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Definitions

  1. A filter is a poset $(I,\leq)$ such that for any $\alpha,\beta\in I$ there is $\gamma\in I$ such that $\gamma\geq\beta,\gamma\geq\alpha$.
  2. A net in a set $X$ is a function from a poset to $X$.
  3. A subnet of a net $(f(\alpha))_{\alpha\in I}$ is a net $(f(g(\beta)))_{\beta\in J}$ where $g:J\to I$ tends to infinity, i.e. for any $\alpha\in I$ we have $\beta_\alpha$ such that $\beta\geq\beta_\alpha\implies g(\beta)\geq\alpha$.
  4. A net $(f(\alpha))_{\alpha\in I}\subseteq X$ where $X$ is a topological space converges to $x\in X$ if for every neighborhood $U$ of $x$ there is $\alpha_U\in I$ such that $\alpha\geq\alpha_U\implies f(\alpha)\in U$.

I will indicate a net $(f(\alpha))_{\alpha\in I}$ by $(x_\alpha)_{\alpha\in I}$ where $x_\alpha=f(\alpha)$.

Lemma

For any infinite filter $I$ there exists $f:\mathbb{N}\to I$ which tends to infinity in the sense of definition 3.

Proof.

Set $\alpha_1=f(1)$ for any $\alpha_1\in I$. We can always find $\beta\neq\alpha_1$ unless the filter is a single point, which is impossible since we assumed it is infinite. So we have either $\beta\geq\alpha_1$ or $\alpha_1\geq\beta$. If the former, set $f(2)=\beta$. If the latter, set $f(1)=\beta,f(2)=\alpha_1$. Better rename. Filter is infinite, so we find a third element distinct from all 3. Order them and set $f(1)$ to be the minimum, $f(2)$ the middle point, $f(3)$ the maximum. If the three elements form a branch (i.e. we have $\alpha,\beta,\gamma$ with $\alpha\leq\beta,\alpha\leq\gamma$ but $\beta,\gamma$ not comparable), then $f(1)=\alpha,f(2)=\beta$, and by definition of filter you find $\gamma':\gamma'\geq\beta,\gamma'\geq\gamma$, and so $\alpha,\beta,\gamma'$ form a totally ordered subset and $f(3)=\gamma'$ makes $f$ increasing from $\{1,2,3\}$ to the filter considered. By such arguments, one can inductively construct $f(n)$ for any $n$, proving the claim.

Lemma

Let $I$ be a finite filter. Then any net from $I$ to any topological space converges.

Proof.

Pick any $a\in I$. If there is $b\geq a$, replace $a$ with $b$. In a finite number of steps, you will have reached a maximal $a$. $(x_\alpha)_{\alpha\in I}$ will converge to $x_a$, since the definition states that for any $U$ neighborhood of $x_a$ we must find $b\in I:c\geq b\implies x_c\in U$, but $b=a$ implies $c\geq b$ equates to $c=b=a$ so we conclude.

In fact, the above shows we wouldn't want to consider finite filters for nets, since if a finite filter has a branch then it has at least two distinct maximal elements $a\neq b$ and so any net for which $x_a\neq x_b$ will converge to two distinct points, impeding uniqueness of limits.

It is known (see here) that $X$ is compact iff every net has a convergent subnet.

Lemma

Any net from an infinite filter has a subnet which is a sequence.

Proof.

We have proven above that there exists $f$ from the naturals to the filter which tends to infinity. We have $(x_\alpha)$ a net. Just set $y_n=x_{f(n)}$ and you have your subnet sequence.

Theorem

Sequential compactness implies compactness.

Proof.

Take a net $(x_\alpha)$. By the lemma above we can find a subnet which is a sequence, so $(x_{f(n)})$. By hypothesis this sequence has a convergent subsequence $(x_{f(g(n))})$. But this is a convergent subnet to $x_\alpha$, hence compactness.

Great! Except there are counterexamples, i.e. spaces which are sequentially compact but not compact. So what did I get wrong here?

