IDs 100-199

$T_5$ 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.

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).

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.

A subcover is trivially, a subcolection with dense union.

Left as an exercise for the reader.

Left as an exercise for the reader.

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.

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.

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

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

Globally implies locally.

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.

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.

Globally implies locally.

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.

Every subspace is finite.