Talk:Sequentially compact space

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Your definition of sequence seems incorrect: a sequence in a set X is not the range of a function from the natural numbers to X; it is a function from the natural numbers to X. Also, is it really necessary to define convergent subsequence as you do? Plclark (talk) 09:01, 10 July 2008 (UTC)Plclark[reply]

I tend to think of a sequence as a set, countable one. Otherwise, it doesn't make sense to say, for example, that a given sequence is bounded. -- Taku (talk) 08:36, 11 July 2008 (UTC)[reply]

For the purposes of this article, it is an important distinction. The property of being sequentially compact and the property that every infinite subset has a limit point are distinct and, in general, inequivalent notions. siℓℓy rabbit (talk) 14:17, 11 July 2008 (UTC)[reply]

You mean how a sequence is defined? To me, the definition of a limit point is usually a tricky one.-- Taku (talk) 16:28, 11 July 2008 (UTC)[reply]

I can't tell from the responses whether I made myself clear in my last message. The main point I was making was that the definition of sequence given here is not consistent with the formal definition of sequence used on wikipedia and throughout mathematics: from the article sequence,

"A more formal definition of a finite sequence with terms in a set S is a function from {1, 2, ..., n} to S for some n ≥ 0. An infinite sequence in S is a function from {1, 2, ...} (the set of natural numbers without 0) to S."

It doesn't make sense to identify a sequence with its range, because then you lose the ordering. You also don't see how many times a given element of the set occurs as an element of the sequence, which is information you need to define the subsequence. Anyway, I will be bold and make the change.

The second point that I made (or tried to) is quite trivial: I just find it hard to believe that the extremely careful "we say that (sn) has a convergent subsequence if there is a subsequence of (sn) that converges" was really necessary: what else could it possibly mean for a sequence to have a convergent subsequence? But on second thought, there's no law against being pedantic in an encyclopedia article. Plclark (talk) 10:36, 13 July 2008 (UTC)Plclark[reply]

I carried through the change. At the moment I have more global concerns about the article -- for instance, is it necessary to have a separate article on sequential compactness or can that material be presented in the article on compactness? Should one have to link to this article to understand the statement of the Bolzano-Weierstrass theorem? I want to think further... Plclark (talk) 10:54, 13 July 2008 (UTC)Plclark[reply]

I agree that viewing a sequence as a range of a function can be problematic. For example, a sequence with repeated terms (1, 1, 1, ...) would be {1} according to this definition, which is really not quite true (not being a countable set for one thing). I have come to think it make more sense to define a sequence as a graph of a function defined on the set of natural numbers; this way you still get the ordering and the sequence is guaranteed to be countable. Put in another way, this definition also means that a sequence of, say, real numbers, is an element of ${\displaystyle \mathbb {R} ^{\mathbf {N} }}$. Since a finite sequence of length n is just an element of ${\displaystyle \mathbb {R} ^{n}}$, this definition is quite intuitive too, in my opinion. (Set-theoretically speaking, since there is no difference between a function and the graph of the function, this definition is the same as one cited above. But, in reality, we tend to distinguishes a function from its graph.) But this definition isn't quite compatible with how we treat sequences in reality. For example, one asks whether a sequence has a limit point; this is related (but not equivalently) to whether the sequence has a convergent subsequence. Here, a sequence is, obviously, thought of a subset of the

As to the point of this article, I think there can be a lot of things to discuss on this topic. For example, in functional analysis, if a compact space is actually sequentially so is an important question. (This point has to be mentioned somehow in the article, I think.) -- Taku (talk) 12:10, 13 July 2008 (UTC)[reply]

The sentence "A limit point compact space is always sequentially compact; the same is true for a compact space." is wrong. The product of uncountably many copies of [0,1] with the product topology is both limit-point compact and compact, but it is not sequentially compact. Oxeador (talk) 08:54, 18 July 2008 (UTC)[reply]

I'm a little confused, what exactly does it mean to have a convergent sequence in a general topological space? Are we talking about filters or nets or something? A nice, tight definition would be helpful 202.36.179.66 (talk) 04:01, 12 November 2009 (UTC)[reply]

A topological space has necessarily a "nearness structure" defined on it. Therefore, ideas such as convergence (which rely on nearness - that is, a sequence "approaching a limit") can be defined in an abstract topological space. Hope this helps. --PST 06:50, 12 November 2009 (UTC)[reply]

Sorry can you be a little more explicit or point me in a direction where I can read more about this? Nearness structures, to my mind, require stronger axioms than those of a general topological space --- those of a uniform space, perhaps. 202.36.179.66 (talk) 20:47, 12 November 2009 (UTC)[reply]