Update

Naturally the thing about finite filters is that no branching can avoid reconnecting. But that is general: if I have two maximal elements, they are either equal or they cannot both be maximal, since for any two distinct elements there is a third above both. Hence the remark on "I wouldn't want to consider finite filters for nets" is just bogus :).

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  • 2
    $\begingroup$ The first lemma is wrong. If $I = \omega_1$, where $\omega_1$ is the first uncountable ordinal, then all maps $f\colon \mathbb{N}\to I$ are bounded. (I haven't looked further, and have no time now, will look after dinner.) $\endgroup$ – Daniel Fischer Apr 24 '16 at 16:04
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    $\begingroup$ People have pointed out that Lemma 1 is false. You asked what's wrong with the proof. In the non-proof of Lemma 1 you construct $f$ just fine. Then you give no indication whatever why it should tend to infinity (in general it doesn't). $\endgroup$ – David C. Ullrich Apr 24 '16 at 16:39
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    $\begingroup$ Without reading, the easiest way to know that a fake proof to a known false statement is to pick a concrete counterexample and apply the proof to it in details (as necessary). So take a compact space which is not sequentially compact (e.g. one point compactification of an uncountable discrete space) and see where your proof fails when trying to apply it there. $\endgroup$ – Asaf Karagila Apr 26 '16 at 7:54
  • $\begingroup$ Well the counterexample here would rather be a sequentially compact but not compact space, like $\omega_1$, which I previously knew next to nothing of, so I couldn't really apply the proof there. Good suggestion anyway, in genera, @Asaf. $\endgroup$ – MickG Apr 26 '16 at 8:12
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    $\begingroup$ Right, the other way around. As I said, without reading. I'm all about abstractions and generalizations, even when it comes to advice. :-) $\endgroup$ – Asaf Karagila Apr 26 '16 at 8:18
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Lemma 1 is wrong, it states that every "filter" (better use "directed set", which is more standard) has a convergent subsequence, essentially. But there are linearly or partially ordered sets that do not have countable cofinality:

Take e.g. $I = \omega_1$, the first uncountable ordinal, then any function $f: \mathbb{N} \rightarrow I$ is bounded above, i.e. there is some $\beta \in \omega_1$ such that $f(n) \le \beta$ for all $n$. And then it cannot converge in your sense 3, because $f$ never gets above $\beta+1 \in \omega_1$.

The rest falls down after that (the lemma on finite index sets is true, but irrelevant).

What is true is that sequential compactness implies countable compactness, but no more: $\omega_1$ in the order topology is sequentially compact but not compact.

And spaces like $\beta\mathbb{N}$ and $\{0,1\}^\mathbb{R}$ are compact but not sequentially compact.

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  • $\begingroup$ To know what $\omega_1$ is. Sorry, my knowledge of ordinals is at this level :). How is the boundedness of any $f$ shown? What is the order topology on $\omega_1$? $\endgroup$ – MickG Apr 24 '16 at 17:11
  • $\begingroup$ en.wikipedia.org/wiki/First_uncountable_ordinal It has links to all those statements. $\endgroup$ – Henno Brandsma Apr 24 '16 at 17:19
  • $\begingroup$ Is it correct to say that the topology induced from $[0,\omega_1)$ on any $[x\omega,(x+1)\omega)$ (therein including $[\omega_1-\omega,\omega_1)$) is discrete, and that taking a cover of $[0,\omega_1-\omega]$ plus the singletons of $[\omega_1-\omega,\omega_1)$ gives me a cover of $[0,\omega_1)$ with no finite subcover? $\endgroup$ – MickG Apr 24 '16 at 17:55
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    $\begingroup$ @MickG No, not quite. But simpler: $[0,\alpha), \alpha < \omega_1$ are all open sets (increasing) that cover $\omega_1$ but it does not have any countable (or finite) subcover, because then the countably many upper limits have an even larger upperbound that is not covered by them. $\endgroup$ – Henno Brandsma Apr 24 '16 at 17:57
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    $\begingroup$ Yes, the intervals $[\alpha\omega, (\alpha+1)\omega)$ are discrete, but there are many more countable ordinals than that (and limits too). So they don't cover all. And $\omega_1 - \omega$ is not an ordinal smaller (as $\omega_1$ is a limit), but sort of undefined. There are a lot of isolated points, but a lot of countable limit ordinals too. $\endgroup$ – Henno Brandsma Apr 24 '16 at 18:03
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This is not quite an answer, but rather a sum-up of what I gathered on ordinals from various sources. First of all we have the comments to Henno's answer (links in the screenshot: Vsauce video in first comment, Wikipedia link, Related answer, post found by me via googling). Then we have this other Math SX post and this third one. Finally, Wikipedia and more Wiki.

