Evident.

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.

Contrapositively, suppose $\{V_n\}$ is a cover with no finite subcover. Then for $W_n = \bigcup_{m \le n} V_n$, each $X \setminus W_n$ is nonempty. So for each $n$, choose $x_n \in X \setminus W_n$. The sequence $\{x_n\}$ cannot have a converging subsequence: If $p \in X$, then $p \in V_m$ for some $m$ and so $x_n \notin V_m$ for all $n > m$.

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

Let $\{K_n\}$ be a countable compact cover of $X$ of which for all $x \in X$, some $K_n$ is a compact nbd of $x$. Let $K$ be compact. For each $x \in K$, let $n_x$ be the least element of which $K_n$ is a compact nbd. Then $\{\text{int}(K_{n_x})\}$ is a cover of $K$, which must have a finite subcover, and so finitely many $K_n$. Meaning $A \subseteq \bigcup_{n \le m} K_n$ for big enough $m$.

So define $K'_m = \bigcup_{n \le m} K_n$. Clearly $\{K'_m\}$ is a countable compact cover of $X$ and any compact is contained in some $K'_m$ by above.

Any closed set is compact, so any closure of a nbd is compact.

Take one nbd from the local basis. Its closure is compact.

By definition.

Indeed.

Contrapositively, suppose $(x_n)$ is a sequence with $x_n \to p$ yet has infinitely many terms. If needed, take an injective subsequence. If $V$ is a nbd of $p$ that is $T_2$, then there’s a $n_0$ with $x_n \in V$ for $n \ge n_0$. So by another subsequence, we can assume $(x_n)$ is an injective converging sequence in a $T_2$ space.

For each $n$, it’s possible to construct a neighborhood $V_n$ which only contains $x_n$ and no other $x_m$, nor $p$: Let $x_n \in V$ and $p \in U$ with $U \cap V = \emptyset$. $x_m \in U$ for all $m \ge n_0$, so apply the Hausdorff condition finitely many times to terms $x_m < n_0$ to construct such $V_n \subseteq V$.

Now note that $x_{2n} \to p$ and $x_{2n+1} \to p$, so if $E = \bigcup V_{2n}$ and $O = \bigcup V_{2n+1}$, then any nbd of $p$ must intersect $E$ and $O$, so $p \in \cl{E} \cap \cl{O}$. Yet by definition, $E \cap O = \emptyset$, so the space is not extremally disconnected.

We just have to argue a $k$-netork is a network: Let $\mathcal{N}$ be a $k$-network. Singletons are compact. So in a $k$-network, for each $x$ in an open set $U$, we can find $\mathcal{N}^*_x \subseteq \mathcal{N}$ with $\{x\} \subseteq \bigcup \mathcal{N}^*_x \subseteq U$ (we don’t need $\mathcal{N}^*_x$ to be finite here, so no need for the axiom of choice, just take every set of $\mathcal{N}$ containing $x$). Thus,

$$U = \bigcup_{x \in U} \{x\} \subseteq \bigcup_{x \in U} \mathcal{N}^*_x \subseteq U \ \implies \ \text{$U$ is a union of sets in $\mathcal{N}$, so it is a network}$$

A subcover is a refinement. A finite subcover is star-finite.

If finitely many intersect a nbd around the point, finitely many will intersect the point.

If true for any covers, then true for countable ones.

If true for any covers, then true for countable ones.

A subcover is a refinement. A finite subcover is star-finite.

This is Paracompact $\implies$ Metacompact, just countable this time.

Let $A$ be an infinite set. Contrapositively, suppose every limit point of $A$ is not an $\omega$-accumulation point. If $p$ were to be a limit point of $A$, there exists a nbd $V$ of $p$ of which $V \cap A = \{ a_1, \dots, a_n \}$ is finite. But now for each $a_k$, there’s $V_k$, a nbd of $p$, so that $a_k \notin V_k$ (by being $T_1$). Then $W = V \cap V_1 \cap \cdots \cap V_n$ is a nbd of $p$ with $W \cap A = \emptyset$. So $A$ is an infinite set with no limit points.

