# Show that in [0,1] with its usual topology there exists a net having no convergent strict subnet

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.

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?
• @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. – Brian M. Scott Jun 28 at 21:41