Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A space $X$ is called radial if, for any $A \subset X$ and any $x \in cl(A)$, there is a transfinite sequence $s=\{a_\alpha: \alpha \in \kappa\} \subset A$ which converges to $x$. What's meaning of "transfinite sequence" here?

Added: If I may ask more, What is difference between redial space and Frechet space? It seems that they are same.

share|improve this question
add comment

3 Answers

up vote 2 down vote accepted

As Paul VanKoughnett has remarked in his answer, a transfinite sequence will be any sequence/function $\langle x_\xi \rangle_{\xi < \alpha}$ where $\alpha$ is an ordinal. Recall that ordinals are topological spaces under the usual order topology, and in essence if $\langle x_\xi \rangle_{\xi < \alpha}$ converges to some point $x$, this is equivalent to the $(\alpha+1)$-sequence obtained by appending $x$ to the end of $\langle x_\xi \rangle_{\xi < \alpha}$ being continuous at $\alpha$. (Of course, the $\alpha$-sequence may fail to be continuous at limit ordinals $< \alpha$.)

I assume that the prototypical example of a radial non-Fréchet space would be the ordinal space $\omega_1 + 1$. Here $\omega_1 \in \overline{ [ 0 , \omega_1 ) }$, however there is no ($\omega$-)sequence in $[0 , \omega_1 )$ converging to $\omega_1$ (since $\omega_1$ has uncountable cofinality). However there is a perfectly good $\omega_1$-sequence converging to $\omega_1$: $x_\xi = \xi$ for all $\xi < \omega_1$.

share|improve this answer
add comment

A general sequence $x_1, x_2, ...$ is indexed by the natural numbers. If we allow our indexing set to change to something bigger, like the reals, then you get a transfinite sequence.

So in your question, $\kappa$ is some general indexing set like the reals or the complexes, or perhaps something bigger still, and the sequence is just a collection of elements that are indexed by $\kappa$.

share|improve this answer
I'm sorry but I just don't think this is true. This definition is useless without a definition of 'transfinite sequence' that's more specific than a map from a 'general indexing set.' 'Transfinite' usually refers to the ordinals, as I've detailed in my answer. –  Paul VanKoughnett Jan 21 '13 at 2:34
@mixedmath: If I may ask, what is difference between the radial space and Frechet space? –  Paul Jan 21 '13 at 2:34
add comment

Presumably, this is a sequence indexed by some ordinal number. To be precise, an ordinal number $\kappa$ (which in the standard construction is identified with the set of all ordinals less than $\kappa$) can be given the order topology, and a transfinite sequence indexed by $\kappa$ in a space $X$ is just a continuous map $\kappa \to X$.

A sequence indexed by a finite ordinal $n$ is just an $n$-tuple of points. Ordinary sequences are indexed by the ordinal $\omega$. A sequence indexed by $\omega + 1$ is an ordinary sequence with a limit point. And so on.

Though I've never seen this definition before, here's what I'd guess is the reasoning behind it. We're used to talking about topological conditions like closure, convergence, and so on in terms of sequences, and in metric spaces or more generally first countable spaces, this is all we need. In 'bigger' spaces, the topology of the space isn't really captured by mere countable sets of points, so we need to replace the familiar notion of 'sequence' with something more general. This turns out to be a thing called a net, which is a sequence indexed by a directed set. So while it's not generally true that every limit point of a set in a topological space is a limit of a sequence in that set, it is true that it's the limit of a net in that set.

Now, directed sets are only partially ordered, so dealing with nets might end up being too general. In a radial space, you only need to deal with nets indexed by well-ordered sets -- that is, with transfinite sequences indexed by ordinals.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.