@JM That only covers the interesting part of the question which is stated at the very end. The main part of the question is concerned with showing properties that immediately follow from the existence.
I first encountered the word "to prune" in the context of "pruned trees" (set theory) and at first I wondered why it meant sort of the exact opposite of what I had in mind when I was thinking of a tree full of prunes (in the French sense)...
@JM It was extended very early. The first citation in the OED is from 1547; the first figurative citation is from 1565. (There is one possible 15th century instance of the word; either it's a figurative use of this verb, or it's a figurative use of another verb prune meaning 'to preen' (of a bird preening its feathers).
@tb I used to point out that my adviser published the first known consistent example of a Dowker space in 1955 and the first known ZFC example in 1971. That's quite a few hours of thinking about the problem.
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@Ilya Huh, I have the longest. Whatever I say and wherever I talk, someone jumps in and asks "I doubt you could do that since you called someone narrow-minded, could you elaborate?".
@anon I know I am supposed to transform it into a geometric series, but I do not see how what you posted helps me? Afer carying out the calculations I do not seem to end up with a geometric series? =( I feel dumb.
@MattN yes, a subspace is said to be complemented if it is the image of a continuous projection. People say $K$-complemented if there's a projection of norm $\leq K$.
@MattN the question is whether it works in an arbitrary Banach space. Hahn-Banach only gives you a projection of norm $n$ or $\sqrt{n}$ if you're a bit more clever but you can't do much better in general.
@Jeff $e^{i\varphi}$ is a unit vector with angle $\varphi$ measured counterclockwise from $1$. Here $\varphi = 1$ which corresponds to an angle of approximately $57.3$ degrees. The real part of $e^i$ is the cosine of $1$ which is $0.54$
yup ... actually, i want to figure out how to simplfy an expression which has complex numbers in x+iy form and also in e^it form and also absolute value bars. :D
@Ilya I was so focused on finding something that worked with $\cos(t)^2$, I guess it is true that in the forest you can not see the trees... Thanks once again =)
@tb sorry, i've been asking about all different parts of this question and mixing up response. i have to get more organized. here is the actualy identity i need to prove (no typos this time: $$\frac{1-|z|^2}{|e^{it}-z|^2}=Re\frac{e^{it}+z}{e^{it}-z}$$
@Jeff Why do you think $z$ has an argument equal to $t$? That isn't necessarily the case. I see no advantage in writing $z$ in polar form; just keep it as $z$.
Don't worry... I won't actually do it, just would like to know how.
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