  • Basically, an ordinal is an order type, so an equivalence class of well-orders under "order isomorphism", where an order isomorphism between well-ordered sets $I,J$ is, I take it, a bijective order-preserving function with order-preserving inverse; this can be alternately viewed with the Von Neumann approach, where an ordinal is a well-ordered set such that every element of it is also a subset;
  • Ordinals can be topologized with the order topology, generated by the subbase of open rays $[0,\alpha)$ for $\alpha$ an ordinal; with this, sets of the form $[\alpha\omega,(\alpha+1)\omega)$ are discrete, whereas if we add in $(\alpha+1)\omega$ (i.e. consider the closed interval $[\alpha\omega,(\alpha+1)\omega]$), it is not discrete since $(\alpha+1)\omega$ is a limit point; those sets are certainly open, but not an open cover, as the comments say;
  • $I=[0,\omega_1)$, with $\omega_1$ the first uncountable ordinal (which exists by well-ordering, since the set of uncountable ordinals must have a least element $\omega_1$), is not compact, since $[0,\alpha)$ for $\alpha\in I$ is an open cover but no finite subcover exists, since if there were one the union of all the upper limits would be $\beta$, an ordinal such that $[0,\beta)$ contains all the subcover, but a countable one; that a union of ordinals is an ordinal comes from here;
  • Just like $\omega-1$, $\omega_1-\omega$ is sort of undefined;
  • $I$ has no countable closed discrete subset; the link's argument is: let $A$ ba such a set; let $A_c=\{\alpha\in I:\alpha\geq\beta\text{ for countably many }\beta\in A\}$; $A_c$ is nonempty (because the union of all $\alpha+1$ for $\alpha\in A$ is an ordinal strictly greater than all of $A$) so by well-ordering (here is the proof of the well-order property of $I$) it has a least element $\beta$; $\beta$ is a limit point of $A$ (why? Will think in a while…), so either $\beta\in A$ and $A$ is not discrete, or $\beta\not\in A$ so $A$ is not closed;
  • The union argument above clearly shows that any $f:\mathbb{N}\to\omega_1$ is bounded above, because the image of $f$ is at most countable, and so is bounded by $\bigcup_n(f(n)+1)$; so no such $f$ can tend to infinity;
  • $\omega_1$ is first countable since the rays that generate its topology are countable (right?); so a sequence in $\omega_1$, having a limit point as seen above (and why $\beta$ is a limit point still escapes me), must have a convergent subsequence to that limit point, since for any $U_n$ of the countable local base at $\beta$ I find $a_n\in U_n$ from the sequence, and since there are countably many "$a_n$"s I can find $k(n)$ an increasing function such that $a_{k(n)}\in U_n$ for all $n$.

Anything wrong here?

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  • $\begingroup$ The order topology also has all right segment $(\alpha, \rightarrow)$ in its subbase, not just the initial segments. $\omega_1 - \omega$ is defined, it's just $\omega_1$, because $\omega + \omega_1 = \omega_1$ (as ordinals), so it's not some ordinal less than $\omega_1$, as people often think. See proofwiki.org/wiki/Ordinal_Subtraction_when_Possible_is_Unique $\endgroup$ – Henno Brandsma Apr 25 '16 at 17:33

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