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I only have difficulties in the final step: Show that {$x_y:y\in I$} has no convergent strict subnet.

My efforts:

With the construction, $I$ is a minimal uncountable well-ordered set. Thus it has the following properties: (1) Every countable subset of $I$ has an upper bound in $I$. (2) $I$ has no largest element. (3) For every $\alpha\in I$, the subset {$x|\alpha Wx$} is uncountable.

Also {$x_y:y\in I$} is a monotonically increasing net in the $W$ sense. Assume {$x_y:y\in I$} has a convergent strict subnet {$x_z:z\in J\subset I,J$ is cofinal in $I$}, say, converging to $v\in J$. Then given any neighborhood of $v$ in the form of $(a,b)$ or $[0, b)$ or $(a, 1]$ in the usual sense there is $\alpha\in J$ such that $\beta\in J$ is in this neighbor if $\alpha W\beta$. If $v\neq 0$, choose a neighborhood $(a,b)$ of $v$. Then I don't know how to continue.

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HINT: For each $n\in\Bbb Z^+$ there is an $\alpha_n\in J$ such that $|x_\beta-v|<\frac1n$ whenever $\alpha W\beta\in J$. Let $\alpha=\sup_n\alpha_n$. Do you see the problem here?

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  • $\begingroup$ I guess the sup is in the W sense. The sup alpha should be smallest element of J excluding the alpha sequence. Then all beta such that alpha W beta would just correspond to a single point v. There are uncountably many such beta. So that's kind of uncountable set corresponding to one point, all in the same set J. So a contradiction. Am I correct? $\endgroup$ – Junk Warrior Jun 28 at 21:23
  • $\begingroup$ @JunkWarrior: Yes, I was taking the supremum with respect to $W$. Yes, you’d have $x_\beta=v$ whenever $\alpha W\beta$, contradicting the fact that the net is injective. $\endgroup$ – Brian M. Scott Jun 28 at 21:41

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