Open sets are, very roughly, points which are near each other. So you can define convergence in a topological space by: ${\displaystyle x_{n}\rightarrow x}$ iff given any open set U containing x, all x_n sufficiently far along lie within U.Katrielalex (talk) 15:51, 17 November 2009 (UTC)[reply]

Is this definition standard? It seems problematic because using this definition of convergence you can no longer infer anything about the continuity of a function. In particular, it is no longer true that a function is continuous if every sequence composed with the function converges to the same limit. Are there any meaningful results about general sequentially compact spaces (with no separation axioms!) using this definition?202.36.179.66 (talk) 03:23, 26 November 2009 (UTC)[reply]

If a ${\displaystyle f:X\to Y}$ is continuous, and if ${\displaystyle {(x_{n})}_{n\in \mathbb {N} }}$ is a sequence converging to ${\displaystyle x\in X}$, ${\displaystyle {(f(x_{n}))}_{n\in \mathbb {N} }}$ converges to ${\displaystyle f(x)}$. The converse, however, does not hold unless the space is first countable; the opening of the article Net may explain this. One purpose of the notion of a "Sequentially compact space" is to provide an alternate way in which one may show certain spaces are compact. For instance, since for metric spaces compactness is equivalent to sequential compactness, and since ${\displaystyle [0,1]^{\omega }}$ is a metric space, one can show that it is compact by merely showing that it is sequentially compact. If you attempt to do this, you will find that sequential compactness is easier to deal with compared to compactness, in concrete cases. Of course, in some cases, compactness in its original form is far more natural (such as proving the compactness of ${\displaystyle [0,1]}$). Hope this helps. --PST 10:27, 26 November 2009 (UTC)[reply]

Thanks for your response. I understand your points, but it seems to me that the notion of sequential compactness is only invoked if you have a space with more structure than a general topological space. I've only come across them in the context of metric spaces. In fact, if the space is not first countable, then the whole notion of a convergent sequence seems problematic. I wondered if you knew of any examples where sequential compactness was useful where the underlying space is non-metrisable or even not first countable. 202.36.179.66 (talk) 22:39, 26 November 2009 (UTC)[reply]

The concept is interesting even when the underlying space is not metrizable! If mathematicians only studied metric spaces, life would be rather dull, since topological spaces exhibit great diversity in complexion, especially in cardinality and the like. Sequential compactness for non-metrizable spaces does include spaces such as ${\displaystyle {\mathbb {R} }^{J}}$ for some uncountable set J, but it also includes more abstract topological spaces. I suppose by "useful" you wish to know the sorts of theorems that can be proved about sequential compactness; Open Problems in Topology I and its sequel should be of use. Outside topology, these compactness conditions have some use in functional analysis. Hope this helps. --PST 02:08, 27 November 2009 (UTC)[reply]

I understand the importance of studying more abstract spaces than metric spaces --- you don't need to talk down to me. I just don't see the point of being more general than is necessary. If nothing interesting can be said about sequential compactness in a general space, then the scope of the article should be narrowed to at least first-countable spaces or something similar. 202.36.179.66 (talk) 03:02, 27 November 2009 (UTC)[reply]

Well, would not the extent to which the equivalence between sequential compactness and compactness fails, be a criterion for lack of first countability or metrizability? How about employing this in understanding general topological spaces? --PST 08:38, 27 November 2009 (UTC)[reply]

Is it common to work with sequences in a general topological space, where all you have are the 3 axioms of topology? My understanding has been that topological spaces are too general to allow meaningful results for sequences -- hence the need for filters and nets. Seeing how sequences in general topological spaces don't seem to be standard, I wonder if the scope of the opening paragraph needs to be narrowed to spaces with more structure. Or perhaps more information added about the problems of sequential compactness in a general topological space, or even analogues to sequential compactness with filters or nets. 121.98.84.152 (talk) 09:52, 17 February 2010 (UTC)[reply]

Why does one need to take uncountable product of [0,1]? Isn't the countable product sufficient (e. g. sequence (0,..0,1,0,...) seems to have no convergent subsequence)? Michall (talk) 17:42, 14 May 2010 (UTC)[reply]

Every sequence of 0s and 1s has (1,1,1,...) or (0,0,0,...) as a subsequence. These are both convergent. Strad (talk) 22:38, 14 May 2010 (UTC)[reply]

Maybe I don't understand something but I don't think we're writing about the same thing: elements of countable product of [0,1] are ω-tuples (sequences) of form (a,b,c,...), so by writing (0,..0,1,0,...) I meant such a sequence (of sequences):

(1,0,0,0,0,...)

