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1. • CommentRowNumber1.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 6th 2010
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexCreated [[sequential compactness]], should probably link to all these compactness variations from [[compact space]]. Not sure if I got the "iff" bit right in the relationship with compactness itself; will check it myself if no-one fixes it in the meantime. I decided that this was the key property in [[manifolds of mapping spaces]] and to stop trying to figure out a Froelicher version of sequentially compact for the time-being.

   Created sequential compactness, should probably link to all these compactness variations from compact space. Not sure if I got the “iff” bit right in the relationship with compactness itself; will check it myself if no-one fixes it in the meantime.

   I decided that this was the key property in manifolds of mapping spaces and to stop trying to figure out a Froelicher version of sequentially compact for the time-being.

2. • CommentRowNumber2.
   • CommentAuthorTim_van_Beek
   • CommentTimeMay 6th 2010
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   Author: Tim_van_Beek
   Format: MarkdownHi Andrew, I think there is in general no implication from compact to sequentially compact and vice versa (which violates my intuition), you can look it up in the book by Helmut Schäfer: "Topological Vector Spaces", chapter 11 "weak compactness", excercise 37 gives examples of a space that is compact but not sequentially compact and one which is sequentially compact but not compact.

   Hi Andrew,

   I think there is in general no implication from compact to sequentially compact and vice versa (which violates my intuition), you can look it up in the book by Helmut Schäfer: "Topological Vector Spaces", chapter 11 "weak compactness", excercise 37 gives examples of a space that is compact but not sequentially compact and one which is sequentially compact but not compact.

3. • CommentRowNumber3.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 6th 2010
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexMy favourite book! Sadly not at work right now so will have to check what Schafer says tomorrow. Wikipedia backs you up on this so I'll take out the statement. Whoops! Sadly, I'm over the "mutiple edits count as one" limit on that page so my topological goof will stand for all time, unless I abuse my powers and edit the database itself.

   My favourite book!

   Sadly not at work right now so will have to check what Schafer says tomorrow. Wikipedia backs you up on this so I’ll take out the statement. Whoops! Sadly, I’m over the “mutiple edits count as one” limit on that page so my topological goof will stand for all time, unless I abuse my powers and edit the database itself.

4. • CommentRowNumber4.
   • CommentAuthorTim_van_Beek
   • CommentTimeMay 6th 2010
   • (edited May 6th 2010)
   • PermaLink
   Author: Tim_van_Beek
   Format: MarkdownItexSince I did not see Andrew's answer #3 in time I wrote: Wikipedia mentions this, too, see <a href="http://en.wikipedia.org/wiki/Compact_space">compact space</a> and then look out for the sentence " Not every compact space is sequentially compact; an example is given by $2^{[0,1]}$, with the product topology (Scarborough & Stone 1966, Example 5.3)."

   Since I did not see Andrew’s answer #3 in time I wrote:

   Wikipedia mentions this, too, see compact space and then look out for the sentence ” Not every compact space is sequentially compact; an example is given by $2^{[0,1]}$, with the product topology (Scarborough & Stone 1966, Example 5.3).”

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeMay 6th 2010
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   Author: Mike Shulman
   Format: MarkdownItexThe link [[sequential compactness]] doesn't exist, did you mean [[sequentially compact]]? Perhaps the name should be [[sequentially compact space]]?

   The link sequential compactness doesn’t exist, did you mean sequentially compact? Perhaps the name should be sequentially compact space?

6. • CommentRowNumber6.
   • CommentAuthorTim_van_Beek
   • CommentTimeMay 6th 2010
   • PermaLink
   Author: Tim_van_Beek
   Format: MarkdownItexSimply added both [[sequential compactness]] and [[sequentially compact space]] as redirects, (plus a warning that compactness does not imply sequential compactness, which is a pitfall, admittedly, that caught me once, too).

   Simply added both sequential compactness and sequentially compact space as redirects, (plus a warning that compactness does not imply sequential compactness, which is a pitfall, admittedly, that caught me once, too).

