Radial/pseudoradial implies Fréchet-Urysohn/sequential for locally countable spaces

Question

In a radial space, every point in the closure of a set is the limit of a transfinite sequence from the set. In a Fréchet-Urysohn space, every point in the closure of a set is the limit of an ($\omega$-length) sequence from the set.

Likewise, in a pseudoradial space, every transfinite-sequentially-closed set is closed, while in a sequential space, every ($\omega$-length-)sequentially-closed set is closed.

It seems evident to me that when a space is locally countable (every point has a countable neighborhood), radial and Fréchet-Urysohn are equivalent. Likewise, given locally countable, pseudoradial and sequential are equivalent. However, is there an elegant way to express a proof more rigorously than "transfinite sequences sure look like countable sequences when they map into a countable set"?

Answer 1

It boils down to the following. (Note: no Hausdorff assumption needed.)

Suppose $X$ is a countable topological space and $A\subseteq X$. If there is a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ of points of $A$ that converges to a point $p\in X$, then there is an ordinary sequence ($\omega$-indexed) of points of $A$ converging to $p$. Here $\lambda$ is some limit ordinal and saying the transfinite sequence converges to $p$ means: for every nbhd $V$ of $p$, there is some $\beta<\lambda$ such that $x_\alpha\in V$ for all $\alpha>\beta$.

By considering the cofinality of $\lambda$ and taking a corresponding suitable cofinal subsequence of $(x_\alpha)_{\alpha<\lambda}$, we can assume $\lambda$ is a regular cardinal and then we can further assume it is of smallest (infinite) cardinality satisfying the conditions.

Now, if one of the $x_\alpha=a\in A$ is repeated cofinally many times, then extracting the subsequence of all the elements with value $a$ gives a constant (transfinite) sequence that converges to $p$. (Note that $a$ need not be equal to $p$.) Then, by definition of convergence, every nbhd of $p$ contains $a$ and the ordinary sequence $(y_n)_{n<\omega}$ with each $y_n=a$ also converges to $p$.

Otherwise, for each $a\in A$ the set of indices $I_a=\{\alpha<\lambda: x_\alpha = a\}$ is not cofinal in $\lambda$, that is, $I_a$ has cardinality smaller than $\lambda$. But there are only countably many values $a\in A$. So we get a partition of $\lambda$ into countably many sets, each of cardinality smaller than $\lambda$. If $\lambda$ were uncountable, that would contradict the fact that $\lambda$ is a regular cardinal. So necessarily $\lambda=\omega$ and our original sequence was an ordinary sequence.

This proof also applies to a locally countable space, by first restricting the sequence to a tail contained in a countable nbhd of $p$.

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[Added later] From the proof above, all that matters is that $A$ is countable, even if $X$ is not. So basically the result is:

If a point $p\in X$ is the limit of a transfinite sequence (repetitions allowed) of points from a countable set $A\subseteq X$, the point is also the limit of an ordinary sequence of points from $A$.

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An attempt at a more handwaving proof would be: (1) if an element is repeated cofinally many times, do the same as the above. (2) otherwise, for each repeated element in the sequence, take its first appearance and remove all the later copies. The resulting subsequence should also converge to $p$. Why? Because it's cofinal in the original sequence. And why is that? It's not completely obvious, but the argument above justifies it rigorously.

[As mentioned in Steven's comment below, this handwaving attempt does not work.]

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(added 11/15/2023)

For more general cardinalities one can use essentially the same argument as above to show for any topological space $X$:

Proposition: Suppose the point $p\in X$ is the limit of a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ of points from a set $A\subseteq X$ of infinite cardinality $\kappa$. (Here $\lambda$ is some limit ordinal.) Then $p$ is also the limit of a transfinite sequence of points of $A$ indexed by some infinite regular cardinal $\mu$ with $\mu\le\min(\kappa,\lambda)$.