Lemma. Every open cover has a $\sigma$-locally finite refinement which is also an open cover.

Proof. Using the axiom of choice, let $\mathcal{U} = (U_\alpha)_{\alpha < \xi}$ be an open cover indexed by some ordinal $\xi$. Let $n \in \N$ be fixed. For each $\alpha < \xi$, define $S_\alpha(n) = \{ x \in X \ : \ B(x, 1/n) \subseteq V_\alpha \}$. Then define $T_\alpha(n) = S_\alpha(n) \setminus \bigcup_{\beta < \alpha} V_\beta$. This way, $\{T_\alpha(n)\}$ is defined to not only be pairwise disjoint, but so that $d(T_\alpha(n), T_\beta(n)) \ge 1/n$ for $\alpha \ne \beta$: If $x \in T_\alpha(n)$ and $y \in T_\beta(n)$, with $\beta < \alpha$, then $B(y, 1/n) \subseteq V_\beta$ and $x \notin V_\alpha$, so $d(x, y) \ge 1/n$.

So let $E_\alpha(n) = \bigcup_{x \in T_\alpha(n)} B(x, 1/3n)$. By definition, $d(E_\alpha(n), E_\beta(n)) \ge 1/3n$. To see why, if $y_1 \in E_\alpha(n)$ and $y_2 \in E_\beta(n)$ with $\beta < \alpha$, then there exist $x_1 \in T_\alpha(n)$ and $x_2 \in T_\beta(n)$ such that $d(y_1, x_1) < 1/3n$ and $d(y_2, x_2) < 1/3n$. By above, $d(x_1, x_2) \ge 1/n$, so we have that $d(y_1, y_2) \ge 1/3n$. By construction, noting the definition of $S_\alpha(n)$, we see that $E_\alpha(n) \subseteq V_\alpha$, so $\mathcal{E}_n = \{E_\alpha(n)\}$ is an open refinement of $\mathcal{U}$. It is locally finite because for any $x \in X$, the open ball $B(x, 1/6n)$ can only intersect at most one $E_\alpha(n)$, since any other element has distance at least $1/3n$.

Finally, take $\mathcal{E} = \bigcup_{n < \omega} \mathcal{E}_n$. By definition it is a $\sigma$-locally finite refinement of open sets, so we just have to prove it is also a cover. For $x \in X$, let $\alpha$ be the minimum ordinal such that $x \in V_\alpha$. As it is open, let $n \in \N$ be big enough so that $B(x, 1/n) \subseteq V_\alpha$. Then $x \in E_\alpha(n)$. $\square$

For each $n \in \N$, let $\mathcal{B}_n = \{ B(x, 1/n) \ : \ x \in X \}$ be the cover of $1/n$-balls, and let $\mathcal{R}_n$ be the $\sigma$-locally finite refinement which is also a cover, by above. Then $\mathcal{R} = \bigcup_{n < \omega} \mathcal{R}_n$ is also $\sigma$-locally finite, and clearly a basis: If $x \in B(x, 1/n)$, then $x \in \mathcal{B}_n$ and so there is some element $R \in \mathcal{R}_n$ such that $x \in R \subseteq B(x, 1/n)$.

Source: Munkres, Topology: Lemma 39.2 (p. 246) + Theorem 40.3 at step 2 (p. 251)

Let $\mathcal{V}$ be a collection of pairwise-disjoint nonempty open sets. Let $D$ be a dense countable set. For each $V \in \mathcal{V}$, $D \cap V \ne \emptyset$ means it must be the nbd of some $x_n \in D$. So let $n \sim V$ if and only if $x_n \in V$. As the collection is pairwise-disjoint, this induces a function $f : \omega \to \mathcal{V}$ with $f(n) = V$ that must be surjective, and so $\text{card}(\omega) \ge \text{card}(\mathcal{V})$.