7. • CommentRowNumber7.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 6th 2010
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexNow I'm worried that the property that I _really_ want isn't equivalent to sequential compactness after all. What I want my space $X$ to satisfy is that a neighbourhood of $\{0\} \times X$ in $\mathbb{R}$ contains a subset of the form $(-\epsilon,\epsilon) \times X$. I thought that this was equivalent to sequential compactness. My argument was as follows: Suppose that $X$ is sequentially compact. Let $U \subseteq \mathbb{R} \times X$ be a subset containing $\{0\} \times X$. Suppose that $U$ does not contain a subset of the form $(-\epsilon,\epsilon) \times X$. Then for each $n \in \mathbb{N}$ we can find some $(t_n, x_n) \in U$ such that $\abs{t_n} \le \frac1n$. The sequence $(x_n)$ in $X$ has a convergent subsequence, say $(x_{n_k}) \to x$. Then $(t_{n_k}, x_{n_k})$ converges in $\mathbb{R} \times X$, but $(t_{n_k}, x_{n_k}) \notin U$ for all $k$ but its limit, $(0,x)$ is in $U$. Hence $U$ is not open. Conversely, assume that neighbourhoods of $\{0\} \times X$ contain slices as claimed. Then let $(x_n)$ be a sequence in $X$. Consider the set $U$ formed by taking $\mathbb{R} \times X$ and removing $(\frac1n, x_n)$. This does not contain a subset of the form $(-\epsilon,\epsilon) \times X$ and so cannot be open in $\mathbb{R} \times X$. ... Ah, and now I see my mistake! The spaces that I want to apply this to have the property that their topology is determined by a family of curves. So as the set $U$ is not open, there must be a curve $c \colon \mathbb{R} \to U$ which detects the fact that it is not open. That is, $c^{-1}(U)$ is not open in $\mathbb{R}$. Now we can find a sequence $(s_k) \to s$ in $\mathbb{R}$ such that $(s_k) \notin c^{-1}(U)$ but $s \in U$. Then as $c(s_k) \notin U$, we must have $c(s_k) = (t_{n_k}, x_{n_k})$ for some $n_k$. By passing to a subsequence if necessary, we can assume that the $n_k$s are strictly increasing. Then as $(s_k) \to s$, $c(s_k) \to c(s)$ and so $x_{n_k} \to p_X c(s)$. Hence $X$ is sequentially compact. So my claim is true for spaces whose topology is generated by a family of curves. There's some fancy name for that, but I can't remember it right now. I've removed the claim from [[sequentially compact]]. Now, what should I do about the correct version? Should it go where I want to use it, or can it go at sequentially compact?

   Now I’m worried that the property that I really want isn’t equivalent to sequential compactness after all.

   What I want my space $X$ to satisfy is that a neighbourhood of $\{0\} \times X$ in $\mathbb{R}$ contains a subset of the form $(-\epsilon,\epsilon) \times X$. I thought that this was equivalent to sequential compactness. My argument was as follows:

   Suppose that $X$ is sequentially compact. Let $U \subseteq \mathbb{R} \times X$ be a subset containing $\{0\} \times X$. Suppose that $U$ does not contain a subset of the form $(-\epsilon,\epsilon) \times X$. Then for each $n \in \mathbb{N}$ we can find some $(t_n, x_n) \in U$ such that $\abs{t_n} \le \frac1n$. The sequence $(x_n)$ in $X$ has a convergent subsequence, say $(x_{n_k}) \to x$. Then $(t_{n_k}, x_{n_k})$ converges in $\mathbb{R} \times X$, but $(t_{n_k}, x_{n_k}) \notin U$ for all $k$ but its limit, $(0,x)$ is in $U$. Hence $U$ is not open.

