Let \(\cal C\) be the open cover of \(\omega_1\) from Kelley's hint. That is, \(\cal C = \{\{\beta : \beta < \alpha\}: \alpha < \omega_1 \}\). We show that \(\cal C\) has no open, locally finite refinement by contradiction. Assume that \(\cal A\) is an open, locally finite refinement of \(\cal C\). We define an increasing function \(f:\omega\to \omega_1\) by induction: \(f(0)=0\). Assume that \(f(n)\) is defined. For each \(\beta \le f(n)\), the set \(D_\beta = \{ x\in \cal A :\beta\in x\}\) is finite since \(\cal A\) is locally finite. It's also true that for each \(x\in D_\beta\), sup\(\{\gamma : \gamma\in x \}\) is less that \(\omega_1\) since \(x\) is a subset of one of the elements of \(\cal C\). Let \(g(\beta)\) be the maximum of \(\{\hbox{sup}(x) :x\in D_\beta\}\). \(g(\beta)\) is also less than \(\omega_1\). Now apply Form 34 to the set \(W = \{g(\beta) : \beta \le f(n)\}\). This is a countable set of ordinals each \(< \omega_1\) and thereforesup \(W < \omega_1\). Let \(f(n+1)\) be the least ordinal greater than sup \(W\). The important property of the function \(f\) is that \[](\forall n\in \omega)(\forall j < n)(\forall x\in\cal A)( f(j)\in x \to f(n)\notin x) \tag{\(*\)}\] The sequence \(\langle f(n) \rangle_{n\in\omega}\) is strictly increasing so its sup (call the sup \(\beta\)) is a limit ordinal. Further, by Form 34, \(\beta< \omega_1\).Since \(\cal A\) is locally finite, there is an (open) neighborhood \(y\) of\(\beta\) such that \(\{x\in \cal A : x\cap y \ne\emptyset\}\) is finite. Since \(y\) is open and \(\beta\) is a limit ordinal, there is an ordinal \(\gamma < \beta\) such that the interval \((\gamma,\beta+1)\subseteq y\). For some natural number \(k\), this interval must contain \(f(n)\) for all \(n \ge k\). For each \(n \ge k\) we consider the sets \(D_{f(n)} = \{x\in \cal A : f(n)\in x\}\). These sets are pairwise disjoint (by (\(*\))), each is non-empty and for each element\(x\in D_{f(n)}\), \(x\cap (\gamma,\beta+1) \ne \emptyset\). Therefore \(\{x\in \cal C : x\cap y \ne\emptyset\}\) is infinite, a contradiction.

To show that Form 34 implies [34 C] (\(\omega_1\) with the order topology has the property that every infinite subset has a limit point in \(\omega_1\).), let \(X\) be an infinite subset of \(\omega_1\). Since \(X\) is well ordered and infinite, it must have an initial segment of order type \(\omega\). Form 34 implies that each such initial segment must be bounded, so the least upper bound of this set is a limit point of \(X\) in \(\omega_1\).

If \(\omega_1\) is regular and contains a denumerable discrete family of open sets, then the set of smallest elements in each set would be an infinite set with no limit point, contradicting [34 C]. Thus, Form 34 implies [34 D] (\(\omega_1\) with the order topology does not contain a countable discrete family of open subsets.).

The uncountable open cover of \(\omega\), \(\cal C= \{\{\beta :\beta < \alpha\}: \alpha < \omega_1\}\), described above does not contain a countable refinement. If it did we could construct a countable cofinal sequence in \(\omega_1\), contradicting Form 34. Therefore, Form 34 implies [34 E] (\(\omega_1\) with the order topology is not weakly Lindelöf.)