Question: Fix \(\alpha \) with \(0<\alpha <1\) and repeat our "middle thirds" construction for the Cantor set except that now, at the nth stage, each of the \({{2}^{n-1}}\) open intervals we discard from [0, 1] is to have length \(\left( 1-\alpha \right){{3}^{-n}}\) (We still want to remove each open interval from the middle of a closed interval in the current level — it is important that the closed intervals that remain turn out to be nested.) The limit of this process, a set that we will name \({{\Delta }_{\alpha }}\), is called a generalized Cantor set and is very much like the ordinary Cantor set. Note that A. is uncountable, compact, nowhere dense, and so on, but has nonzero outer measure. Indeed, check that m*( \({{\Delta }_{\alpha }}\) ) = \(\alpha \)

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