   Conversely, assume that neighbourhoods of $\{0\} \times X$ contain slices as claimed. Then let $(x_n)$ be a sequence in $X$. Consider the set $U$ formed by taking $\mathbb{R} \times X$ and removing $(\frac1n, x_n)$. This does not contain a subset of the form $(-\epsilon,\epsilon) \times X$ and so cannot be open in $\mathbb{R} \times X$.

   …

   Ah, and now I see my mistake! The spaces that I want to apply this to have the property that their topology is determined by a family of curves. So as the set $U$ is not open, there must be a curve $c \colon \mathbb{R} \to U$ which detects the fact that it is not open. That is, $c^{-1}(U)$ is not open in $\mathbb{R}$. Now we can find a sequence $(s_k) \to s$ in $\mathbb{R}$ such that $(s_k) \notin c^{-1}(U)$ but $s \in U$. Then as $c(s_k) \notin U$, we must have $c(s_k) = (t_{n_k}, x_{n_k})$ for some $n_k$. By passing to a subsequence if necessary, we can assume that the $n_k$s are strictly increasing. Then as $(s_k) \to s$, $c(s_k) \to c(s)$ and so $x_{n_k} \to p_X c(s)$. Hence $X$ is sequentially compact.

   So my claim is true for spaces whose topology is generated by a family of curves. There’s some fancy name for that, but I can’t remember it right now. I’ve removed the claim from sequentially compact. Now, what should I do about the correct version? Should it go where I want to use it, or can it go at sequentially compact?

8. • CommentRowNumber8.
   • CommentAuthorEric
   • CommentTimeMay 7th 2010
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   Author: Eric
   Format: MarkdownItex>So my claim is true for spaces whose topology is generated by a family of curves. Curvaceous? :)

   So my claim is true for spaces whose topology is generated by a family of curves.

   Curvaceous? :)

9. • CommentRowNumber9.
   • CommentAuthorAndrew Stacey
   • CommentTimeMay 7th 2010
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   Author: Andrew Stacey
   Format: MarkdownItexI use that for Froelicher spaces, but I think that there's a general notion in topology for when the topology is generated by some subfamily. We certainly have "compactly generated" if the family is compact topological spaces. While I'm happy to invent fun terms when there isn't an existing one, if there is one already in use then I think that I should use that! And if there is one for the _topological_ notion, then it would still be correct to use "curvaceous" for the Froelicher one since it's not guaranteed that the topology generated by the smooth curves is the same as that generated by the continuous ones! (It may be so, I haven't thought about it)

   I use that for Froelicher spaces, but I think that there’s a general notion in topology for when the topology is generated by some subfamily. We certainly have “compactly generated” if the family is compact topological spaces. While I’m happy to invent fun terms when there isn’t an existing one, if there is one already in use then I think that I should use that!

   And if there is one for the topological notion, then it would still be correct to use “curvaceous” for the Froelicher one since it’s not guaranteed that the topology generated by the smooth curves is the same as that generated by the continuous ones! (It may be so, I haven’t thought about it)

10. • CommentRowNumber10.
    • CommentAuthorTim_van_Beek
    • CommentTimeMay 7th 2010
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    Author: Tim_van_Beek
    Format: MarkdownItex<blockquote> <p> I use that for Froelicher spaces, but I think that there's a general notion in topology for when the topology is generated by some subfamily. </p> </blockquote> All I ever heard of is final or inductive topology as it is defined by wikipedia <a href="http://en.wikipedia.org/wiki/Final_topology">here</a>, I suppose you mean something that is more specific? Until someone remembers or fills us in I would suggest to create a subpage to [[inductive limit]] like "example: inductive topology" and there "inductive topology with respect to a collection of curves", what do you think?

    I use that for Froelicher spaces, but I think that there’s a general notion in topology for when the topology is generated by some subfamily.

    All I ever heard of is final or inductive topology as it is defined by wikipedia here, I suppose you mean something that is more specific?

    Until someone remembers or fills us in I would suggest to create a subpage to inductive limit like “example: inductive topology” and there “inductive topology with respect to a collection of curves”, what do you think?

