Let $A = f^{-1}(0)$ and $B = g^{-1}(0)$ be disjoint cozero sets, where $f,g : X \to \R$ are continuous. Then the function

$$h(x) = \frac{\abs{f(x)}}{\abs{f(x) + g(x)}}$$

is continuous, since the denominator would only be zero at $x$ if $x \in A \cap B$. $h$ maps $X$ to $[0, 1]$. Furthermore, $A = h^{-1}(0)$ and $B = h^{-1}(1)$. Note that $\{h^{-1}(t)\}_{0 \le t \le 1}$ is an uncountable family, yet $X$ is countable, so $h^{-1}(r) = \emptyset$ for some $r$. Then $h^{-1}([0, r))$ and $h^{-1}((r, 1])$ are clopen disjoint sets of $X$ which separate $A$ and $B$.

A space is a subspace of itself.

The space is a countable union of finite sets.

$X$ 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.

If $X$ itself is not dense, then some point is isolated.

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 $p$ 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 $U \cap V = X$.

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.

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

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.

The only family of disjoint nonempty open sets is empty, which is countable.

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

$X = \bigcup \emptyset$ where $\emptyset$ is countable and each $A \in \emptyset$ is nowhere dense.

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.

The space cannot be a partition of more than one nonempty closed set.

By definition, let $X$ be a countable union of increasing compact sets $\{K_n\}$. Now define the sequence of sets

$$L_n = \{ x \in X \ : \ \cl{\{x\}} \cap K_n \ne \emptyset \}$$

Clearly by construction, $K_n \subseteq L_n$ for each $n$, $\{L_n\}$ is increasing, and $X = \bigcup_{n < \omega} L_n$. Now let $\mathcal{U}$ be an open cover of $L_n$. Since it also covers $K_n$, let $\{U_1, \dots, U_m\}$ be a countable subcover of $K_n$. For any $x \in L_n$, there is some $y \in \cl{\{x\}} \cap K_n$. Thus, $y \in U_j$ for some $1 \le j \le m$, but that implies $x \in U_j$. So this is also a finite subcover of $L_n$, proving each $L_n$ is compact.

By contraposition, suppose $K$ is not contained in any $L_n$. So choose $x_n \in X \setminus L_n$ for each $n$, so that $\cl{\{x_n\}} \cap K_n = \emptyset$. Then define $V_n = X \setminus \bigcup_{k=n}^\infty \cl{\{x_k\}}$. We see that $\cl{\{x_k\}}$ will only intersect $K_m$ for $k < m$, so that $V_n \cap K_m = K_m \setminus \bigcup_{k=n}^{m-1} \cl{\{x_k\}}$, which is open in $K_m$. By the $k_{\omega,1}$-space property, $\{V_n\}$ is an open cover of $K$: for any $x$, $x \in K_n$ for some $n$ and so $x \notin \cl{\{x_n\}} \implies x \in V_n$. Note that it is also increasing, which means if it had a finite subcover, we’d have $K \subseteq V_n$ for some $n$, which would imply $x_n \notin K$. So $K$ is not compact.

Source: M W on stack exchange, awesome trick

If $K \cap A$ is closed for every compact $K$, in particular $K_n \cap A$ is closed for each $\{K_n\}$ stipulated by the $k_{\omega,1}$-space property, proving $A$ is closed.

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.

Globally implies locally.

Every metric is a pseudometric.

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 $y \notin B(x, r)$.

This is just “(Pseudometrizable ∧ $T_0$) $\implies$ Metrizable” (T265) but for local nbds.

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.

By definition.

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 $\card{X} \le \card{\mathcal{P}(\mathcal{N})} = 2^{\aleph_0} = \mathfrak{c}$.

By definition.

By definition.

By definition.

By definition.

By definition.

By definition.

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

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.

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 $X$.

Groups/monoids are nonempty by definition.

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 intersection of open sets is open, including countable ones.

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

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.

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.

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 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)

$\kappa < 2^\kappa$ is true for any cardinal.

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

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 $U \cap V = \emptyset$.

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 $B = f^{-1}\left( \bigcap_{n=1}^\infty [0, 1/n) \right) = f^{-1}(0)$.

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

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.