Let $\mathcal{N} = \{ \{x\} \ : \ x \in X \}$ be the network of singletons. The open sets don’t matter: If any compact $K$ is necessarily finite, then $N^* = \{ \{x\} \ : \ x \in K \} \subseteq \mathcal{N}$ is finite and $K \subseteq N^*$. Since $X$ is countable, so is $\mathcal{N}$.

This is, in essence, one of the directions of the Nagata-Smirnov theorem, which in its usual form, states

$$\text{Metrizable $\iff$ ($T_3$ ∧ Has a $\sigma$-locally finite basis)}$$

Note that if $Y$ is the Kolmogorov quotient of $X$, then $Y$ is $T_3$ and the $\sigma$-locally finite basis of $X$ induces one on $Y$ via the surjective open map $p : X \to Y$. And if $Y$ is metrizable, its metric induces a pseudometric on $X$. So it suffices to prove the $\impliedby$ direction of the metrization theorem above and assume the space is $T_3$.

Lemma 1. (Regular ∧ Has a $\sigma$-locally finite basis) $\implies$ (Normal ∧ $G_\delta$ space)

Proof. Let $\mathcal{B} = \bigcup_{n < \omega} \mathcal{B}_n$ be the basis where each $\mathcal{B}_n$ is locally finite. We use the following facts about locally finite families:

1. If $\mathcal{A}$ is locally finite, then $\mathcal{A}' \subseteq \mathcal{A}$ is locally finite.

2. If $\mathcal{A}$ is locally finite, $\cl{\bigcup_{A \in \mathcal{A}} A} = \bigcup_{A \in \mathcal{A}} \cl{A}$.

(1) is trivial and for (2), $\supseteq$ always holds in general. To prove $\subseteq$, if $x$ is such that every nbd $V$ of $x$ intersects $\bigcup \mathcal{A}$, note that for any fixed nbd $U$, it intersects only finitely many $\{A_1, \dots, A_n\} \subseteq \mathcal{A}$. So because any $U \cap V$ intersects at most those sets, $x \in \cl{\bigcup_{i=1}^n A_i} = \bigcup_{i=1}^n \cl{A_i} \subseteq \bigcup_{A \in \mathcal{A}} \cl{A}$.

To prove $G_\delta$, we’ll prove something a little stronger: Every open set is a countable union of closures of open sets (so in particular, it is $F_\sigma$). Let $V$ be an open set. Define $\mathcal{C}_n = \{ B \in \mathcal{B}_n \ : \ \cl{B} \subseteq V \}$. $\mathcal{C}_n$ is locally finite by (1). If we define $U_n = \bigcup_{B \in \mathcal{C}_n} B$, then by (2), $\cl{U_n} = \bigcup_{B \in \mathcal{C}_n} \cl{B} \subseteq V$. We wish to prove $V = \bigcup_{n < \omega} \cl{U_n}$, and we’ve just proven $\supseteq$. For $\subseteq$, let $x \in V$. As the space is regular, there is an open nbd $U$ of $x$ such that $x \in \cl{U} \subseteq V$. $\mathcal{B}$ is a basis, so there is some $B \in \mathcal{B}_n$ such that $x \in B \subseteq U$, and so $\cl{B} \subseteq V$ implies $x \in \cl{U_n}$.

To prove normality, let $A,B$ be closed disjoint. By above, let $\{V_n\}$ be open sets with $X \setminus A = \bigcup_{n < \omega} \cl{V_n}$ and $\{U_n\}$ be open sets with $X \setminus B = \bigcup_{n < \omega} \cl{U_n}$. We now construct the separation in a similar fashion to (T26). Define

$$U = \bigcup_{n < \omega} \left(U_n \setminus \bigcup_{k \le n} \cl{V_k}\right) \qquad V = \bigcup_{n < \omega} \left(V_n \setminus \bigcup_{k \le n} \cl{U_k}\right)$$

Then $A \subseteq U$, $B \subseteq V$, and $U \cap V = \emptyset$. $\square$

Lemma 2. If $X$ is $T_1$ and If $f : X \to [0, 1]^I$ is a map such that

1. $f_\alpha : X \to [0, 1]$ is continuous for each $\alpha \in I$.

