contradicts $x_n \notin K_n$.

We first show the space is Alexandrov. Note that if $\{V_i\}$ is any family of open sets, and $K$ is any compact subspace, $K \cap V_i$ is open in $K$ and also finite. Thus, by removing duplicates, the family $\{K \cap V_i\}$ is finite and so $\bigcap_{i} (K \cap V_i)$ is a finite intersection of open sets. Therefore, $K \cap \bigcap_{i} V_i$ is open for every compact $K$ and so by the $k_1$-space property, $\bigcap_{i} V_i$ is open. Being Alexandrov, let $V$ be the smallest nbd of $x$. Then it must be compact (T284), which is finite.

$X$ must be $T_1$: The constant sequence $x_n = x$ could not converge to $y \ne x$ (T226) and so there is a nbd of $y$ not containing $x$, and vice versa. If $f : [0, 1] \to X$ is a path, $f([0, 1])$ is connected, so it can’t be a finite union of multiple closed sets, hence it is either a singleton or infinite. Take $\{y_n\}$ countably infinite and for each $n$ choose $t_n \in [0, 1]$ so that $f(t_n) = y_n$. Because $[0, 1]$ is sequentially compact, $t_{n_k} \to t$ for some subsequence and then $y_{n_k} \to f(t)$ by continuity of $f$, yet $\{y_{n_k}\}$ isn’t eventually constant, a contradiction. Thus, $f([0, 1])$ is a singleton and the path is constant.

It suffices to show finite sets have isolated points: If $A$ is any set, take $V \subseteq A$ a finite nbd, then $W$ the nbd of an isolated point of $V$. Then it must be isolated in $A$ since $V$ is open $\implies$ $W$ is open.

By induction, the isolated point of $\{x_1\}$ is trivial and $\{x_1, x_2\}$ has an isolated point by $T_0$. For the set $F = \{x_1, \dots, x_{n+1}\}$, let $x_j$ with nbd $U$ such that $U \cap \{x_1, \dots, x_n\} = \{x_j\}$. If $x_{n+1} \notin U$, then $x_j$ is the isolated point of $F$. Otherwise, let $y \in \{x_j, x_{n+1}\}$ with nbd $V \cap \{x_j, x_{n+1}\} = \{y\}$. Then $y$ is the isolated point of $F$ since $(U \cap V) \cap F = \{y\}$.

Let $f : X \to A$ be a continuous map with $f|_A = \text{id}_A$. Define $h : X \to X^2$ as $h(x) = (x, f(x))$. If $U \times V \subseteq X^2$, then $h(U \cap f^{-1}(V)) \subseteq U \times V$, so $h$ is continuous. Since $X$ is $T_2$, the diagonal $\Delta \subseteq X^2$ is closed, and $h^{-1}(\Delta) = \{ x \in X \ : \ f(x) = x \} = A$ is closed.

Let $A \subseteq X$ such that there is some continuous $f : X \to A$ with $f|_A = \text{id}_A$. To prove $A$ is closed, we use the $k_1$-space property, and prove $K \cap A$ is closed in $K$ for each compact $K \subseteq X$. $K$ is closed, so $f^{-1}(K)$ is closed, and so is $K' = f^{-1}(K) \cap K$. Then $f(K') = K \cap A$ is compact, so it is closed.

We wish to show that for every $x$ and $V$ nbd of $x$, there exists a clopen set $B$ with $x \in B \subseteq V$. Define $U_0 = V$. By regularity, there some $x \in U_1$ and $U_0^C \subseteq U_1^*$ such that $U_1 \cap U_1^* = \emptyset$. Inductively, given $x \in U_n$, there exists $x \in U_{n+1}$ and $U_n^C \subseteq U_{n+1}^*$ with $U_{n+1} \cap U_{n+1}^* = \emptyset$. Thus, we construct a decreasing sequence $(U_n)$ such that $x \in B \coloneqq \bigcap_{n < \omega} U_n$ and $\cl{U_{n+1}} \subseteq U_n$ for all $n$. $B$ is open by P-space, and closed because $B = \bigcap_{n < \omega} \cl{U_n}$.