11. • CommentRowNumber11.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 7th 2010
    • (edited May 7th 2010)
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexI guess the point I would want to emphasise is that the family of sources is fixed, once and for all. In my case, it's $\mathbb{R}$, but more general situations are possible. I'm not sure that "final topology" captures this. As an example of what I'm trying to get at, in Kriegl and Michor's book then they talk about various alternative topologies on a locally convex tvs (section 4.7): * $k E$, the Kelley-fication: final topology induced by the inclusions of all subsets being compact for the locally convex topology * $s E$, final topology induced by the inclusions of all continuous curves; equivalently, convergent sequences * $c^\infty E$, final topology induced by the inclusions of all smooth curves They talk about a lctvs being **sequentially generated** if $s E = E$. However, the equivalence of sequences and curves doesn't hold for all _topological_ spaces so "sequentially generated" is not what I'm looking for here.

    I guess the point I would want to emphasise is that the family of sources is fixed, once and for all. In my case, it’s $\mathbb{R}$, but more general situations are possible. I’m not sure that “final topology” captures this.

    As an example of what I’m trying to get at, in Kriegl and Michor’s book then they talk about various alternative topologies on a locally convex tvs (section 4.7):

    • $k E$, the Kelley-fication: final topology induced by the inclusions of all subsets being compact for the locally convex topology
    • $s E$, final topology induced by the inclusions of all continuous curves; equivalently, convergent sequences
    • $c^\infty E$, final topology induced by the inclusions of all smooth curves

    They talk about a lctvs being sequentially generated if $s E = E$. However, the equivalence of sequences and curves doesn’t hold for all topological spaces so “sequentially generated” is not what I’m looking for here.

12. • CommentRowNumber12.
    • CommentAuthorTim_van_Beek
    • CommentTimeMay 7th 2010
    • PermaLink
    Author: Tim_van_Beek
    Format: MarkdownItex<blockquote> <p> I guess the point I would want to emphasise is that the family of sources is fixed, once and for all. </p> </blockquote> Hm. And you are sure that there already is a similar concept? How about this: A subcategory of Top is $\mathbb{R}$-detectable, if for every object X a subset $U \subset X$ is open iff for all arrows r: $\mathbb{R} \to U$ the set $r^{-1}(U)$ is open?

    I guess the point I would want to emphasise is that the family of sources is fixed, once and for all.

    Hm. And you are sure that there already is a similar concept?

    How about this: A subcategory of Top is $\mathbb{R}$-detectable, if for every object X a subset $U \subset X$ is open iff for all arrows r: $\mathbb{R} \to U$ the set $r^{-1}(U)$ is open?

13. • CommentRowNumber13.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 7th 2010
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexI like it, though I wonder if it's necessary to be able to talk about arbitrary subcategories. Wouldn't it be enough to have the full subcategory of all objects with that property?

    I like it, though I wonder if it’s necessary to be able to talk about arbitrary subcategories. Wouldn’t it be enough to have the full subcategory of all objects with that property?

14. • CommentRowNumber14.
    • CommentAuthorTim_van_Beek
    • CommentTimeMay 7th 2010
    • (edited May 7th 2010)
    • PermaLink
    Author: Tim_van_Beek
    Format: MarkdownItexSure, if you don't see the need to restrict the sets of arrows...as you said before, you would like to stress that the family of domains is fixed, so it would suffice to say: $Top_{\mathbb{R}}$ is the maximal full subcategory of Top that is $\mathbb{R}$-detectable and has $\mathbb{R}$ as object.

    Sure, if you don’t see the need to restrict the sets of arrows…as you said before, you would like to stress that the family of domains is fixed, so it would suffice to say:

    $Top_{\mathbb{R}}$ is the maximal full subcategory of Top that is $\mathbb{R}$-detectable and has $\mathbb{R}$ as object.

