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For metric spaces, knowledge of the convergence of sequences determines the topology completely. A set is closed in the metric topology if and only if it is closed under the limit of convergent sequences operation. Put another way, a map between metric spaces is continuous if and only if it preserves limits of sequences.

For general topological spaces, this result no longer holds. Spaces for which it does hold are called sequential spaces. Spaces are usually not sequential because they are not first countable. In such spaces, there are "not enough" numbers in the countable index set to extend into these very large spaces and reach every limit point. For example, if $\omega_1$ is the first uncountable ordinal, then $\omega_1+1$ is a topological space under the order topology, and $\omega_1$ is a limit point, but no sequence of ordinals reaches $\omega_1$. The topology of this space is not characterized by limits of sequences.

This defect is usually fixed by replacing the notion of a sequence with the more general notion of a net, which is like a sequence but indexed by an arbitrary directed set, instead of the natural numbers. Since the net's index set is arbitrary, it can be of any cardinality and therefore nets can reach those faraway limit points in non first countable spaces, and the result is restored: a function between topological spaces is continuous iff it preserves limits of nets. A set is closed iff it is closed under limit of nets.

But if the problem is that the cardinality of the space is too high for sequences to do the job, the most conservative generalization that suggests itself to me is to just increase the cardinality of the index set; the transfinite sequence, indexed by an arbitrary ordinal. So my question is: is this strong enough? Is it just a matter of convenience? Nets require less machinery, and are natural because e.g. the neighborhoods of a point constitute a net, whereas transfinites require a detour into unrelated areas of set theory. But nets are not totally ordered, and there may be times it would be nice to have a totally ordered index set for our sequence-like object.

So can we say that a set in an arbitrary topological space is closed iff it's closed under taking limits of arbitrary convergent transfinite sequences? Is there a space with a set which is closed under limit of transfinite sequence, but not limit of net?

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@Asaf, a directed set might not have true cofinality. –  azarel Jan 19 '12 at 8:04
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3 Answers 3

The answer to your last question is yes. $\beta\omega$, the Čech-Stone compactification of $\omega$, is such a space. No sequence in $\omega$, transfinite or otherwise, converges unless it’s eventually constant, but each point in $\beta\omega\setminus\omega$ is the limit of a net in $\omega$. Specifically, if $p\in\beta\omega\setminus\omega$, we can regard $p$ as a free ultrafilter on $\omega$. Let $I=\{\langle n,A\rangle:n\in A\in p\}$, and let $\nu:I\to\omega:\langle n,A\rangle\mapsto n$; then $\nu$ is a net that converges to $p$.

A more elementary example is the Arens-Fort space, which is a countable Hausdorff space that is not first countable at its one non-isolated point. Let $X$ be this space, let $p$ be the one non-isolated point, and let $D$ be the set of isolated points. Then no sequence in $D$ of any length converges to $p$, but the net in $D$ built in the usual way from the nbhds of $p$ does converge to $p$.

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Brian: If I may ask, what is the situation for separable Banach spaces with a weak topology? –  Mark Jan 19 '12 at 15:24
    
@Mark: My functional analysis is really rusty, but if I’m not mistaken, weakly convergent sequences in $\ell^1$ are strongly convergent, so every sequence weakly converging to a point in the open unit ball $B$ must eventually be in $B$. If $\ell^1$ with the weak topology were sequential, this would make $B$ open in the weak topology, which it isn’t. –  Brian M. Scott Jan 19 '12 at 17:27
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An example is also given in: E. S. Wolk, On the inadequacy of cofinal subnets and transfinite sequences, American Mathematical Monthly 89 (1982), 310-311. However, I don't have a copy of Wolk's paper with me or have access to JSTOR, so I don't know if it's related to any of Brian M. Scott's examples. (So, how I did I even know about this paper? I saw it cited in a 2001 sci.math post of mine . . .) –  Dave L. Renfro Jan 19 '12 at 17:57
    
@Dave: The example turns out to be $(\omega_1+1)\times(\omega+1)$: $\langle\omega_1,\omega\rangle$ is a cluster point of $\omega_1\times\omega$, but no transfinite sequence in $\omega_1\times\omega$ converges to $\langle\omega_1,\omega\rangle$. –  Brian M. Scott Jan 19 '12 at 18:18
    
@Brian: Thanks, but I'm familiar with that. I was actually asking whether (it is known if) every separable Banach space is "ordinally sequential", in the sense that closed sets are the sets closed under "transfinite limits" (in the sense of ziggurism) of elements from the set? –  Mark Jan 19 '12 at 19:00
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The spaces characterized by the property that a subset is closed if and only if it is closed w.r.t. limits of transfinite sequences are called pseudoradial spaces.

I would like to give an example of a spaces that is not pseudoradial together with sketch of a proof. (I hope I am not doing something very easy in too complicated way.) I would also like to mention a few references and some properties of these spaces.


Before presenting the example, one short remark. Pseudoradial spaces are represented by the convergence of net on well-ordered nets. Obviously, it is sufficient to take the nets on ordinals. We can go a little further - regular cardinals are sufficient. Indeed, if we have a cofinal subset of an ordinal, we can use this cofinal subset to get another convergent net.