2. For every $x_0$ and nbd $U$ of $x_0$, there is some $\alpha \in I$ such that $f_\alpha(x_0) > 0$ and $f_\alpha(X \setminus U) = \{0\}$.

Then $f$ is an embedding (a homeomorphism to its image).

Proof. As we’re taking $[0, 1]^I$ with the product topology, (1) implies $f$ is continuous. Since $X$ is $T_1$, for $x \ne y$, let $x \in U$ with $y \notin U$ and so for some $\alpha$, $f_\alpha(x) > 0$ while $f_\alpha(y) = 0$ by (2). Now let $V$ be open in $X$. For each $f(x) = y \in f(V)$, let $\alpha$ such that $f_\alpha(x_0) > 0$ and $f_\alpha(X \setminus V) = \{0\}$. Then $W = \pi_\alpha^{-1}((0, \infty))$ is a nbd of $f(x)$, so that $W \cap f(X) \subseteq f(V)$. This proves $f$ is open. $\square$

Now we’re ready to prove: ($T_3$ ∧ Has a $\sigma$-locally finite basis) $\implies$ Metrizable.

Let $\mathcal{B} = \bigcup_{n < \omega} \mathcal{B}_n$ be the basis where each $\mathcal{B}_n$ is locally finite. Let $I = \{ (n, B) \in \N \times \mathcal{B} \ : \ B \in \mathcal{B}_n \}$. By Lemma 1, and (T257), we see that each $X \setminus B$ is a cozero set, so there exists a continuous map $f : X \to [0, 1]$ with $X \setminus B = f^{-1}(0)$. Using the axiom of choice, for each $(n, B) \in I$, choose a function $f_{(n, B)} : X \to [0, 1/n]$ such that $f_{(n, B)}^{-1}(0) = B$. Then clearly the map $f : X \to [0, 1]^I$ with $f_{(n, B)}$ as each coordinate satisfies the statements in Lemma 2, since for $x_0 \in U$, there is some $B \in \mathcal{B}_n$ with $x_0 \in B \subseteq U$, and $g = f_{(n, B)}$ satisfies $g(x_0) = 0$ and $g(X \setminus U) > 0$.

However, $f : X \to [0, 1]^I$ is only a homeomorphism to $f(X)$ as a subspace of the product topology on $[0, 1]^I$. We need to show it is as well as a subspace of the uniform topology on $[0, 1]^I$, which is the one induced by the metric $\rho(x, y) = \sup_{i \in I} \abs{x_i - y_i}$. We already know it is injective, and if $V$ is open, $f(V)$ is open in the product topology, so it is also open in the metric topology as it is finer. We merely have to show $f$ is still continuous.

Let $x_0 \in X$ and $\varepsilon > 0$. Choose $n_0$ big enough so that $1/n_0 \le \varepsilon$. For $n \ge n_0$, the range of $f_{(n, B)}$ lies in $[0, 1/n]$, so $\abs{f_{(n, B)}(x) - f_{(n, B)}(x_0)} < \varepsilon$ is immediate. For each $n < n_0$, we know some nbd $U(n)$ of $x_0$ intersects finitely many $\{B_1, \dots, B_m\} \subseteq \mathcal{B}_n$. This means $f_{(n, B)}(U) = \{0\}$ for every other $B \in \mathcal{B}_n$, yet $f_{(n, B_i)}(x_0) > 0$ for each $1 \le i \le m$. By continuity, choose a $V_i(n) \subseteq U(n)$ nbd of $x_0$ such that for all $x \in V_i(n)$, $\abs{f_{(n, B_i)}(x_0) - f_{(n, B_i)}(x)} < \varepsilon$. Then define $V(n) = \bigcup_{i=1}^m V_i(n)$ and $V = \bigcap_{n < n_0} V(n)$, which is open, and has the property that for any $x \in V$, we have as we wished, $\rho(f(x), f(x_0)) < \varepsilon$.