Suppose $\{F_n\}$ is a decreasing sequence of compact sets with $\bigcap F_n = \emptyset$. For each $n$, we have $F_n = \bigcap_{k < \omega} V_{nk}$ where $\{V_{nk}\}$ are open and decreasing. Thus, define $U_m = \bigcap_{n \le m} V_{nm}$. Clearly $F_m = \bigcap_{n \le m} F_n \subseteq \bigcap_{n \le m} V_{nm} = U_m$. For any $x$, there exists some $n$ with $x \notin F_n$, which means there is some $k$ with $x \notin V_{nk}$. But then $x \notin V_{nk'}$ for all $k' \ge k$, so just take $k' \ge \max \{k, n\}$ and $x \notin U_{k'} \subseteq V_{nk'}$. So $\bigcap U_m = \emptyset$.

We first show that if $A$ is countable and $x \notin A$, then $x \notin \cl{A}$. Just enumerate $A = \{x_n\}$ and let $U_n$ be a nbd of $x$ with $x_n \notin U_n$. Then $V = \bigcap U_n$ is a nbd of $x$ (using P-space) and $V \cap A = \emptyset$. This proves every countable set is closed and discrete, since for any $x \in A$, $B = A \setminus \{x\}$ is countable with $x \notin B$, so there must be a nbd $V$ of $x$ with $V \cap B = \emptyset \implies V \cap A = \{x\}$.

The space is zero-dimensional (T351). Contrapositively, if $A = \{x_n\}_{n=1}^\infty$ were to be a countably infinite set, being discrete, let $U_n$ be a nbd of $x_n$ with $A \cap U_n = \{x_n\}$, then choose $x \in B_n \subseteq U_n$ where $B_n$ is clopen. Being a P-space, note that $\tilde{B}_n \coloneqq B_n \setminus \bigcup_{m \ne n} B_m$ is also clopen with $x_n \in \tilde{B}_n$, and $\{\tilde{B}_n\}_{n=1}^\infty$ are pairwise-disjoint. $\tilde{B}_0 \coloneqq X \setminus \bigcup_{n < \omega} \tilde{B}_n$ is also clopen. Then $\{\tilde{B}_n\}$ is a partition of clopen sets of $X$ and we can define $f(x) = n$ for $x \in \tilde{B}_n$. This function must be continuous, so the space is not pseudocompact.

Let $\tau$ be the current topology of $X$ and $\tau' \subseteq \tau$ a metrizable topology on $X$ with metric $d$. If $x \ne y$, then $d(p, x) + d(p, y) > 0$ for all $p \in X$. So define

$$f(p) = \frac{d(p, x)}{d(p, x) + d(p, y)}$$

Then $f : (X, \tau') \to [0, 1]$ is continuous with $f(x) = 0$ and $f(y) = 1$. Since $\tau' \subseteq \tau$, $f : (X, \tau) \to [0, 1]$ is also continuous.

Consider $X$ with such a separable metrizable topology. Let $A$ be a countable dense subset. To prove the result, we wish to find a function $f : X \to \mathcal{P}(A)^{\N}$ which is injective, so that $\text{card}(X) \le \text{card}(\mathcal{P}(A)^{\N}) = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} = \mathfrak{c}$. Such a function is $f(x) = (f_1(x), f_2(x), \dots)$ where each $f_n : X \to \mathcal{P}(A)$ is defined as $f_n(x) = B(x, 1/n) \cap A$. As $A$ is dense, $f_n(x) \ne \emptyset$ for all $n$ and all $x$. For $x \ne y$, let $n > 2d(x, y)$ so that $B(x, 1/n) \cap B(y, 1/n) = \emptyset$, and so $f_n(x) \cap f_n(y) = \emptyset \implies f(x) \ne f(y)$.

