$T_5$ is $T_2$ (hence, $T_1$) by definition.

In particular, $X$ is $T_1$ and normal, which implies $T_2$ (shown in T99), and so it must be $T_5$ by definition.

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).

By definition.

By definition.

Shrink the cover twice.

Let $\mathcal{R} = \{R_j\}$ be a point-countable open refinement of some open cover $\{U_i\}$. Contrapositively, suppose no countable subcover exists. Recursively, define $x_0$ as any point and $\mathcal{R}_0$ the countable collection of sets in $\mathcal{R}$ containing $x$. Then $X \setminus \mathcal{R}_0$ is nonempty and we can choose $x_1$ from that. Inductively, let $\mathcal{R}_n \subseteq \mathcal{R}$ of the sets containing $x_n$, so that $\bigcup_{j=0}^n \mathcal{R}_j$ is countable and so $x_{n+1} \in X \setminus (\bigcup_{j=0}^n \mathcal{R}_j)$.

The sequence $Y = \{x_n\}$ is infinite and has no cluster point: If $x$ were to be, take $x \in R_j$ and the minimum $n$ such that $x_n \in R_j$. By construction, it is the only element of $Y$ in $R_j$. For any sequence with no cluster point, we can construct a countable cover with no finite subcover. Take $U_0$ nbd of $x_0$, let $m$ be the least element of which $x_m \notin U_0$ and $U_1$ a nbd of $x_m$. Then if $m$ is the least element with $x_m \notin \bigcup_{k \le n} U_k$, let $U_{n+1}$ a nbd of $x_m$. Finally, if $V = X \setminus \bigcup U_n$, then $\{V\} \cup \{U_n\}$ is a countable cover of $X$ with no finite subcover.

If only singletons are connected and $X$ has a basis of connected sets, then the basis must be all singletons.

If $x \ne y$ are indistinguishable, then $\overline{\{x\}} \cap \overline{\{y\}} = \emptyset$. So no two distinct points are distinguishable.

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.

By contradiction, suppose $x \ne y$ are topologically indistinguishable. Then $Y = X \setminus \{x, y\}$ must be disconnected, since $\{x, y\}$ is connected and $Y$ has at least two points. Then there are open sets $U,V$ such that $Y \subseteq U \cup V$ and $U \cap V \cap Y = \emptyset$. Note that $U$ can’t contain any of $x$ or $y$, otherwise it must contain both and $U$ and $V$ are a separation of $X$, which should be connected. Same goes for $V$. So $\{x, y\} = (U \cup V)^C$. But notice $U \cup \{x\}$ is connected: If $A,B$ is a separation by open sets with $x \in A$, then $y \in A$ and the pair $A \cup V$,$B \cap U$ is a separation of $X$. Through a similar reasoning, $V \cup \{y\}$ is connected. But then $X$ is not biconnected.

I’m not sure why this is here. $T_5$ implies $T_1$ (T100) and completely normal (T336). Completely normal implies normal (T36). $T_1$ and normal implies $T_4$ (T99).

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.

By definition.

By definition.

Clear from their definitions.

Clear from their definitions.

Let $f : X \to Y \subseteq \R$ be a homeomorphism. $\R$ is $T_1$, so $Y \simeq X$ is as well. Define order on $X$ by $x < y$ iff $f(x) < f(y)$. If $U$ is a nbd of $x$, then $f(x) \in f(U)$ and $f(U) = V \cap Y$, where $I = (f(x) - \delta, f(x) + \delta) \subseteq V$ for some $\delta > 0$. Let $J = f^{-1}(I) \subseteq U$. Suppose $a,b \in J$ and $a < z < b$. Then $f(a) < f(z) < f(b)$ with $f(a),f(b) \in I$, so $f(z) \in I$ and $z \in J$ as we wished, to prove $J$ is order-convex.

A single set is trivially a countable union.

If each compact has a finite subcover, a countable union of them will have a countable subcover.

Take any open cover. Apply Lindelöf, then apply countably paracompact.

Take any open cover. Apply Lindelöf, then apply countably metacompact.

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 $\tau = \tau'$.

By definition.

A subcover is trivially, a subcolection with dense union.

A GO-space has a basis of order-convex neighborhoods. A LOTS space has a basis containing all open intervals. So by contraposition, suppose $A$ is some order-convex neighborhood of a point $x$ that contains no open intervals. Then it must be closed. If $y \in \cl{A}$ and $y < x$, then there is some $z \in (y, x) \cap A$. But then either $(z, x) \subseteq A$ or $(x, z) \subseteq A$. The same holds if $y > x$. But then $A$ is a proper nonempty clopen set, and $X$ is not connected (unless $X = \{x\}$, but then the result is trivial).

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).

It suffices to show a basis is a $k$-network: If $K \subseteq U$ with $U$ open and $K$ compact, then $U = \bigcup \mathcal{B}$, where $\mathcal{B}$ is a set of basis elements. $K$ being compact means there exists a finite subset $\mathcal{B}^* \subseteq \mathcal{B}$ with $K \subseteq \bigcup \mathcal{B}^* \subseteq U$.

Left as an exercise for the reader.

Left as an exercise for the reader.

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 $x$.

Every set is clopen.

By definition.

