# An application of the Pressing Down Lemma (Part 1)

For each $$sigma$$ and $$alpha$$ let $$P = W_sigma – alpha$$ which is still stationary. Now for each $$f^* in P$$, let $$g_sigma(f^*)$$ be $$f(n)$$ where $$n$$ is least such that $$f(n) ge alpha$$. Now using pressing down lemma and the pigeonhole principle, you can find some fixed $$n$$ and some fixed $$gamma ge alpha$$ such that $${ f^* in P: f(n) = gamma}$$ is stationary. Now if $$nle |sigma|$$, you are done. Else to fill the gap between $$|sigma|$$ and $$n$$ you can use repeated applications of the pressing down lemma, to get the desired $$theta$$.

EDIT: [This edit will try to complete the below answer you provided.][Disclaimer: I am using your notation.]

As in your answer, let $$S = {f^* in P: f(m) = gamma}$$ and suppose $$m gt |sigma|$$. By the fact that $$P subseteq W_sigma$$, we have $$f||sigma| = sigma$$, for any $$f^* in S$$. First we inductively choose a finite sequence of stationary sets $$langle S_0, dots, S_{m-|sigma|-1}rangle$$ and a finite sequence of ordinals $$langle beta_0, dots, beta_{m-|sigma|-1}rangle$$, such that $$S_0 subseteq S$$, $$S_{i+1} subseteq S_i$$, for $$i lt m-|sigma|-1$$. Also we make sure that for each $$f^* in S_i$$, $$f(i+|sigma|) = beta_i$$.

This can be done easily, using the pressing down lemma. For the base case $$i = 0$$, consider $$g(f^*) = f(|sigma|)$$ and by the pressing down lemma you have some stationary $$S_0 subseteq S$$ and some ordinal $$beta_0$$ such that $$g”S_0 = {beta_0}$$. At the $$i$$th step just look at $$g(f^*) = f(i+|sigma|)$$, and construct $$S_i$$ and $$beta_i$$ as above.

So we wish to build a $$theta in chi^{m+1}$$ that satisfies the conditions in the question. First let $$theta||sigma| = sigma$$ and $$theta(m) = gamma$$. Now for $$|sigma| le i lt m$$, let $$theta(i) = beta_{i-|sigma|}$$. Now you can see that $$W_theta$$ is stationary as it contains $$S_{m-|sigma|-1}$$. And also because of $$gamma$$ you have $$theta not in cup_{ninomega} alpha^n$$.