(0,1,0,0,0,...)

(0,0,1,0,0,...)

...

I don't think it has a chance of having a convergent subsequence (or does it?). Michall (talk) 23:50, 14 May 2010 (UTC)[reply]

I think that you may be confusing the box and product topologies on ${\displaystyle [0,1]^{\omega }}$. In the box topology on ${\displaystyle [0,1]^{\omega }}$, no subsequence of the sequence you quote converges, which is probably what you intended to say. To see this, note that if some subsequence of ${\displaystyle \{x_{n}\}_{n\in \mathbb {N} }}$ converges (in the box topology), where ${\displaystyle x_{i}=(0,\cdots ,0,1,0,\cdots )}$ and 1 appears as the ith coordinate of x_i, the point to which it converges must be a sequence consisting entirely of 1's and 0's (denote this sequence by y). Furthermore, if some coordinate of y, suppose the jth coordinate of y, is a 1, then the neighborhood ${\displaystyle [0,1]\times \cdots \times [0,1]\times ({\frac {1}{2}},1]\times [0,1]\times \cdots }$ (where ${\displaystyle ({\frac {1}{2}},1]}$ is the jth term in the cartesian product), contains precisely 1 term of the sequence; a contradiction. Therefore, the sequence must consist entirely of 0's, but this is also impossible as it is easy to see that the neighbourhood ${\displaystyle [0,{\frac {1}{2}})\times \cdots \times [0,{\frac {1}{2}}]\times \cdots }$ contains no term of the sequence.

However, in the product topology on ${\displaystyle [0,1]^{\omega }}$, the sequence itself converges to ${\displaystyle (0,\cdots ,0,\cdots )}$. Any neighborhood of this point in the product topology must contain some open basis set of the form ${\displaystyle \prod _{i\in \mathbb {N} }U_{i}}$ where ${\displaystyle U_{j}=[0,1]}$ for all but finitely many j; choose k such that for ${\displaystyle i>k}$k}"/>, ${\displaystyle U_{i}=[0,1]}$. Then if we denote the sequence by ${\displaystyle \{x_{n}\}_{n\in \mathbb {N} }}$, U will contain every x_i for ${\displaystyle i>k}$k}"/>, and since U was arbitrary, we see that the sequence converges to ${\displaystyle (0,\cdots ,0,\cdots )}$.

In fact, a very "neat argument" yields that ${\displaystyle [0,1]^{\omega }}$ equipped with the product topology is sequentially compact; that is, any sequence in this space has a convergent subsequence (again, in the product topology). The argument I quote does not use the compactness of ${\displaystyle [0,1]^{\omega }}$. (I apologize for the lack of reference or proof here; I do know the argument but do not have the time to post it here at the moment. However, I also know that the argument can be found in Counterexamples in Topology by Steen and Seebach (as referenced in this very article); since I do not have the book at hand, I cannot precisely tell you its location, and I apologize again for that.) Of course, Tychonoff's theorem, and the fact that ${\displaystyle [0,1]^{\omega }}$ is metrizable under the product topology (indeed, it is not hard to see that the metric defined by ${\displaystyle D(x,y)={\mbox{lub}}\{{\frac {d'(x_{i},y_{i})}{i}}\}}$ induces the topology in question, where ${\displaystyle d'(a,b)={\mbox{min}}\{|a-b|,1\}}$ and "lub" is an abbreviation for "least upper bound"; a reference is Theorem 9.5 on page 123 of Topology, A First Course by James Munkres (this corresponds to the section entitled "The Metric Topology")), give an alternative argument for the sequential compactness of ${\displaystyle [0,1]^{\omega }}$. PST 03:34, 15 May 2010 (UTC)[reply]

Thanks a lot for your effort. I should have attended Topology I lectures more often (or just read the definition of product topology, as I indeed did identify it with the box topology). Michall (talk) 09:54, 15 May 2010 (UTC)[reply]

I must be missing something. A sequence in a compact space is a net and any net has a convergent subnet; therefore the sequence has a convergent subsequence. Therefore compactness should imply sequential compactness. What am I missing here? I think I should remove this statement. ThomasSteinke 02:03, 7 July 2010 (UTC) —Preceding unsigned comment added by Thomassteinke (talk • contribs) OK, I figured it out, it comes down to how a subnet is defined. A subnet of a sequence is not necessarily a subsequence. ThomasSteinke 04:13, 8 July 2010 (UTC)[reply]