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I am compiling a list of 'swindles' in the style of the Eilenberg-Mazur swindle. I've already got some swindles in K-theory, the Mazur Swindle and the proof of the Cantor–Bernstein–Schroeder theorem. Can you tell me any more?

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up vote 3 down vote accepted

The Pelczynski decomposition technique in Banach space theory.

Here's one version:

Let $X$ and $Y$ be Banach spaces such that $X$ is isomorphic to a complemented subspace of $Y$ and $Y$ is isomorphic to a complemented subspace of $X$. Suppose further that $X$ is isomorphic to the $\ell_p$-sum of $X$ with $1 \leq p \lt \infty$: $$ X \cong \bigoplus_{n=1}^\infty X = \left\{(x_n) \subset X : \sum_{n=1}^\infty \lVert x_n\rVert^p \lt \infty\right\}.$$ Then $X$ and $Y$ are isomorphic.

Proof. The hypothesis on $X$ implies that $X \cong X \oplus X$ and the hypotheses on complementation can be written as $X \cong Y \oplus U$ and $Y \cong X \oplus V$. Thus, $$ Y \cong X \oplus V \cong X \oplus X \oplus V \cong X \oplus Y. $$ Combining this with the "swindle" $$ X \cong \bigoplus_{n=1}^\infty X \cong \bigoplus_{n=1}^\infty (Y \oplus U) \cong Y \oplus \bigoplus_{n=1}^\infty (U \oplus Y) \cong Y \oplus X, $$ we get $X \cong Y \oplus X \cong X \oplus Y \cong Y$. $\hskip{1cm}\blacksquare$

This can be used to show that every infinite-dimensional complemented subspace of $\ell_p$ is isomorphic to $\ell_p$: one shows using basic sequence techniques that a closed and infinite-dimensional subspace $Y$ of $\ell_p$ contains a complemented subspace isomorphic to $\ell_p$. The above argument with $X = \ell_p$ then yields the result. This is usually expressed by saying that "$\ell_p$ is prime".

There are many variants of this argument and lots of applications. See chapter 2 of the book Topics in Banach space theory by Albiac and Kalton for a good exposition.

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Excellent, thankyou. This is just the kind of thing I am looking for. – tharris Dec 23 '12 at 13:44
You are welcome. I'm glad you liked it. May I suggest that you un-accept this answer for the time being? It might increase the chances that you get more answers and examples - I'd be interested to see more of them! – Martin Dec 23 '12 at 19:57
Done. Thanks for the tip, I'm new to this :) – tharris Dec 23 '12 at 22:47
No, nothing coming. If you send me some contact details i can email you the list when I've written it up. – tharris Dec 24 '12 at 15:58
Too bad... I would have recommended to be a bit more patient, holidays and all, you know :) Why don't you wait a bit longer and then you may think about trying to ask this on MathOverflow. This question seems to be reasonably on-topic there. Especially after you haven't got that much feedback here. – Martin Dec 25 '12 at 10:42

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