15. • CommentRowNumber15.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 11th 2010
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexBack to sequential compactness itself. I'm now confused by the claim that compactness does **not** imply sequential compactness. Compactness implies that every net has a convergent subnet. So if I take a sequence, that has a convergent subnet. But surely a subnet of a sequence is just a subsequence, so doesn't this imply sequential compactness? Where am I going wrong, here?

    Back to sequential compactness itself. I’m now confused by the claim that compactness does not imply sequential compactness. Compactness implies that every net has a convergent subnet. So if I take a sequence, that has a convergent subnet. But surely a subnet of a sequence is just a subsequence, so doesn’t this imply sequential compactness? Where am I going wrong, here?

16. • CommentRowNumber16.
    • CommentAuthorMike Shulman
    • CommentTimeMay 11th 2010
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    Author: Mike Shulman
    Format: MarkdownItexActually, a subnet of a sequence need not be a subsequence! A subnet of some net is just some other net equipped with a cofinal map to the net you started from. But just because the original net is indexed on $\mathbb{N}$, the subnet need not be. It's perhaps arguable that using the term "subnet" for a cofinal map of directed sets creates more confusion than it is worth, but it's pretty standard.

    Actually, a subnet of a sequence need not be a subsequence! A subnet of some net is just some other net equipped with a cofinal map to the net you started from. But just because the original net is indexed on $\mathbb{N}$, the subnet need not be.

    It’s perhaps arguable that using the term “subnet” for a cofinal map of directed sets creates more confusion than it is worth, but it’s pretty standard.

17. • CommentRowNumber17.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 11th 2010
    • (edited May 11th 2010)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexSo the issue is that subnets might not be "sub", i.e., injective? Edit: never mind -- I see that's not quite the issue at hand. Anyway, then reword "sequential compactness" to say that every sequence has a convergent subsequence. Is there a claim that compactness does not imply sequential compactness in that sense?

    So the issue is that subnets might not be “sub”, i.e., injective? Edit: never mind – I see that’s not quite the issue at hand.

    Anyway, then reword “sequential compactness” to say that every sequence has a convergent subsequence. Is there a claim that compactness does not imply sequential compactness in that sense?

18. • CommentRowNumber18.
    • CommentAuthorTim_van_Beek
    • CommentTimeMay 11th 2010
    • PermaLink
    Author: Tim_van_Beek
    Format: MarkdownItexMike wrote: <blockquote> <p> Actually, a subnet of a sequence need not be a subsequence! </p> </blockquote> Interesting! But in the context of [[TVS]] a sequence and all it's subsequences are actually nets by definition. You refer to a more general definition - so that's not the pitfall... Andrew wrote: <blockquote> <p> Compactness implies that every net has a convergent subnet. </p> </blockquote> Now that's the pitfall, isn't it? Actually compactness implies that every net has a <b>cluster</b> point, and in many situations that implies that there is a convergent subnet. It does not in the example that I tried to explain in the warning section of [[sequential compactness]]: The sequence there has a cluster point (the function that is identical zero), but does not have a convergent subsequence :-) For further information it could be worthwhile to look at a proof of the Eberlein-Smulian theorem, for example in the book "Series and Sequences in Banach Spaces" by Joseph Diestel.

    Mike wrote:

    Actually, a subnet of a sequence need not be a subsequence!

    Interesting! But in the context of TVS a sequence and all it’s subsequences are actually nets by definition. You refer to a more general definition - so that’s not the pitfall…

    Andrew wrote:

    Compactness implies that every net has a convergent subnet.

    Now that’s the pitfall, isn’t it? Actually compactness implies that every net has a cluster point, and in many situations that implies that there is a convergent subnet. It does not in the example that I tried to explain in the warning section of sequential compactness: The sequence there has a cluster point (the function that is identical zero), but does not have a convergent subsequence :-)

    For further information it could be worthwhile to look at a proof of the Eberlein-Smulian theorem, for example in the book “Series and Sequences in Banach Spaces” by Joseph Diestel.

