Latest proofs

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

A space is a subspace of itself.

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 $k$-netork is a network: This was done in (T11).

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.

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,

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

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 $p \in \overline{E} \cap \overline{O}$. Yet by definition, $E \cap O = \emptyset$, so the space is not extremally disconnected.

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.

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 $n > m$.

It’s impossible to have an infinite strictly decreasing sequence of open sets if there are only finitely many possible open sets.

Let $x \sim y$ iff they are indistinguishable. The equivalent classes $[x]$ form a basis for a topology which must be finer than $X$ (if $U$ is a nbd of $x$, it must contain all $y \sim x$). It suffices to show each $[x]$ is an open set. But $Y = X \setminus [x]$ is compact, and any points of $Y$ are distinguishable from $x$, so it’s possible to find $x \in U$ and $Y \subseteq V$ nbds with $U \cap V = \emptyset$ (this is analogous to the result that for $T_2$ spaces, any point $x$ and a compact $K$ can be separated)

$X$ is a countable union of finite open sets. $X$ hereditarily connected, so this basis is linearly ordered and it’s possible to enumerate them as $\{B_n\}$ where $n < m \implies B_n \subset B_m$. Therefore, $B_n \nearrow X$ and $X$ is a union of finite sets.

$T_0$ ensures any two points are distinguishable. This is an if and only if.

By definition, $T_1$ is $T_0$ and $R_0$.

Immediate by definition.

By contraposition, if $d(x, y) = 0$, then they both have the same neighborhoods.

Define $d(x, y) = 1$ if $x,y$ belong to the same set in the partition, and 0 otherwise. This is a pseudometric.

If each local basis $\mathcal{V}_x$ is countable and $X$ is countable, then $\mathcal{B} = \bigcup_{x \in X} \mathcal{V}_x$ is a countable basis.

Given $p \in \overline{A}$, choose $D \subseteq A$ countable with $p \in \overline{D}$. Suppose $(x_\alpha)_{\alpha < \lambda}$ is some transfinite sequence in $D$ with $x_\alpha \to p$. We now construct a sequence in $D$ which converges to $p$ and we’re done (thanks to PatrickR).

If some $y \in D$ is cofinal in $(x_\alpha)$, this means any neighborhood $U$ of $x$ must contain $y$, so we can construct a constant sequence $y_n = y$ with $y_n \to p$. If no $y \in D$ is cofinal, then each $I_y = \{ \alpha < \lambda \ : \ x_\alpha = y \}$ is finite and $\lambda = \bigcup_{y \in D} I_y$ is a countable union of finite sets, so it is a countable ordinal. We can assume $\lambda$ to be a regular cardinal, so it is already a sequence.

Suppose a set $A$ is sequentially closed. Let $p \in X$ and $(x_\alpha)_{\alpha < \lambda} \subseteq A$. Take $V$ a countable nbd of $p$. If $x_\alpha \to p$, then there is an ordinal $\mu$ for which $x_\alpha \in V$ for all $\alpha \ge \mu$. The set of ordinals $\mu \le \alpha < \lambda$ is isomorphic to some $\lambda'$. So we have $(y_\alpha)_{\alpha < \lambda'} \subseteq V$, still with $y_\alpha \to V$.

From a converging transfinite sequence within a countable set, we can extract a $\omega$-sequence that still converges to the same value, this was done in (T211). So we construct a sequence $(z_n)$ with $z_n \to p$, and so $p \in A$ as we wished to prove.

Any radially closed set must be, in particular, sequentially closed.

Fréchet Urysohn is stronger because it asserts there is a sequence, a.k.a a transfinite sequence of length $\omega$.

Essentially the same proof as Fréchet Urysohn $\implies$ Sequential (T184). Just swap “sequential closure” with “radial closure”.

Any bijective function is a homeomorphism in discrete space. Just take the permutation that swaps $a$ and $b$.

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.

Globally implies locally.

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.

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.

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 $x_n \to p$.

Globally implies locally.

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

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

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.

Every set is clopen.

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 $x$.

A subcover is trivially, a subcolection with dense union.

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.

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).

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

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

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.

By definition.

By definition.

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).

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.

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

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