By one of the equivalent definitions.

Being a W-space is hereditary.

A space is a subspace of itself (and is homeomorphic to itself).

We first prove that for every decreasing sequence $\{F_n\}$ of closed sets with $\bigcap_{n < \omega} F_n = \emptyset$, there exists a sequence $\{A_n\}$ of closed $G_\delta$-sets such that $\bigcap_{n < \omega} A_n = \emptyset$ and $F_n \subseteq A_n$ for all $n$. Let $\{F_n\}$ be such a sequence. Using Ishikawa’s characterization, there exists a sequence of open sets $\{U_n\}$ with $F_n \subseteq U_n$ and $\bigcap_{n < \omega} U_n = \emptyset$. That means $F_n \cap F'_n = \emptyset$ where $F'_n = U_n^C$. By Urysohn’s lemma, let $f_n : X \to [0, 1]$ with $f(F_n) \subseteq \{0\}$ and $f(F'_n) \subseteq \{1\}$. Then define $V_{nk} = f_n^{-1}([0, 1/k))$, which proves $A_n = \bigcap_{k=1}^\infty V_{nk} = f^{-1}(0)$ is a closed $G_\delta$-set with $F_n \subseteq A_n$. Now to show $\{A_n\}$ has empty intersection, we just have to show $A_n \subseteq U_n$ for each $n$. But $x \notin U_n \implies f(x) = 1 \implies f(x) \ne 0 \implies x \notin A_n$.

Now let $\{U_n\}$ be a countable open cover. Define $F_n = X \setminus \bigcup_{k \le n} U_n$. Then $\{F_n\}$ is decreasing and must have empty intersection. Let $\{A_n\}$ be the closed $G_\delta$-sets with empty intersection such that $F_n \subseteq A_n$. Then each $B_n = X \setminus A_n$ is an $F_\sigma$-set, so we write $B_n = \bigcup_{k < \omega} F'_{nk}$. Note that by the way we constructed the $V_{nk}$ above, we could ensure that $V_{nk} \supseteq \cl{V_{n(k+1)}}$. So here, we can assume $\{F'_{nk}\}_{k < \omega}$ is increasing with $F'_{nk} \subseteq \text{int}(F'_{n(k+1)})$. Define $H_{nk} = \text{int}(F'_{nk})$. Then $B_n = \bigcup_{k < \omega} H_{nk}$ and $F'_{nk} \subseteq B_n \subseteq \bigcup_{k \le n} U_n$.

If we define $V_n = U_n \setminus \bigcup_{k < n} F'_{nk}$, each $V_n$ is open. For each $x$, let $n$ be the first index with $x \in U_n$ and so $x \notin \bigcup_{k < n} U_k \implies x \notin F'_{nk}$ for each $k < n$, so that $x \in V_n$, proving $\{V_n\}$ is an open refinement of $\{U_n\}$. Finally, since $\bigcap_{n < \omega} A_n = \emptyset$, there must be some $n$ with $x \notin A_n$, meaning $x \in F'_{n(k-1)} \subseteq H_{nk}$ for some $k$. Then for $i > \max \{n, k\}$, $H_{nk} \subseteq F'_{ni}$ so that $H_{nk} \cap V_i = \emptyset$. So $H_{nk}$ is a nbd of $x$ which intersects at most $V_n$ with $n < i$, proving it is locally finite.

Source: C. H. Dowker, On Countably Paracompact Spaces, Theorem 2 (d) ⇒ (a)

As the partition forms a basis for $X$, it must refine any open cover, and each element of the partition is a clopen set.

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

Denote $x \sim y$ to mean $x$ and $y$ are indistinguishable. For each $x$, let $W(x)$ be its smallest nbd. Clearly $x \sim y \implies y \in W(x)$ by definition, and for each $y \not\sim x$, there is a nbd $V$ of $x$ with $y \notin V$, so $y \notin W(x)$. Furthermore, this proves that $\sim$ is an equivalence relation, since $x \sim y \iff W(x) = W(y)$ and the sets $[x]_{\sim} = W(x)$ form a partition of $X$.

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 $x$.

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 $f : X \to Y \subseteq \R^n$ 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 $X$.

By definition.

By definition.

The partition would be $\{X\}$ itself, the trivial partition.

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.

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 $f(p) = p$.

If $X$ is disconnected, let $A$ be a clopen set with $p \in A$ and $q \notin A$. Then the function $f : X \to X$ defined as $f(x) = q$ at $x \in A$ and $f(x) = p$ at $x \notin A$, then $f$ is continuous yet has no fixed point.

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

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.

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

Initially, for each $x \in X$, if a sequence $(y_n)$ does not converge to $x$, then there exists a nbd $V$ of $x$ such that for all $n_0 \in \omega$, there exists some $n > n_0$ such that $y_n \notin V$. This means there are infinitely many $n$ for which $y_n \notin V$, and so we can construct a subsequence $(y_{n_k})$ such that (1) it doesn’t have a converging subsequence either and (2) $y_{n_k} \notin V$ for all $k$.