19. • CommentRowNumber19.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 11th 2010
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    Author: Andrew Stacey
    Format: MarkdownItexOkay, so the issue is more subtle than I thought and needs a bit of careful thought on my behalf (which, once I've figured it out, I shall put on the relevant pages!). Just to confirm, then, that the wikipedia page on nets is wrong in this regard: <http://en.wikipedia.org/wiki/Net_%28mathematics%29> since, to my reading, it claims that limits of nets are what I think they are and then says that "every net has a convergent subnet" is equivalent to compactness.

    Okay, so the issue is more subtle than I thought and needs a bit of careful thought on my behalf (which, once I’ve figured it out, I shall put on the relevant pages!). Just to confirm, then, that the wikipedia page on nets is wrong in this regard: http://en.wikipedia.org/wiki/Net_%28mathematics%29 since, to my reading, it claims that limits of nets are what I think they are and then says that “every net has a convergent subnet” is equivalent to compactness.

20. • CommentRowNumber20.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 11th 2010
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexJust playing catch-up here with this discussion: I didn't know compactness did not imply sequential compactness. But wikipedia cites Steen and Seebach, Counterexamples in Topology, and I see that that's Tim's example. Interesting! However (Tim), are you quite sure of the claim that "every net has a convergent subnet" is inequivalent to compactness? Because I think it is equivalent, and I think I'd be able to write out a proof.

    Just playing catch-up here with this discussion: I didn’t know compactness did not imply sequential compactness. But wikipedia cites Steen and Seebach, Counterexamples in Topology, and I see that that’s Tim’s example. Interesting!

    However (Tim), are you quite sure of the claim that “every net has a convergent subnet” is inequivalent to compactness? Because I think it is equivalent, and I think I’d be able to write out a proof.

21. • CommentRowNumber21.
    • CommentAuthorTim_van_Beek
    • CommentTimeMay 11th 2010
    • (edited May 11th 2010)
    • PermaLink
    Author: Tim_van_Beek
    Format: MarkdownItexHm, that's really interesting. Wikipedia is not the only page that contradicts me, I'm outmatched! However, I will simply cite Helmut Schäfer, who states in the introduction to his book: compactness = every filter has a cluster point = every ultrafilter converges. And I'm still convinced of the counterexample (yes, I see that's the one in Steen and Seebach, I probably saw it there first). But, no, I'm not sure anymore about my claim.

    Hm, that’s really interesting. Wikipedia is not the only page that contradicts me, I’m outmatched!

    However, I will simply cite Helmut Schäfer, who states in the introduction to his book: compactness = every filter has a cluster point = every ultrafilter converges.

    And I’m still convinced of the counterexample (yes, I see that’s the one in Steen and Seebach, I probably saw it there first). But, no, I’m not sure anymore about my claim.

22. • CommentRowNumber22.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 11th 2010
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    Author: Andrew Stacey
    Format: MarkdownItexGot to run now, but I had a go at constructing a convergent subnet! Have a look to see if anyone agrees with me. (Bit rushed so may be lacking some details, and may have misunderstood some concepts)

    Got to run now, but I had a go at constructing a convergent subnet! Have a look to see if anyone agrees with me.

    (Bit rushed so may be lacking some details, and may have misunderstood some concepts)

23. • CommentRowNumber23.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 11th 2010
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexDoesn't quite work - the ordered set isn't directed, but it _feels_ close.

    Doesn’t quite work - the ordered set isn’t directed, but it feels close.

24. • CommentRowNumber24.
    • CommentAuthorTim_van_Beek
    • CommentTimeMay 11th 2010
    • (edited May 11th 2010)
    • PermaLink
    Author: Tim_van_Beek
    Format: MarkdownItexI wrote <blockquote> <p> Actually compactness implies that every net has a cluster point, and in many situations that implies that there is a convergent subnet. </p> </blockquote> Rewind and delete, that is wrong. Andrew, Mike and Todd are right, compact = every net has a cluster point = every net has a subnet that is convergent. As Mike said (my mistake, sorry, I misunderstood you): A sequence does not need to have a convergent <i>subsequence</i> because not every subnet is a subsequence. So, although I do not see how, there has to be a way to construct a convergent subnet in the counterexample.