Note: This applies in particular when $A$ is the set of values of the transfinite sequence. Also, if $\mu$ is finite, there is a constant (ordinary) sequence with value in $A$ converging to $p$.

Answer 2

Not exactly an answer to the original question, but more a postscript to PatrickR's answer. The observation:

If a point $p\in X$ is the limit of a transfinite sequence (repetitions allowed) of points from a countable set $A\subseteq X$, the point is also the limit of an ordinary sequence of points from $A$.

has an immediate corollary:

If $X$ is countably tight and radial, then $X$ is Fréchet-Urysohn.

Proof.

Given a set $S\subseteq X$ and a point $p\in \overline{S}$, since $X$ is countably tight there is some countable subset $D\subseteq S$ with $p\in\overline{D}$. Since $X$ is radial there is a transfinite sequence in $D$ converging to $p$. Taking $A=D$ in the quoted statement from PatrickR's answer yields the result.

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Remark.

We cannot do the same trick to show that pseudoradial countably tight spaces are sequential, as this is not the case. For a counter-example, consider the space $X=\mathbb R\cup\{\infty\}$, where $\mathbb R$ has the Euclidean topology, and open neighborhoods of $\infty$ are precisely sets of the form $U\cup\{\infty\}$ for $U\subseteq \mathbb R$, where $U$ is open and cocountable.

$X$ is pseudoradial.

If $A\subseteq X$ is radially closed and $\infty\in \overline{A}\backslash A$, then since $A$ is closed in $\mathbb R$ (by pseudoradiality of $\mathbb R$), $A$ must be uncountable, as otherwise $\{\infty\}\cup \mathbb R\backslash A$ would be a neighborhood of $\infty$ missing $A$.

Therefore there is a subset $B\subseteq A$ with $|B|=\aleph_1$. Letting $\alpha\mapsto b_\alpha$ be a bijection from $\omega_1$ to $B$, we see that since $(b_\alpha)$ eventually leaves every countable set, $b_\alpha\to \infty$, contradicting $\infty\notin A$.

We conclude that if $\infty\in \overline{A}$, then $\infty\in{A}$. For $x\in \mathbb R$ we also have $x\in A\iff x\in\overline{A}$, again by pseudoradiality of $\mathbb R$. Therefore $A$ is closed.

$X$ is countably tight.

If $A\subseteq X$, then let $E$ be a countable dense subset of $A\cap \mathbb R$, and let $D=E$, unless $\infty\in A$, in which case let $D=E\cup\{\infty\}$. Then $\overline{D}=\overline{A}$, so for all $p\in \overline{A}$, $D$ witnesses the countable tightness condition. (Note that since we chose $D$ independently of $p$, we have actually proved a stronger condition, namely hereditary separability.)

$X$ is not sequential.

$\mathbb R$ is sequentially closed.

To see this, suppose $(x_i)$ is a sequence in $\mathbb R$. If $(x_i)$ is unbounded, then it has a subsequence $(y_j)$ “converging to $\pm\infty$” (in the usual sense, not to the actual point $\infty\in X$), so that $\{y_j\mid j\in \mathbb N\}$ is countable and closed, whereby $\{\infty\}\cup \mathbb R\backslash \{y_j\mid j\in \mathbb N\}$ is a neighborhood of $\infty$ that $(x_i)$ frequently leaves. Similarly, if $(x_i)$ is bounded, it has a subsequence $(y_j)$ converging to some $y\in\mathbb R$, so that $\{y_j\mid j\in \mathbb N\}\cup \{y\}$ is countable and closed, and therefore $(x_i)$ frequently leaves the neighborhood $\{\infty\}\cup \mathbb R\backslash \left(\{y_j\mid j\in \mathbb N\}\cup \{y\}\right)$ of $\infty$. In either case, the sequence $(x_i)$ cannot converge to $\infty$.

But $\mathbb R$ is not closed, hence $X$ is not sequential.