Let us consider the following example. Each arrow in the picture bellow represents a convergent sequence. I.e., this is a topological space homeomorphic to $\{0\}\cup\{\frac1n;n\in\mathbb N\}$ taken as a subspace of real line. Equivalently, this is precisely the ordinal $\omega+1$ taken with the order topology.

We take all these sequences and identify the some of the points as in the picture. (I.e., we make a quotient space of sum of these spaces.) Let us call the resulting space $S_2$. Then we take the subspace of this space as shown in the picture. This subspace will be called $S_2^-$. (I've taken the notation $S_2$ and $S_2^-$ from this paper: Franklin S.P., Rajagopalan M., On subsequential spaces, Topology Appl. 35 (1990), 1-19. But you can notice that this space is very very similar to Arens-Fort space mentioned in Brian's answer.)

Example

Now we want to show that $S_2^-$ is not pseudoradial.

Note that the space $S_2^-$ has only one non-isolated point. Let us call it $\omega$ . So we ask whether there is a transfinite sequence, consisting only of points different from $\omega$, which converges to $\omega$.

First, let us show that this is not possible for a regular cardinal $\alpha>\omega$. Suppose that $(x_\eta)_{\eta<\alpha}$ is an $\alpha$-sequence of points of $S_2^-\setminus\{\infty\}$, which converges to $\infty$. Let us denote $n_\eta$ the "column" to which $x_\eta$ belongs. In we use the notation the notation from the picture bellow $n_\eta$ is the first coordinate of ordered pair $x_\eta$.

We can see that $n_\eta$ converges to $\omega$. (E.g. by noticing that $(x,y)\mapsto x$ and $\omega\to\omega$ is a quotient map from $S_2^-$ to $\omega$ with order topology.)

Now this is not possible, since the we would be able to construct an increasing $\alpha$-sequence converging to $\omega$ and using this sequence we would be able to show that cofinality of $\alpha$ is $\omega$.

Example

So the only possibility is to take a sequence in the usual sense, i.e., a sequence of length $\omega$. Perhaps with a little handwaving, but it is more-or-less clear that general situation is similar to the situation when the $n$-term of the sequence is in the $n$-th column. So we have a sequence $x_n=(n,y_n)$. Obviously $\{\omega\}\cup\bigcup\limits_{n\in\omega} \{n\}\times(y_n,\infty)$ is a neighborhood of the point $\omega$ containing no terms of this sequence.


Pseudoradial spaces were introduced by H. Herrlich. Quotienten geordneten Räume und Folgenkonvergenz. Fund. Math., 61:79–81, 1967; pdf. They were later studied by A.V. Arhangelskii and many others.

The class of pseudoradial spaces is closed under the formation of closed subspaces, quotients and topological sums. They are a coreflective subcategory of the category Top of all topological spaces. This means that for each topological space we have pseudoradial coreflection; a pseudoradial space which is, in some sense, close to this space. The pseudoradial coreflection is obtained simply by taking sets closed under limits of transfinite sequences as closed sets in a new topology on the same set. (E.g. the pseudoradial coreflection of $S_2^-$ is discrete.)

The same thing can be done with any class $\mathbb P$ of directed sets instead of ordinals. This is called $\mathbb P$-net spaces in P. J. Nyikos. Convergence in topology. (In M. Hušek and J. van Mill, editors, Recent Progress in General Topology, pages 537–570, Amsterdam 1992. North-Holland.) The properties of pseudoradial spaces which I mentioned in the preceding paragraph are true for $\mathbb P$-net spaces, too.

Interestingly, if we take the linearly ordered sets, we obtain the same class of spaces as from well-ordered sets, see James R. Boone: A note on linearly ordered net spaces. Pacific J. Math. Volume 98, Number 1 (1982), 25-35; link.

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Edit: this argument appears to be wrong, but what's wrong with it?

I think Asaf's comment answers this. He says: under AC, any directed set $D$ has a cofinal well ordered set.

To see this, apply Zorn's lemma to well-ordered subsets of $D$, partially ordered by inclusion, deduce that a maximal element must be cofinal.

Now, closure under nets implies closure under transfinite sequences, since well-ordered sets are in particular directed. In the other direction if you're not closed under nets, say $f: D \rightarrow X$ converges to something outside of your subset, then extract a cofinal well ordered set $W$ from $D$ to produce your flawed transfinite sequence.

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The problem is that the maximal well-ordered subset $\mathcal C$ is not really cofinal, i.e., there can be a $x\in D$ such that $x\nleq y$ for $y\in \mathcal C$. This is possible because $\mathcal C\cup \{x\}$ is not necessarily a chain. –  azarel Jan 19 '12 at 8:20
    
That would do it. Cheers! –  uncookedfalcon Jan 19 '12 at 8:31
    
For example, the collection of finite subsets of a fixed uncountable set, ordered by subset, is directed but has no cofinal linearly ordered subset. –  JDH Jan 19 '12 at 17:40
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