    I wrote

    Actually compactness implies that every net has a cluster point, and in many situations that implies that there is a convergent subnet.

    Rewind and delete, that is wrong. Andrew, Mike and Todd are right, compact = every net has a cluster point = every net has a subnet that is convergent.

    As Mike said (my mistake, sorry, I misunderstood you): A sequence does not need to have a convergent subsequence because not every subnet is a subsequence. So, although I do not see how, there has to be a way to construct a convergent subnet in the counterexample.

25. • CommentRowNumber25.
    • CommentAuthorTim_van_Beek
    • CommentTimeMay 11th 2010
    • PermaLink
    Author: Tim_van_Beek
    Format: MarkdownItexNow that I think I understand what Mike said... :-) ... I added the definition of subnet to [[net]] and a short explanation to [[sequentially compact]].

    Now that I think I understand what Mike said…

    :-)

    … I added the definition of subnet to net and a short explanation to sequentially compact.

26. • CommentRowNumber26.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 12th 2010
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexI've fixed my example of a convergent subnet in $I^I$, conditional on a cluster point being given. It would be nice to explain how to find such a cluster point of this particular sequence; I'm beginning to suspect that this would depend explicitly on the axiom of choice (or some weaker version perhaps). (To be clear, given that everything is so explicit in this example then it would be nice to trace along the process of finding a cluster point to make it very clear where such an axiom is needed.)

    I’ve fixed my example of a convergent subnet in $I^I$, conditional on a cluster point being given. It would be nice to explain how to find such a cluster point of this particular sequence; I’m beginning to suspect that this would depend explicitly on the axiom of choice (or some weaker version perhaps). (To be clear, given that everything is so explicit in this example then it would be nice to trace along the process of finding a cluster point to make it very clear where such an axiom is needed.)

27. • CommentRowNumber27.
    • CommentAuthorTodd_Trimble
    • CommentTimeMay 12th 2010
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexHere's a first pass: let $X$ be the space $I^I$, and let $X_n = \{a_m: n \leq m\}$. Clearly any finite intersection of the $X_n$ is nonempty, and therefore any finite intersection of their closures is nonempty. Hence by compactness, the intersection of all their closures, $$\bigcap_n \widebar{X_n}$$ is nonempty. Any point in this intersection is a cluster point. Of course, this finite intersection property is an equivalent formulation of compactness, and AC enters in showing that $I^I$ is compact to begin with. (For the Tychnoff theorem for compact Hausdorff spaces, you can get away with ever so slightly less than full AC. Basically you need that every filter, for example the filter generated by any collection satisfying the finite intersection property, can be extended to an ultrafilter; that's called the [[ultrafilter theorem]], and as a principle it's weaker than AC. Such an ultrafilter will converge to a *unique* point if the spaces are Hausdorff -- so no choice there -- any choice is concentrated in the ultrafilter theorem itself. Anyway, that point of convergence will be a cluster point of the collection you started with, the one that generates a filter.)

    Here’s a first pass: let $X$ be the space $I^I$, and let $X_n = \{a_m: n \leq m\}$. Clearly any finite intersection of the $X_n$ is nonempty, and therefore any finite intersection of their closures is nonempty. Hence by compactness, the intersection of all their closures,

    $$\bigcap_n \widebar{X_n}$$

    is nonempty. Any point in this intersection is a cluster point. Of course, this finite intersection property is an equivalent formulation of compactness, and AC enters in showing that $I^I$ is compact to begin with. (For the Tychnoff theorem for compact Hausdorff spaces, you can get away with ever so slightly less than full AC. Basically you need that every filter, for example the filter generated by any collection satisfying the finite intersection property, can be extended to an ultrafilter; that’s called the ultrafilter theorem, and as a principle it’s weaker than AC. Such an ultrafilter will converge to a unique point if the spaces are Hausdorff – so no choice there – any choice is concentrated in the ultrafilter theorem itself. Anyway, that point of convergence will be a cluster point of the collection you started with, the one that generates a filter.)