Suppose $(x_n)$ is a sequence with $x_n \to x$ and $x_n \to y$. $\omega + 1$ with the order topology is a compact space. Define $f,g : \omega + 1 \to X$ as $f(n) = g(n) = x_n$, $f(\omega) = x$, $g(\omega) = y$. Then $f$ and $g$ are continuous. By $k_2$-Hausdorff, the set $F = \{ \alpha \in \omega + 1 \ : \ f(\alpha) = g(\alpha) \}$ is closed. Clearly by definition $\omega \subseteq F$, yet $\omega$ is not closed, since $\omega \in \cl{\omega}$. So $F = \omega + 1$ and $x = y$.

Let $p$ be the dispersion point. Clearly if $x \ne y$ are points in $X \setminus \{p\}$ then there is no path from $x$ to $y$ (disconnected implies path disconnected). So the only possible non-constant paths are from $p$ to $x \ne p$. Let $f : [0, 1] \to X$ with $f(0) = p$ and $f(1) = x$. Being $T_1$, we see that $f^{-1}(p)$ is closed, so let $t = \sup f^{-1}(p)$ and then $f(t) = p$. But for any $s > t$, the path $f|_{[s, t]}$ lies within $X \setminus \{p\}$ so it must be constant, proving $f(s) = x$ for all $s > t$, and so $f$ is discontinuous at $t$: Take a nbd of $p$ not containing $x$, yet every open interval containing $t$ must contain some $s > t$.

Similar proof to (T427). Let $\{K_n\}$ be an increasing sequence of compact sets which cover $X$. Let $\mathcal{U}$ be an open cover. For each $n$, there exists a finite subset $\mathcal{F}_n \subseteq \mathcal{U}$ such that $\mathcal{F}_n$ covers $K_n$. Then define $\mathcal{R}_0 = \mathcal{F}_0$ and $\mathcal{R}_{n+1} = \{ U \setminus K_n \ : \ U \in \mathcal{F}_n \}$. Clearly by construction, $\mathcal{R} = \bigcup_{n < \omega} \mathcal{R}_n$ must be an open refinement of $\mathcal{U}$. And for each $x$, let $n$ be the smallest index so that $x \in K_n$. This immediately implies $x \notin U$ for any $\mathcal{R}_k$ with $k > n$, and so $x$ intersects at most the nbds in $\bigcup_{k \le n} \mathcal{R}_n$, which is finite.

Firstly, note that a compact LOTS space must have minimum and maximum elements. If it didn’t have a maximum, the open intervals $({\leftarrow}, x)$ would be an open cover with no finite subcover. Similarly for the minimum. So let $x_0$ be the minimum and $y_0$ the maximum elements.

We first see that $A = \{ x \in X \ : \ f(x) > x \}$ is open. Given $x \in A$, if there is some $x < z < f(x)$, then take $U = ({\leftarrow}, z)$ and $V = (z, {\rightarrow})$. If there is no such $z$, let $U = ({\leftarrow}, f(x))$ and $V = (x, {\rightarrow})$. In both cases, we have $x \in U$, $f(x) \in V$, and every element of $V$ is strictly greater than every element of $U$. Let $W = f^{-1}(V) \cap U$. Then $y \in W \implies f(y) > y$ and so $x \in W \subseteq A$. Through a similar proof, $B = \{ x \in X \ : \ f(x) < x \}$ is open as well. Note that $A \cap B = \emptyset$, $y_0 \notin A$, and $x_0 \notin B$. Because $X$ is connected, this means $A \cup B \ne X$, and so there must be some $p \notin A \cup B$, so $f(p) = p$.

$X$ is a countable union of finite open sets. $X$ hereditarily connected, so this basis is linearly ordered and it’s possible to enumerate them as $\{B_n\}$ where $n < m \implies B_n \subset B_m$. Therefore, $B_n \nearrow X$ and $X$ is a union of finite sets.

If not discrete, there exists a non-isolated point $x$. Let $V$ be its injectively path connected nbd, which must contain some $y \ne x$. Let $f : [0, 1] \to X$ be an injective path from $x$ to $y$. Then $f([0, 1/2))$ and $f((1/2, 1])$ are connected and disjoint.