Let $X = \{x_n\}$. Contrapositively, let $\mathbf{y} = (y_n)$ be a sequence with no converging subsequence. Let $V_0$ be a nbd of $x_0$ and construct a subsequence $\mathbf{y}_0$ of $\mathbf{y}$ such that every term in $\mathbf{y}_0$ lies outside $V_0$. Inductively, given $\mathbf{y}_n$, choose $V_{n+1}$ a nbd of $x_{n+1}$, and let $\mathbf{y}_{n+1}$ be a subsequence of $\mathbf{y}_n$ which lies entirely outside of $V_{n+1}$. Note that by construction, each $\mathbf{y}_n$ is a subsequence of $\mathbf{y}_k$ for any $k < n$, so all of its terms lie outside of $\bigcup_{k \le n} V_k$. This way, we’ve constructed an open cover $\{V_n\}$ of $X$ with no finite subcover, and so $X$ is not compact.

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.

Let $\{K_n\}_{n=1}^\infty$ be a cover of compact nbds with $K_n \subseteq \text{int}(K_{n+1})$. Let $\mathcal{U}$ be any open cover. For each $n$, there is a finite $\mathcal{F}_n \subseteq \mathcal{U}$ which covers $K_n$. Now we use the fact that each $K_n$ is closed by KC, and so $U_n = \text{int}(K_n) \setminus K_{n-1}$ are all open (to define it at $n=1$, denote $K_0 = \emptyset$). Thus, each $U_n$ is covered by the finite sets $\mathcal{F}_n$, and if we define $\mathcal{R}_n = \{ U_n \cap F \ : \ F \in \mathcal{F}_n \}$, we get that $\mathcal{R} = \bigcup_{n=1}^\infty \mathcal{R}_n$ is an open refinement, where for each $x$, take the smallest $n$ for which $x \in \text{int}(K_n)$ so that $x \in U_n$ and so $x$ intersects at least one and at most all (finitely many) elements of $\mathcal{R}_n$.

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 $s > t$.

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 $x = y$.

Vacuously true: there are no disjoint subsets with at least two points each.

Let $K \subseteq X$ compact and $x \notin K$. Then $K \cup \{x\}$ is compact and $T_2$. $T_2$ spaces separate points from compact sets, so there exists a nbd $V$ of $x$ with $V \cap K = \emptyset$. So $K$ is closed.

Let $x \in U \subseteq K$ and $y \in V \subseteq K'$ with $U,V$ open, $K,K'$ compact, and $x \ne y$. Then $L = K \cup K'$ is compact, so it is $T_2$. Thus, let $x \in W$ and $y \in W'$ be nbds such that $W \cap W' \cap L = \emptyset$. Then $(U \cap W) \cap (V \cap W') = \emptyset$ separates $x$ and $y$ by open sets in $X$.

The space has a coarser metrizable topology, and any coarser topology of a separable space is separable as well (a dense set remains dense in coarser topologies).

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.

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 $V \cap W = \emptyset$.

Let $H = \{x_n\}$ and $H \subseteq U$ open. Then $x_n \notin F = U^C$ for each $n$, so by regularity, let $x \in V_n$ and $F \subseteq U_n$ with $V_n \cap U_n = \emptyset$. Then $A \subseteq V = \bigcup_{n < \omega} V_n$ and $F \subseteq W = \bigcap_{n < \omega} U_n$ with $V \cap W = \emptyset$. But that implies $\cl{V} \cap F = \emptyset$, so $H \subseteq V \subseteq \cl{V} \subseteq U$.

If you remove “countable” from the definition of pseudonormal, that’s precisely one property of normal spaces. If $H$ is closed and $H \subseteq U$ is open, then $H \cap U^C = \emptyset$, so there are open sets $H \subseteq V$ and $U^C \subseteq V'$ with $V \cap V' = \emptyset$. In particular, $\cl{V} \subseteq U$.

Contrapositively, let $f : X \to \R$ is continuous and non-constant. So $f(x) = \alpha$ and $f(y) = \beta$ with $\alpha \ne \beta$. Clearly $A_1 = f^{-1}(\alpha)$ and $A_2 = f^{-1}(\beta)$ are disjoint. They are zero sets, since $A_1 = g^{-1}(0)$ for $g(x) = f(x) - \alpha$ and similarly, $A_2$ is a zero set. So $A_1 \subseteq B_1$ and $A_2 \subseteq B_2$ with $B_1,B_2$ disjoint clopen sets, by strongly zero-dimensional. In particular, $x \in B_1 \subset X$, so the space is not connected.