By definition.

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.

By definition.

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 $y$.

By definition.

By definition.

$T_6$ is $T_1$ and perfectly normal. Then it is completely normal (T156). $T_1$ and completely normal is $T_5$ (T101).

It suffices to show that for $x \in W$ with $W$ open, there is some regular open set $O$ with $x \in O \subseteq W$. By regularity, let $x \in U$ and $W^C \subseteq V$ open sets with $U \cap V = \emptyset$. In particular, $\cl{U} \cap W^C = \emptyset$, so $\cl{U} \subseteq W$. Then just pick $O = \text{int}(\cl{U})$.

Normal and $T_2$ is the definition of $T_4$.

Any compact set is relatively compact.

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

Let $f : V \to \R^n$ be an embedding with $V$ nbd of $x$. If $n = 0$ for every choice of $x$, the space is discrete with two or more points, so clearly not hyperconnected. If $n \ge 1$ for some $x$, then $f(V)$ is an open set with infinitely many points, so by taking any two points with disjoint nbds $U$ and $V$ ($\R^n$ is $T_2$), we have $f^{-1}(U)$ and $f^{-1}(V)$ disjoint, open, nonempty.

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.

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 $X$.

By definition.

Let $U$ be a cozero set. Then by perfect normality, let $f : X \to [0, 1)$ be continuous with $f^{-1}(\{0\}) = \cl{U}$ (since $\emptyset$ is closed, so we chose it as the set where $f^{-1}(\{1\}) = \emptyset$, this is in fact why every closed set is a zero set in perfectly normal spaces). It follows that $V = f^{-1}((0, 1))$ is cozero with $U \cap V = \emptyset$ and $X = \cl{U} \cup V \subseteq \cl{U \cup V}$.

Globally implies locally.

By definition.

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.

Let $p \in \cl{A}$ and choose $B$ a countable basis nbd of $p$. Then $D = A \cap B$ is a countable subset of $A$ such that if $V$ is any open nbd of $p$, $B \cap V$ is open, so $A \cap (B \cap V) = D \cap V \ne \emptyset$.

That’s right, “countable” does not mean infinitely countable.

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 topology is a subset of $\mathcal{P}(X)$, so there are finitely many open sets.

By definition, $\aleph_1$ is the smallest cardinality greater than $\aleph_0$. Assuming the continuum hypothesis, $\aleph_1 = \mathfrak c$.

By definition, $\aleph_1$ is the smallest uncountable cardinal.

It suffices to prove Weak Hausdorff. Thus, let $K$ be compact $T_2$ and $f : K \to X$ continuous. Using the $k_2$-space property, we wish to show that for any other $K'$ compact $T_2$ and $g : K' \to X$ continuous, that $g^{-1}(f(K))$ is closed. Let $\Delta$ be the diagonal of $X^2$. If we define the function $f \times g : K \times K' \to X^2$ as $(f \times g)(x, y) = (f(x), g(x))$, then the set $L = (f \times g)^{-1}(\Delta)$ is closed, by the $k_2$-Hausdorff property ($K \times K'$ is compact $T_2$). If $\pi_2 : K \times K' \to K'$ is the projection map, since $K$ is compact, $\pi_2$ is a closed map. But $\pi_2(L) = g^{-1}(f(K))$.

Globally implies locally.

Let $L$ be compact. Using the $k_2$-space property, we wish to show that for any $K$ compact $T_2$ and any $f : K \to X$ continuous, that $f^{-1}(L)$ is closed. Our objectives are to show (1) $X$ is $T_1$ (2) $f(K)$ is compact $T_2$ (3) prove $f^{-1}(L)$ is closed.

(1) The inclusion map $\text{id} : \{x\} \to X$ is continuous and $\{x\}$ is clearly compact $T_2$, so apply Weak Hausdorff.

(2) $f(K)$ is compact since $f$ is continuous. Note that $T_2$ spaces separate compact sets, and since every closed set in $K$ is compact, $K$ is $T_4$. Take $x \ne y \in f(K)$. $\{x\}$ and $\{y\}$ are closed in $X$ by (1), so $A = f^{-1}(\{x\})$ and $B = f^{-1}(\{y\})$ are closed and can be separated by open nbds $A \subseteq U$ and $B \subseteq V$. Note that $K \setminus U$ is compact, so $f(K \setminus U)$ is closed and $U' = f(K) \setminus f(K \setminus U)$ is a nbd of $x$. Similarly, $V' = f(K) \setminus f(K \setminus V)$ is a nbd of $y$ and $U' \cap V' = \emptyset$.

(3) $L' = f(K) \cap L$ is compact. Since $f(K)$ is $T_2$, $L'$ is closed in $f(K)$ But $f(K)$ is closed in $X$ by Weak Hausdorff, so $L'$ is closed. Then clearly $f^{-1}(L) = f^{-1}(L')$ is closed.

Take any $x$ and a nbd $U$ which is $T_2$. We wish to find a neighborhood $V$ not containing some other $y$. If $y \notin U$, just take $V = U$. Otherwise, take $x \in V \subseteq U$ with $y \notin V$ since $U$ is $T_1$. But then $V$ is also open in $X$, so we’re done.

By definition.

Every subspace is finite.

By definition.