Source: Munkres, Topology: Theorems 39.1 (for Lemma 1), 34.2 (for Lemma 2), 40.3 (main result)

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.

By definition.

It suffices to show the space is normal. We first show regularity. Let $A$ be closed and $x \notin A$. Let $K$ be a compact nbd of $x$. $A \cap K$ is compact, and $T_2$ spaces separate points from compact sets. So let $x \in U$ and $A \cap K \subseteq V$ with $U \cap V = \emptyset$. Choose $W = U \cap \text{int}(K)$ and $W' = V \cup (X \setminus K)$. Then $W \cap W' = \emptyset$ are open with $x \in W$ and $A \subseteq W'$.

From this point, since the space is $\sigma$-compact, it’s possible to use other theorems of $\pi$-base to prove normality: click here. It feels a bit like cheating, so we instead provide a more direct approach.

Let $A$ and $B$ be closed sets. By regularity, for each $x \in A$, there are nbds $x \in U_x$ and $B \subseteq V_x$ which are disjoint. Thus, $\cl{U_x} \cap B = \emptyset$. If we define $\mathcal{U}_x$ as the nbds of $x$ with this property, then it’s nonempty and $\mathcal{U} = \bigcup_{x \in X} \mathcal{U}_x$ is an open cover of $A$. $A$ is closed and $X$ is $\sigma$-compact, which clearly implies it is Lindelöf (T122), so $A$ is Lindelöf and we can extract a countable subcover $\{U_n\}$. We can do the exact same to $B$, finding a countable open cover $\{V_n\}$ such that $\cl{V_n} \cap A = \emptyset$ for each $n$. We then define

$$U = \bigcup_{n < \omega} \left(U_n \setminus \bigcup_{k \le n} \cl{V_k}\right) \qquad V = \bigcup_{n < \omega} \left(V_n \setminus \bigcup_{k \le n} \cl{U_k}\right)$$

$U$ and $V$ are open by construction. For each $x \in A$, $x \in U_n$ for some $n$ and since $\cl{V_k} \cap A = \emptyset$, we have $x \in U$. So $A \subseteq U$ and through a similar argument, $B \subseteq V$. Now we show $U \cap V = \emptyset$. Assume $x \in U \cap V$, which means $x \in U_n \setminus \bigcup_{k \le n} \cl{V_k}$ and $x \in V_m \setminus \bigcup_{k \le m} \cl{U_k}$. If $n \le m$, then $x \in U_n$ and $x \notin \cl{U_n}$ generates a contradiction, and $n \ge m$ would imply $x \in V_m$ yet $x \notin \cl{V_m}$.

Denote $x \sim y$ to mean $x,y$ are indistinguishable. Let $Y = X/{\sim}$ be the Kolmogorov quotient and $q : X \to Y$ the surjective map. Note that it is continuous and an open map, therefore a closed map, as it is surjective. This means two things.

1. If $x \notin A$ with $A \subseteq X$ closed, then $q(x) \notin q(A)$ with $q(A)$ closed, and a function $f : Y \to [0, 1]$ separating $q(x)$ and $q(A)$ induces a continuous function $g : X \to [0, 1]$ separating $x$ and $A$. This proves $Y$ completely regular $\implies X$ completely regular.

2. If $x \in V \subseteq K$ with $V$ open and $K$ compact, then $q(x) \in q(V) \subseteq q(K)$, where $q(V)$ is open as $q$ is open, and $q(K)$ compact because $q$ is continuous. So $X$ weakly locally compact $\implies Y$ weakly locally compact.

Since $X$ is $R_1$, then $Y$ is $T_2$. Joining facts (1) and (2), it suffices to prove (Weakly locally compact ∧ $T_2$) $\implies$ $T_{3 \frac 1 2}$.