Given $x \ne y$, let $f : X \to [0, 1]$ such that $f(x) = 0$ and $f(y) = 1$. Then $V_n = f^{-1}([0, 1/n))$ are all open. Then $x \in B = \bigcap_{n=1}^\infty V_n$ is open in a P-space, and $B$ is closed, since $B = f^{-1}\left( \bigcap_{n=1}^\infty [0, 1/n) \right) = f^{-1}(0)$.

Start by showing regularity. Let $x \notin A$ and $A$ closed. Construct the pairs $(U, V)$ of nbds such that $x \in U$ and $y \in V$ for some $y \in A$ with $U \cap V = \emptyset$. Then the sets $V$ form an open cover of $A$. Closed sets are Lindelöf, so take $\{(U_n, V_n)\}$ a countable subcollection with $\{V_n\}$ still covering $A$. Then $x \in U \coloneqq \bigcap_{n < \omega} U_n$ is open in a P-space, and $A \subseteq V \coloneqq \bigcup_{n < \omega} V_n$, with $U \cap V = \emptyset$.

Now let $A,B$ be closed and disjoint. Construct the pairs $(U, V)$ of nds such that $A \subseteq U$ and $x \in V$ for some $x \in B$ with $U \cap V = \emptyset$. Similarly, we take $\{(U_n, V_n)\}$ such that $B \subseteq V \coloneqq \bigcup_{n < \omega} V_n$ and $A \subseteq U \coloneqq \bigcap_{n < \omega} U_n$, which is open in a P-space, and $U \cap V = \emptyset$.

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.

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 $\bigcap U_m = \emptyset$.

Let $x \notin A$ with $A$ closed. Being $T_1$, $\{x\}$ is a countable closed set, and $\{x\} \subseteq U = A^C$ is open. So let $V$ be open with $x \in V \subseteq \cl{V} \subseteq A^C$. Then $A \subseteq \cl{V}^C = W$ with $V \cap W = \emptyset$.

$\leq$ means $\lt$ or $=$.

A locally-finite collection in a Lindelöf space is countable, seen in the proof of (T387), and so the network is a countable union of countable networks, hence, countable.

A locally-finite collection in a Lindelöf space is countable: If $\mathcal{A}$ is locally finite, the sets $\mathcal{U}$ which intersect finitely many elements of $\mathcal{A}$ is an open cover, so if $\mathcal{U}^*$ is a countable subcover, then every element of $\mathcal{A}$ intersects some $U \in \mathcal{U}^*$, yet only finitely many intersect each $U$, so $\mathcal{A}$ is countable.

Let $\mathcal{U}$ be an open cover. Let $\mathcal{F}$ be the collection of finite subsets of $\mathcal{U}$ and define a function $v(F) = \bigcup F$ on $\mathcal{F}$. Then $\{v(F)\}_{F \in \mathcal{F}}$ is an $\omega$-cover, since $\mathcal{U}$ is a cover of any finite (compact) set $K$, which has a finite subcover, meaning $K \subseteq \bigcup F$ for some finite $F \subseteq \mathcal{U}$. Let $\mathcal{A} \subseteq \mathcal{F}$ be countable so that $\{v(F)\}_{F \in \mathcal{A}}$ is a countable $\omega$-subcover. Then $\bigcup \mathcal{A} \subseteq \mathcal{U}$ is a countable subcover.

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 $B = \bigcap_{n < \omega} \cl{U_n}$.

Any intersection of open sets is open, including countable ones.

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

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

Let $\mathcal{B}$ be a basis of clopen sets and let $\mathcal{U}$ be an open cover. For each $U \in \mathcal{U}$, define $\mathcal{B}_U = \{ B \in \mathcal{B} \ : \ B \subseteq U \}$. Then $\mathcal{V} = \bigcup_{U \in \mathcal{U}} \mathcal{B}_U$ is an open refinement of clopen sets. By Lindelöf, let $\{B_n\}$ be a countable subrefinement of $\mathcal{V}$. If $B^*_n = B_n \setminus \bigcup_{k < n} B_k$, then $\{B^*_n\}$ are all clopen and pairwise disjoint, so it is a clopen refinement which partitions $X$.

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.

If an open refinement is a partition, its elements are pairwise-disjoint, so trivially star-finite.

Let $\mathcal{N}$ be a countable network. For each $x \in X$, define $f(x) = \{ N \in \mathcal{N} \ : \ x \in N \}$. If $x \ne y$, by $T_0$ and without loss of generality, assume $x \in U$ and $y \notin U$ for some open $U$. Since $U$ is a union of network elements, there is some $N$ with $x \in N$ yet $y \notin N$. Then $f : X \to \mathcal{P}(\mathcal{N})$ is injective, and so $\card{X} \le \card{\mathcal{P}(\mathcal{N})} = 2^{\aleph_0} = \mathfrak{c}$.