I am trying to prove this fact, but there is one thing that shows up that makes me unable to finish the proof, do you have any tips on how to finish the proof?

A space X satisfies the sequence lemma if for every subspace A of X.:

If $x \in \bar{A}$ there is a sequence of elements $a_n\in A$ converging to x.

The space $S_\Omega$ is a space that does not contain a larges element, it is uncountable, it has a smallest element $a_0$, and it is well-ordered. And even though it is uncountable, for all $k \in S_{\Omega}$, the set $\{x < k| x \in S_\Omega\}$ is countable. The toplogy on $S_\Omega$ is the order topology. (It is called the minimal uncountable well-ordered set).

I want to show that $S_\Omega$ satisfies the sequence lemma.

attempt at proof:

Assume that $x \in \bar{A} \subset S_\Omega$, and lets just assume for simplicity that $x \ne a_0$.

Let (a,b), be a basis element around x, then since $x \in \bar{A}$, this basis element must contain $a_1\in A$. Depending on wheter $a_1$ is bigger or smaller than x, then look at the intervals $(a,a_1)$ or $(a_1,b)$.(If $a_1$ happens to be x, we are done, because then we just get a constant sequence in x.

If we continue this process, we get a sequence $\{a_n\}$ in A. But I do not think we can be sure that it converges to x. Because if it converges to x, then for any basis element around x, we must have that the sequence at one time, stays within this basis element. I think something like this can happen, even though we are not in the real numbers: we want a sequence convering to 0, first look at (-2,2), choose 1, then look at (-2,1), choose 0.9, then look at (-2,0.9), choose 0.89, then look at (-2,0.89), choose 0.889, etc. so we converge to 0.8888..., instead of 0.

Any tips? I can't just divide the intervals, because we do not have a metric on $S_\Omega$. Since we have the countable property it seems I have to use that, but I am not quite sure how to when I do not have a metric.

  • $\begingroup$ What makes you think it does satisfy the sequence lemma? Actually it doesn't. (Well, it satisfies what you said the sequence lemma is, but your version can't be the actual lemma - surely there's a requirement that $a_n\ne x$...) $\endgroup$ – David C. Ullrich Oct 5 '15 at 18:11
  • $\begingroup$ @DavidC.Ullrich I am not sure why that would be the case, here is a picture of the sequence lemma from my book: s3.postimg.org/uyf2zq029/sequence.png It is the "converse" part I am interested in here, since the other way holds in all topological spaces. $\endgroup$ – user119615 Oct 5 '15 at 18:18
  • $\begingroup$ @DavidC.Ullrich $a_n \in A$ while $x \in \overline{A}$. And by the way, $S_{\Omega}$ satisfies the sequence lemma. $\endgroup$ – Crostul Oct 5 '15 at 18:19
  • $\begingroup$ @Crostul If so then the sequence lemma is not what he said it was. Oh. Maybe saying $S_\Omega$ satisfies the sequence lemma means that if $A\subset S_\Omega$ then etc? If so fine. I assumed that what the OP wrote was the definition of "$A$ satisfies the sequence lemma" - I guess not. $\endgroup$ – David C. Ullrich Oct 5 '15 at 18:33
  • $\begingroup$ @DavidC.Ullrich He wrote "for a subspace $A$ of $X$" $\endgroup$ – Crostul Oct 5 '15 at 18:41

Using the standard notation: $0$ is the smallest element of $S_\Omega$ and $x+1$ is the smallest element larger than $x$:

Say $x\in\overline A\setminus A$. The fact that $(0,x+1)$ is open shows that $x$ cannot be in the closure of the set of elements of $A$ larger than $x$, hence $x$ is in the closure of the set of elements of $A$ smaller than $x$.

That set is countable; say $(a_n)$ is an enumeration of the elements of $A$ smaller than $x$. Now for each $j$, let $b_j$ be the maximum of $a_1,\dots,a_j$. So $b_j\in A$, $b_j<x$, $b_{j+1}\ge b_j$, and for every $a$ in $A$ with $a<x$ there exists $j$ with $b_j\ge a$.

It follows that $b_j\to x$. Suppose $y<x$. There exists $a\in A$ with $y<a<x$. There exists $j$ with $a\le b_j<x$, and hence $y<b_k<x$ for all $k\ge j$.

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Let $x \in \overline{A}$, and call $S_x = \{ s \in S_{\Omega} : s<x \}$. $S_x$ is countable, so the family of open neighbourhoods of $x$ $$\{ (a,x+1) \}_{a \in A \cap S_x}$$ is countable. Index this family by $\mathbb{N}$, say $\{ U_n \}_{n \ge 1}$. Now, for all $n \ge 1$, you have that $U_n$ intersects $A$: pick some $x_n \in A \cap U_n$; the sequence $\{ x_n \}_n \subseteq A$ converges to $x$.

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  • $\begingroup$ Thank you very much, but can you please explain why the sequuence converges to x? If I have an arbitrary basis element $(z_1,z_2)$ that contains x, why will there be an N such that the entire sequence is contained in this basis element, when $n \ge N$ ? $\endgroup$ – user119615 Oct 5 '15 at 18:36
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    $\begingroup$ @user119615 That sequence need not converge to $x$. It has a subsequence converging to x$, but proving that is essentially the same as the original problem. See the other answer that just appeared... $\endgroup$ – David C. Ullrich Oct 5 '15 at 21:35

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