We first prove locally compactness. To do that, we’ll prove regularity and then then the result follows from (T246). Let $x \notin A$ with $A$ closed. Let $x \in V \subseteq K$ with $V$ open and $K$ compact, by weakly locally compact. Note that $T_2$ spaces separate points from compact sets, so let $x \in U$ and $A \cap K \subseteq U'$ with $U \cap U' = \emptyset$. As $K$ is closed, we see that $A \subseteq U' \cup (X \setminus K)$, and $(U \cap V) \cap (U' \cup (X \setminus K)) = \emptyset$, which separates $x$ from $A$.

Since $Y$ is locally compact and $T_2$, its one-point compactification $Y^* = Y \cup \{\infty\}$ is compact $T_2$, which must be $T_4$, since $T_2$ spaces separate compact sets and any closed sets of $Y^*$ are compact. Normal spaces are completely regular (T37), so $Y^*$ is completely regular. Let $x \in Y$ and $A \subseteq Y$ closed. Then $A$ is also closed in $Y^*$, so let $f : Y^* \to [0, 1]$ with $f(A) = \{0\}$ and $f(x) = 1$, and $f|_Y : Y \to [0, 1]$ is still continuous with the same properties.

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

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

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

By definition.

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

Let $x \ne y$. Since $X$ is $T_2$, let $x \in U$ and $y \in V$ nbds with $U \cap V = \emptyset$. This means $x \notin \overline{V}$ and $y \notin \overline{U}$, so being regular, let $W_x,W_V$ be disjoint nbds with $x \in W_x$ and $\overline{V} \subseteq W_V$ and similarly $W_y,W_U$ disjoint nbds with $y \in W_y$ and $\overline{U} \subseteq W_U$. Then $O_x = W_x \cap U$ and $O_y = W_y$ are the nbds of $x$ and $y$ of which $\overline{O_x} \cap \overline{O_y} = \emptyset$ (remember that $\overline{O_x} \subseteq \overline{W_x} \cap \overline{U}$).

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 space is a subspace of itself.

Let $A$ be a closed set and $x \notin A$. We argue $\overline{\{x\}} \cap A = \emptyset$: Every point of $\overline{\{x\}}$ is indistinguishable from $x$, as the space is $R_0$, so $y \in \overline{\{x\}} \cap A$ would imply $x \in \overline{A} \subseteq A$. So by Urysohn’s lemma, let $f : X \to [0, 1]$ with $f(A) = \{0\}$ and $f(\overline{\{x\}}) = \{1\}$. In particular, $f(x) = 1$.

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 $[0, 1]$.

An injective path is a path.

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 $y$.

If it has a dispersion point… it has… a point.

Singletons are clopen.

Let $C$ be a connected component, and $x$ an isolated point of $C$. $X$ is $T_1$ and $C$ is closed, so $\{x\}$ is closed in $C$, and take $W$ the nbd of $x$ with $W \cap C = \{x\}$. Then $C \setminus \{x\}$ and $W$ is a separation of $C$.

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 $D$ 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 $\text{card}(X) \le \text{card}(2^{2^D}) = 2^\mathfrak{c}$

Every open set is connected: If $V$ were not to be, then $V = U_1 \cup U_2$ with $U_1,U_2$ disjoint and open in $V$, but then they must be open in $X$ as well.

The space is not a singleton.

By definition, it suffices to show the space is $\sigma$-compact. Let $\mathcal{K}$ be the set of all compact neighborhoods. So $\{\text{int}(K)\}_{K \in \mathcal{K}}$ forms an open cover. Let $\mathcal{C} \subseteq \mathcal{K}$ be countable such that $\{\text{int}(K)\}_{K \in \mathcal{C}}$ is a countable subcover. Then $X = \bigcup_{K \in \mathcal{C}} K$.

Let $A$ be closed and $x \notin A$. Then $V \cap A = \emptyset$ for some nbd of $x$. Let $B \subseteq V$ be a clopen nbd of $x$. Then $\chi_B$ (indicator function of $B$) is continuous, as any preimage is one of $\{\emptyset, B, B^c, X\}$, with $\chi_B(A) = \{0\}$ and $\chi_B(x) = 1$.

By definition.

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.