Talk:Subsets of Equidecomposable Subsets are Equidecomposable

Throughout, I have been using $$\subseteq$$ for "is a subset of" and $$\subset$$ for "is a proper subset of", i.e. $$A \subset B \iff A \subseteq B \land A \ne B$$.

This is consistent with the notation for $$\le$$ and $$<$$, for example, and (for that and possibly other reasons), I believe it's "better" to use $$\subset$$ as its context above.

As there is generally confusion as to exactly what $$\subset$$ may mean in a context, it might be better to use $$\subseteq$$ unless $$\subsetneq$$ is specifically meant instead.

In this context I don't know which is meant! --Matt Westwood 22:28, 29 January 2009 (UTC)


 * You're absolutely right. Non-proper subset works just fine here.  There's no reason we can't speak of decompositions of all of $$\R^n \ $$, and if $$A, B \ $$ are equidecomposable, then $$S = A \ $$ clearly yields the subset $$T = B \ $$ such that $$S, T \ $$ are equidecomposable, and the theorem is trivial.  Zelmerszoetrop 00:15, 30 January 2009 (UTC)