28. • CommentRowNumber28.
    • CommentAuthorMike Shulman
    • CommentTimeMay 12th 2010
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    Author: Mike Shulman
    Format: MarkdownItexI expect pretty much anything phrased in terms of nets or filters to require AC, or specifically the ultrafilter theorem. I knew that compactness is equivalent to every net having a convergent subnet, but I hadn't ever really thought about how that particular sequence, which "obviously" converges to nothing at all, must have a convergent subnet. That's pretty wild.

    I expect pretty much anything phrased in terms of nets or filters to require AC, or specifically the ultrafilter theorem.

    I knew that compactness is equivalent to every net having a convergent subnet, but I hadn’t ever really thought about how that particular sequence, which “obviously” converges to nothing at all, must have a convergent subnet. That’s pretty wild.

29. • CommentRowNumber29.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 12th 2010
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexI just asked this as a question on MO to see if anyone there can shed any light on the relationship between compactness, sequentially compactness, and the strength of the axiom needed to find this example. Probably everyone here will see it anyway, but just in case someone misses it, [here's the link](http://mathoverflow.net/questions/24437/is-compact-implies-sequentially-compact-consistent-with-zf).

    I just asked this as a question on MO to see if anyone there can shed any light on the relationship between compactness, sequentially compactness, and the strength of the axiom needed to find this example. Probably everyone here will see it anyway, but just in case someone misses it, here’s the link.

30. • CommentRowNumber30.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 13th 2010
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexI'd like to make it more explicit, if possible. I was thinking along the following lines: try to define the family of "partial cluster points" as $\{(b_t)_{t \in T}\}$ with $T \subseteq [0,1]$ and $(b_t)$ being a cluster point on $[0,1]^T$. Order this by restriction. Then chains have maxima so one can apply Zorn's lemma. The tricky bit is showing that a maximal element is a cluster point on $[0,1]$. Namely, that if $(b_t)$ is a partial cluster point on $T \subseteq [0,1]$ with $T$ proper then it can be extended. I'm not convinced that this is automatic, since the values at different points $t \in [0,1]$ are not independent (for example, with the sequence given, the value at $t$ determines the value at $1-t$). I'd sacrifice axiomatic strength for clarity in this example. So I'd rather use the full Zorn's lemma and have it really nice and explicit than try to look for the minimum necessary axioms.

    I’d like to make it more explicit, if possible. I was thinking along the following lines: try to define the family of “partial cluster points” as $\{(b_t)_{t \in T}\}$ with $T \subseteq [0,1]$ and $(b_t)$ being a cluster point on $[0,1]^T$. Order this by restriction. Then chains have maxima so one can apply Zorn’s lemma. The tricky bit is showing that a maximal element is a cluster point on $[0,1]$. Namely, that if $(b_t)$ is a partial cluster point on $T \subseteq [0,1]$ with $T$ proper then it can be extended. I’m not convinced that this is automatic, since the values at different points $t \in [0,1]$ are not independent (for example, with the sequence given, the value at $t$ determines the value at $1-t$).

    I’d sacrifice axiomatic strength for clarity in this example. So I’d rather use the full Zorn’s lemma and have it really nice and explicit than try to look for the minimum necessary axioms.

31. • CommentRowNumber31.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 18th 2010
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexSimplified the compact space that isn't sequentially compact to an uncountable product of $\{0,1\}$s. Added the proof that a countable family of sequentially compact spaces is sequentially compact. Presumably, this needs countable choice at some point (at least for non-Hausdorff spaces).

    Simplified the compact space that isn’t sequentially compact to an uncountable product of $\{0,1\}$s. Added the proof that a countable family of sequentially compact spaces is sequentially compact. Presumably, this needs countable choice at some point (at least for non-Hausdorff spaces).